|
Number of plants |
Number of houses |
|
0-2 |
1 |
|
2-4 |
2 |
|
4-6 |
1 |
|
6-8 |
5 |
|
8-10 |
6 |
|
10-12 |
2 |
|
12-14 |
3 |
Which method did you use for finding the mean and why?
|
Number of plants |
Number of houses (fi) |
Class mark (xi) |
fixi |
|
0-2 |
1 |
1 |
1 |
|
2-4 |
2 |
3 |
6 |
|
4-6 |
1 |
5 |
5 |
|
6-8 |
5 |
7 |
35 |
|
8-10 |
6 |
9 |
54 |
|
10-12 |
2 |
11 |
22 |
|
12-14 |
3 |
13 |
39 |
|
Total |
Σfi = 20 |
Σfixi = 162 |
= 162
Hence, the mean number of plants per house = 8.1
We have used the direct method for finding the mean because numerical values of xiand fi are small.
Consider the following distribution of daily wages of 50 workers of a factory:
|
Daily wages (in Rs.) |
Number of workers |
|
100-120 |
12 |
|
120-140 |
14 |
|
140-160 |
8 |
|
160-180 |
6 |
|
180-200 |
10 |
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency 'f'.
|
Daily pocket allowance (in Rs.) |
11-13 |
13-15 |
15-17 |
17-19 |
19-21 |
21-23 |
23-25 |
|
No. of childrens |
7 |
6 |
9 |
13 |
f |
5 |
4 |
|
Daily pocket allowance C.l. |
No of children (fi) |
Mid-value xi |
fixi |
|
11-13 |
7 |
12 |
84 |
|
13-15 |
6 |
14 |
84 |
|
15-17 |
9 |
16 |
144 |
|
17-19 |
13 |
18 |
234 |
|
19-21 |
f |
20 |
20f |
|
21-23 |
5 |
22 |
110 |
|
23-25 |
4 |
24 |
96 |
|
Σfi = 44 + f |
Σfixi = 752 + 20f |
18 (44 + f) = 752 + 20f
792 + 18f = 752 + 20f 
18f - 20f = 752 - 792 
-2f = -40
f = 20Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
|
Number of heart beats per minute |
65-68 |
68-71 |
71-74 |
74-77 |
77-80 |
80-83 |
83-86 |
|
Number of women |
2 |
4 |
3 |
8 |
7 |
4 |
2 |
A (assumed mean) = 75.5 and h =3
In a retail market, fruit venders were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
|
No. of Mangoes |
50-52 |
53-55 |
56-58 |
59-61 |
62-64 |
|
No. of boxes |
15 |
110 |
135 |
115 |
25 |
The table below shows the daily expenditure on food of 25 household in a locality.
|
Daily Exp. (in Rs.) |
100-150 |
150-200 |
200-250 |
250-300 |
300-350 |
|
No. of house hold |
4 |
5 |
12 |
2 |
2 |
Find the mean daily expenditure on food by a suitable method.
A (assumed mean) = 225 and h = 50
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
|
Concentration of SO2 (in ppm) |
0.00-0.04 |
0.04-0.08 |
0.08-012 |
0.12-0.16 |
0.16-0.20 |
0.20-0.24 |
|
Frequencey |
4 |
9 |
9 |
2 |
4 |
2 |
Find the mean concentration of SO2 in the air.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
|
Number of days |
0-6 |
6-10 |
10-14 |
14-20 |
20-28 |
28-38 |
38-40 |
|
Number of students |
11 |
10 |
7 |
4 |
4 |
3 |
1 |
|
Number of days |
Number of students (fi) |
Class mark (xi) |
fixi |
|
0-6 |
11 |
3 |
33 |
|
6-10 |
10 |
8 |
80 |
|
10-14 |
7 |
12 |
84 |
|
14-20 |
4 |
17 |
68 |
|
20-28 |
4 |
24 |
96 |
|
28-38 |
3 |
33 |
99 |
|
38-40 |
1 |
39 |
39 |
|
Total |
Σfi = 40 |
Σfixi = 499 |
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
|
Literacy rate (in %) |
45-55 |
55-65 |
65-75 |
75-85 |
85-95 |
|
Number of cities |
3 |
10 |
11 |
8 |
3 |
The following table shows the ages of the patients admitted in a hospital during a year.
Find the mode and mean of the data given below. Compare and interpret the measures of central tendency.
|
Age (in yrs) |
No. of patients |
|
5-15 |
6 |
|
15-25 |
11 |
|
25-35 |
21 |
|
35-45 |
23 |
|
45-55 |
14 |
|
55-65 |
5 |
Case I: Finding the mode
Here, the maximum class frequency is 23 and the class corresponding to frequency is 35-45.
So, the modal class is 35-45.
Thus, we have
Modal class = 35 - 45
l = 35
fi = 23, f0 = 21, f2 = 14
and h = 10
Now, substituting these values in the formula of mode, we get
Case II : Finding the median
Now,
Hence, the mode of the given data is 36.818 while mean is 35.373.
Interpretation : Maximum number of patients admitted in the hospital are of the age 36.818 years, while an average age of patients admitted to the hospital is 35.373 yrs.
The following data gives the information on the observed life-times (in hours) of 225 electrical components.
|
Life time (in hours) |
Frequency |
|
0-20 |
10 |
|
20-40 |
35 |
|
40-60 |
52 |
|
60-80 |
61 |
|
80-100 |
38 |
|
100-120 |
29 |
Determine the modal lifetimes of the components.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:
|
Expenditure (in Rs.) |
Number of families |
|
1000-1500 |
24 |
|
1500-2000 |
40 |
|
2000-2500 |
33 |
|
2500-3000 |
28 |
|
3000-3500 |
30 |
|
3500-4000 |
22 |
|
4000-4500 |
16 |
|
4500-5000 |
7 |
|
Expenditure (in Rs.) |
Number of families |
|
1000-1500 |
24 |
|
1500-2000 |
40 |
|
2000-2500 |
33 |
|
2500-3000 |
28 |
|
3000-3500 |
30 |
|
3500-4000 |
22 |
|
4000-4500 |
16 |
|
4500-5000 |
7 |
Here, the maximum class frequency is 30 and the class corresponding to frequency is 3000-3500.
So, the modal class = 1500-2000.
Thus, we haveModal class = 1500-2000
l = 1500
f1 = 40, f0 = 24, f2 = 33 and h = 500
Now, substituting these values in the formula of mode, we get
Hence, the modal monthly expenditure of the families is Rs. 1847.83.
Finding Mean:
Here, we have 
Now, 
= 3250 - 587.50 = 2662.50
Hence, the mean monthly expenditure is Rs. 2662.50.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
|
Number of students per teacher |
Number of states/U.T. |
|
15-20 |
3 |
|
20-25 |
8 |
|
25-30 |
9 |
|
30-35 |
10 |
|
35-40 |
3 |
|
40-45 |
0 |
|
45-50 |
0 |
|
50-55 |
2 |
|
Number of students per teacher |
Number of states/U.T. |
|
15-20 |
3 |
|
20-25 |
8 |
|
25-30 |
9 |
|
30-35 |
10 |
|
35-40 |
3 |
|
40-45 |
0 |
|
45-50 |
0 |
|
50-55 |
2 |
Here, the maximum class frequency is 10 and the class corresponding to frequency is 30-35.
So, the modal class = 30-35
Thus, we have
modal class = 30-35, l = 30, f1 = 10, f0 = 9, f2 = 3 and h = 5
Now, substituting these values in the formula of mode, we get
Finding Mean:
Here, we have 
Now, 
Interpretation : Most stales/U.T have a student teacher ratio of 30.6 and on an average, this ratio is 29.2.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
|
Runs scored |
Number of batsman |
|
3000-4000 |
4 |
|
4000-5000 |
18 |
|
5000-6000 |
9 |
|
6000-7000 |
7 |
|
7000-8000 |
6 |
|
8000-9000 |
3 |
|
9000-10000 |
1 |
|
10000-11000 |
1 |
|
Runs scored |
Number of batsman |
|
3000-4000 |
4 |
|
4000-5000 |
18 |
|
5000-6000 |
9 |
|
6000-7000 |
7 |
|
7000-8000 |
6 |
|
8000-9000 |
3 |
|
9000-10000 |
1 |
|
10000-11000 |
1 |
Here, maximum class frequency is 18 and the class corresponding to frequency is 4000-5000. So, the modal class = 4000-5000.
Now,
Thus, we have Modal class = 4000-5000, l = 4000, f1 = 18, f0 = 4, f2 = 9 and h = 1000
Now, substituting these values in the formula of mode, we get
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :
|
Numberof cars |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
70-80 |
|
Frequency |
7 |
14 |
13 |
12 |
20 |
11 |
15 |
8 |
|
Number of cars |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
70-80 |
|
Frequency |
7 |
14 |
13 |
12 |
20 |
11 |
15 |
8 |
Here, maximum class frequency is 20 and the class corresponding to frequency is 40-50.
So, the modal class = 40-50.
Thus, we havemodal class = 40-50, l = 40, f1 = 20, f0 = 12, f2 = 11 and h = 10
Now, substituting these values in the formula of mode, we get
Hence, the mode of the data is 44.7 cars.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
|
Monthly consumption of electricity (in units) |
Number of consumers |
|
68-85 |
4 |
|
85-105 |
5 |
|
105-125 |
13 |
|
125-145 |
20 |
|
145-165 |
14 |
|
165-185 |
8 |
|
185-205 |
4 |
Finding Median:
|
Monthly consumption of electricity (in units) |
Number of consumers |
Cumulative frequency |
|
68-85 |
4 |
4 |
|
85-105 |
5 |
9 |
|
105-125 |
13 |
22 |
|
125-145 |
20 |
42 |
|
145-165 |
14 |
56 |
|
165-185 |
8 |
64 |
|
185-205 |
4 |
68 |
|
n = 68 |
th observation = 34th observation.
= 34, cf = 22, f = 20 and h =20
Finding Mean:
Finding Mode :
Here, the maximum class frequency is 20 and the class corresponding to frequency is 125-145.
So, the modal class = 125-145.
Thus, we have, Modal class = 125-145, l = 125, f1 = 20, f0 = 13, f2= 14, and h = 20
Now, substituting these values in the formula of mode, we get
Comparison: On comparison, we find that the three measures are approximately the same in this case.
If the median of the distribution given below is 28.5, find the values of x and y.
|
Class interval |
Frequency |
|
0-10 |
5 |
|
10-20 |
x |
|
20-30 |
20 |
|
30-40 |
15 |
|
40-50 |
y |
|
50-60 |
5 |
|
Total |
x = 60 |
Forming the cumulative frequency table, we have
|
Class Interval (C.I.) |
Frequency |
c.f. |
|
0-10 |
5 |
5 |
|
10-20 |
x |
5 + x |
|
20-30 |
20 |
25+ x |
|
30-40 |
15 |
40 + x |
|
40-50 |
y |
40 + x + y |
|
50-60 |
5 |
45 + x + y |
|
Total |
60 |
45 + x + y = 60
x + y = 15
57 = 65 - x
x = 65 - 57
x = 8
8 + y = 15
y = 15 - 8 = 7A life insurance agent found the folio wing data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 yrs. onwards but less than 60 years.
|
Age (in yrs.) |
No. of policy holders |
|
below 20 |
2 |
|
below 25 |
6 |
|
below 30 |
24 |
|
below 35 |
45 |
|
below 40 |
78 |
|
below 45 |
89 |
|
below 50 |
92 |
|
below 55 |
98 |
|
below 60 |
100 |
|
Age (in yrs) |
No. of policy holders |
c.f. |
|
below 20 |
2 |
2 |
|
20-25 |
4 |
6 |
|
25-30 |
18 |
24 |
|
30-35 |
21 |
45 |
|
35-40 |
33 |
78 |
|
40-45 |
11 |
89 |
|
45-50 |
3 |
92 |
|
50-55 |
6 |
98 |
|
55-60 |
2 |
100 |
|
n = 100 |
or 50 th observation, which lies in class 35 - 45.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre and the data obtained is represented in the following table:
|
Length (in nun) |
Number of leaves |
|
118-126 |
3 |
|
127-135 |
5 |
|
136-144 |
9 |
|
145-153 |
12 |
|
154-162 |
5 |
|
163-171 |
4 |
|
172-180 |
2 |
Find the median length of the leaves.
We shall first convert the given data to continuous classes. Then, the data become
|
Length (in mm) |
Number of leaves |
Cumulative frequency |
|
117.5-126.5 |
3 |
3 |
|
126.5-135.5 |
5 |
8 |
|
135.5-144.5 |
9 |
17 |
|
144.5-153.5 |
12 |
29 |
|
153.5-162.5 |
5 |
34 |
|
162.5-171.5 |
4 |
38 |
|
171.5-180.5 |
2 |
40 |
|
n = 40 |
observation = 20th observation.
=20, cf = 17, f=12 and h =10
The following table gives the distribution of the life-time of 400 new lamps.
|
Life-time (hrs.) |
No. of lamps |
|
1500-2000 |
14 |
|
2000-2500 |
56 |
|
2500-3000 |
60 |
|
3000-3500 |
86 |
|
3500-4000 |
74 |
|
4000-4500 |
62 |
|
4500-5000 |
48 |
Forming the cumulative frequency table, we get
|
Life-time (hrs.) |
No. of lamps (f) |
c.f |
|
1500-2000 |
14 |
14 |
|
2000-2500 |
56 |
70 |
|
2500-3000 |
60 |
130 |
|
3000-3500 |
86 |
216 |
|
3500-4000 |
74 |
290 |
|
4000-4500 |
62 |
352 |
|
4500-5000 |
48 |
400 |
|
n = 400 |
We have n = 400, so, 
th observation = 200th observation.
So, median lies in the group of 3000-3500. i.e.,Median class = 3000 - 3500
Now, we have Median class = 3000-3500
l = 300, 
= 200, cf = 130, f = 86 and h = 500
Substituting these values in the formula of Median, we get
= 3000 + 406.97 = 3406.97
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
|
Number of letters |
1-4 |
4-7 |
7-10 |
10-13 |
13-16 |
16-19 |
|
Number of surnames |
6 |
30 |
40 |
16 |
4 |
4 |
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
|
Number of letters |
Number of surnames |
Cumulative frequency |
|
1-4 |
6 |
6 |
|
4-7 |
30 |
36 |
|
7-10 |
40 |
76 |
|
10-13 |
16 |
92 |
|
13-16 |
4 |
96 |
|
16-19 |
4 |
100 |
|
x = 100 |
We have n = 100, so 
th observation = 50th observation.
So, median lies in the group of 7-10.
i.e. Median class = 7 - 10
Now, we have Median class = 70 - 10 l = 7, 
, cf = 36, f = 40 and h =3
Now, substituting these values in the formula of median, we get
Hence, the median number of letters in the surnames is 8.05.
The distribution below gives the weights of 30 students of a class. Find the Median weight of the students.
|
Weight (in k.g) |
40-45 |
45-50 |
50-55 |
55-60 |
60-65 |
65-70 |
70-75 |
|
No. of students (f) |
2 |
3 |
8 |
6 |
6 |
3 |
2 |
Forming the cumulative frequency table we get,
|
Weight (in k.g) |
No. of students (f) |
c.f |
|
40-45 |
2 |
2 |
|
45-50 |
3 |
5 |
|
50-55 |
8 |
13 |
|
55-60 |
6 |
19 |
|
60-65 |
6 |
25 |
|
65-70 |
3 |
28 |
|
70-75 |
2 |
30 |
|
n = 30 |
th observation = 15th observation.
=15, cf = 13, f = 6 and h = 5
The following distribution gives the daily income of 50 workers of a factory.
|
Daily income (in Rs.) |
100-120 |
120-140 |
140-160 |
160-180 |
180-200 |
|
Number of workers |
12 |
14 |
8 |
6 |
10 |
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Cf distribution table
|
Daily income (in Rs.) |
Number of workers |
|
Less than 120 |
12 |
|
Less than 140 |
26 |
|
Less than 160 |
34 |
|
Less than 180 |
40 |
|
Less than 200 |
50 |
Now, we plot the points : (120,12), (140, 26), (160,34), (180,40), (200,50).
Durine the medical check-up of 35 students of a class, their weight were recorded as follows:
|
Weight (in kg) |
Number of students |
|
Less than 38 |
0 |
|
Less than 40 |
3 |
|
Less than 42 |
5 |
|
Less than 44 |
9 |
|
Less than 46 |
14 |
|
Less than 48 |
28 |
|
Less than 50 |
32 |
|
Less than 52 |
35 |
During a less than type ogive for the given data. Hence obtain the median weight from the graph - and verify the result by using the formula.
Here, 
Locatae 17.5 on the y-axis. From this point, draw a line parallel to the x-axis cutting the curve at a point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis determines the median of the given data as 46.5 kg.
Median weight by using the formula :
|
Weight (in k.g) |
No. of students |
Cumulative frequency |
|
0-38 |
0 |
0 |
|
38-40 |
3 |
3 |
|
40-42 |
2 |
5 |
|
42-44 |
4 |
9 |
|
44-46 |
5 |
14 |
|
46-48 |
14 |
28 |
|
48-50 |
4 |
32 |
|
50-52 |
3 |
35 |
|
n = 35 |
We have n = 35. So 
th observation = 17.5th observation.
So, median lies in the group of 46-48 i.e.Median class = 46-48
Now, we have Median class = 46 -48
, cf = 14, f = 14 and h = 2
The following table gives production yield per hectare of wheat of 100 farms of a village.
|
Production yield (in kg/ha) |
50-55 |
55-60 |
60-65 |
65-70 |
70-75 |
75-80 |
|
Number of farms |
2 |
8 |
12 |
24 |
38 |
16 |
Change the distribution to a more than type distribution and draw its ogive.
More than type distribution
|
More than 50 |
100 |
|
More than 55 |
98 |
|
More than 60 |
90 |
|
More than 65 |
78 |
|
More than 70 |
54 |
|
More than 75 |
|
Here n = 40 
Locate 20 on the y-axis, From this point, draw a line parallel to the x-axis. The point of intersection of this perpendicular with the x-axis gives the median of the data.
Hence, the median marks obtained by the student of the class = 54 marks.
, where the symbols have their usual meanings.
The wickets taken by a bowler in 10 cricket matches are as follows:
2, 6, 4, 5, 0, 2, 1, 3, 2, 3
Find the mode of the data.
The mode for a grouped or continuous frequency distribution is given by
where the symbols have their usual meanings.
The wickets taken by a bowler in 10 cricket matches are as follows:
3 6 4 5 0 2 1 3 2 3
Find the mode of the data
Calculate the mode of the following data.
4, 6, 7, 9, 12, 11, 13, 9, 13, 9, 9, 7, 8.
Let us form the frequency distribution table of the given data by arranging it in ascending or in descending order.
|
x |
4 |
6 |
7 |
8 |
9 |
11 |
12 |
13 |
|
f |
1 |
1 |
2 |
1 |
4 |
1 |
1 |
2 |
We observe that the value of 9 occurs most frequently i.e., 4 times in the given set of observations.
So, the mode is 9.
The wicket taken by a bowler in 10 cricket matches arc as follows :
2, 6, 4, 5, 0, 2, 11, 3, 2, 3.
Find the mode of the data.
Let us form frequency distribution table.
|
x |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
|
f |
1 |
1 |
3 |
2 |
1 |
1 |
1 |
We observe that the value of 2 occurs most frequently i.e., 3 times in the given data.
So, the mode is 2.
Thus, the mode is defined as the most frequently occurring values.
Calculate median of the following distribution.
|
Marks (x) |
No. of students |
|
10 |
2 |
|
20 |
8 |
|
30 |
16 |
|
40 |
26 |
|
50 |
20 |
|
60 |
16 |
|
70 |
7 |
|
80 |
4 |
Forming the cumulative frequency table, we have
|
Marks (x) |
No. of students |
Cum. frequencies (c.f.) |
|
10 |
2 |
2 |
|
20 |
8 |
10 |
|
30 |
16 |
26 |
|
40 |
26 |
52 |
|
50 |
20 |
72 |
|
60 |
16 |
88 |
|
70 |
7 |
95 |
|
80 |
4 |
99 |
i.e., 49.5 is 52 and the value of x (variable) corresponding to 52 is 40 therefore, median = 40 Hence, median = 40.
Calculate the median for the following:
|
Heights (am) |
120 |
121 |
122 |
123 |
124 |
125 |
|
No. of students |
8 |
12 |
17 |
14 |
13 |
6 |
Forming the cumulative frequency table, we have
|
Height (x) |
No. of students (f) |
c.f. |
|
120 |
8 |
8 |
|
121 |
12 |
20 |
|
122 |
17 |
37 |
|
123 |
14 |
51 |
|
124 |
13 |
64 |
|
125 |
6 |
70 |
Here, 
i.e., 35 is 37 and the value of x (variable) corresponding to 37 is 122.Write the median class of the following distribution.
|
Classes |
Frequency |
|
0—10 |
4 |
|
10—20 |
4 |
|
20—30 |
8 |
|
30-40 |
10 |
|
40—50 |
12 |
|
50—60 |
8 |
|
60—70 |
4 |
|
Classes |
f |
c.f |
|
0—10 |
4 |
4 |
|
10—20 |
4 |
8 |
|
20—30 |
8 |
16 |
|
30—40 |
10 |
26 |
|
40—50 |
12 |
38 |
|
50—60 |
8 |
46 |
|
60—70 |
4 |
50 |
What is the lower limit of the modal class of the following frequency distribution ?
|
Age (in years) |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
|
Number of patients |
16 |
13 |
6 |
11 |
27 |
18 |
Modal class = 40-50
Lower Limit = 40
Solution not provided.
Ans. 45 Marks
The marks obtained by 40 students of class X of a certain school in Science paper consisting of 10 marks are presented in the table below. Find the mean marks obtained by the students.
|
Marks obtained(xi) |
5 |
6 |
7 |
8 |
9 |
|
Number of students (fi) |
4 |
8 |
14 |
11 |
3 |
|
Marks Obtained (xi) |
Number of students (fi) |
fixi |
|
5 |
4 |
20 |
|
6 |
8 |
48 |
|
7 |
14 |
98 |
|
8 |
11 |
88 |
|
9 |
3 |
27 |
|
Σfi = 40 |
Σfixi = 281 |
Consider the following distribution of daily wages of 50 workers of a factory.
|
Daily wages |
100-120 |
120-140 |
140-160 |
160-180 |
180-200 |
|
No. of workers |
12 |
14 |
8 |
6 |
10 |
Find the mean daily wages of the workers.
|
Daily wages (C.I.) |
No. of workers (fi) |
Class-Mark (xi) |
fi xi |
|
100 - 120 |
12 |
110 |
1320 |
|
120 - 140 |
14 |
130 |
1820 |
|
140 - 160 |
8 |
150 |
1200 |
|
160 - 180 |
6 |
170 |
1020 |
|
180 - 200 |
10 |
190 |
1900 |
|
Σfi = 50 |
Σfi xi = 7260 |
Find the mean of the following frequency distribution.
|
Class-Interval |
0-8 |
8-16 |
16-24 |
24-32 |
32-40 |
|
Frequency |
8 |
10 |
15 |
9 |
8 |
|
Class-Interval (C.I.) |
Frequency (fi) |
Class-mark (xi) |
(fixi) |
|
0 - 8 |
8 |
4 |
32 |
|
8 - 16 |
10 |
12 |
120 |
|
16 - 24 |
15 |
20 |
300 |
|
24 - 32 |
9 |
28 |
252 |
|
32 - 40 |
8 |
36 |
288 |
|
Σfi = 50 |
Σfixi = 992 |
Now,
Hence, (he mean of frequency distriution is 19.84.
If the values of x or f are large, the calculation of mean by the direct method is quite lengthy and time consuming so, to minimize the time involved in calculation, we prefer assumed mean method.
Find the value of P, if the mean of the following distribution is 18.
|
x: |
13 |
15 |
17 |
19 |
20 + p |
23 |
|
f: |
8 |
2 |
3 |
4 |
5p |
6 |
|
xi |
fi |
fixi |
|
13 |
8 |
104 |
|
15 |
2 |
30 |
|
17 |
3 |
51 |
|
19 |
4 |
76 |
|
20 + p |
5p |
5p (20 + p) |
|
23 |
6 |
138 |
|
Σfi = 23 + 5p |
Σfixi = 399 + 5p2 + 100p |
18(5p + 23) = 5p2 + 100p + 399
90p + 414 = 5p2 + 100p + 399
5p2 + 10p - 15 = 0
p2 + 2p - 3 = 0
p2 + 3p - p - 3 = 0
p(p + 3) - 1 (p + 3) = 0
(p - 1) (p + 3) = 0
p - 1 = 0 or p + 3 = 0The following table shows markes scored by 140 students in a examinations.
|
Marks |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
|
No. of Sutdents |
20 |
24 |
40 |
36 |
20 |
|
Marks (C.I.) |
No. of students (fi) |
Class mark (xi) |
di = xi - 25 |
fidi |
|
0-10 |
20 |
5 |
- 20 |
-400 |
|
10-20 |
24 |
15 |
-10 |
-240 |
|
20-30 |
40 |
25=A |
0 |
0 |
|
30-40 |
36 |
35 |
10 |
360 |
|
40-50 |
20 |
45 |
20 |
400 |
|
Σfi = 140 |
Σfidi = 120 |
Now,
Hence, the mean marks obtained by 140 students is 25.86.
Find the mean of the following data
|
Class-lnterval |
15-25 |
25-35 |
35-45 |
45-55 |
55-65 |
|
Frequency |
60 |
35 |
22 |
18 |
15 |
|
C.I. |
fi |
Mid-point (xi) |
di = xi - 40 |
fidi |
|
15-25 |
60 |
20 |
-20 |
-1200 |
|
25-35 |
35 |
30 |
-10 |
-350 |
|
35-45 |
22 |
40 = A |
0 |
0 |
|
45-55 |
18 |
50 |
10 |
180 |
|
55-65 |
15 |
60 |
20 |
300 |
|
Σfi = 150 |
Σfidi = -1070 |
The following table represents the wages of 50 workers in a factory.
|
Wages (in Rs.) |
1450 |
1475 |
1500 |
1525 |
1550 |
|
No. of workers |
12 |
13 |
7 |
10 |
8 |
|
Wages (Rs.) (xi) |
No. of workers (fi) |
di = xi-500 |
fidi |
|
1450 |
12 |
-50 |
-600 |
|
1475 |
13 |
-25 |
-325 |
|
1500 = A |
7 |
0 |
0 |
|
1525 |
10 |
25 |
250 |
|
1550 |
8 |
50 |
400 |
|
Σfi = 50 |
Σfidi=-275 |
The following table represents the expenditure on the usage of water by 70 families in a locality. Find the average expenditure per family.
|
Exp. water (in Rs.) |
15-20 |
20-25 |
25-30 |
30-35 |
35-40 |
40-45 |
45-50 |
50-55 |
|
No. of families |
7 |
8 |
7 |
8 |
10 |
15 |
7 |
8 |
The following table shows marks scored by 140 students in an examination.
|
Marks |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
|
No. of students |
20 |
24 |
40 |
36 |
20 |
Calculate mean marks by Step Deviation Method.
The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequency f1 and f2.
|
C.I. |
0-20 |
20-40 |
40-60 |
60-80 |
80-100 |
100-120 |
Total |
|
f |
5 |
f1 |
10 |
f2 |
7 |
8 |
50 |
30 + f1 + f2 = 50
f1 + f2 = 20
314 = -2f1 + 2f2 + 306 
2f1- 2f2 = -8
f1 = 8The mean of the following frequency distribution is 132 and the sum of observations is 50. Find the missing frequency f1 and f2.
|
Class |
0-40 |
40-80 |
80-120 |
120-160 |
160-200 |
200-240 |
|
Freq. |
4 |
7 |
f1 |
12 |
f2 |
9 |
A (assumed mean) = 100
f1 + f2 + 32 = 50
f1 + f3 = 18
132 = 100 + 
132 = 100 + 
132 = 
132 x 5 = 8f2 + 596
660 = 8f2 + 596
8f2 = 64
f2 = 8
f1 = 10The mean of the following frequency table is 53. But the frequencies f1 and f2 in the classes 20-40 and 60-80 arc missing. Find the missing frequencies.
|
Age (in years) |
0-20 |
20-40 |
40-60 |
60-80 |
80-100 |
Total |
|
Number of people |
15 |
f1 |
21 |
f2 |
17 |
100 |
|
Age (in yrs) class interval |
Mid values xi |
Frequency fi |
fi.xi |
|
0-20 |
10 |
15 |
150 |
|
20-40 |
30 |
f1 |
30f1 |
|
40-60 |
50 |
21 |
1050 |
|
60-80 |
70 |
f2 |
70f2 |
|
80-100 |
90 |
17 |
1530 |
|
Σfi = f1 + f2 + 53 |
Σfixi = 2730 + 30f1 + 70f2 |
15 + fi + 21 + f2 + 17 = 100
f1 + f2 + 53 = 100
f1 + f2 = 47
53 = 
53 x 100 = 2730 + 30f1 + 70f2 
30f1 + 70f2 = 5300 - 2730
30f1 + 70f2 = 2570
3f1 + 7f2 = 257
3f1 + 3f2 - 3f1-7f2 = -116
-4f2 = -116
f2 = 29
f1 + 29 = 47
f1= 1= 47 - 29
f1 = 18Find the mode of the distribution from the following data.
|
Class |
Frequcncy |
|
10-15 |
3 |
|
15-20 |
2 |
|
20-25 |
10 |
|
25-30 |
7 |
|
30-35 |
20 |
|
35-40 |
5 |
|
40-45 |
8 |
Here, the maximum class frequency is 20 and the class corresponding to this frequency is 30-35, so the modal class is 30-35.
Thus, we have
Modal class = 30-35
l (lower limit of modal class) 30
Class size (h) = 5
Frequency (f1) of the modal class = 20
Frequency (f0) of class preceding the modal class = 7
and Frequency (f2) of the class succeeding the modal class = 5
Henc, mode of the given data is 32.321.
Now, substitute the values in the formula of mode, we get
Determine the value of mode from the following frequency distribution table.
|
Murks (C.I.) |
No. of students (f) |
|
0-10 |
5 |
|
10-20 |
12 |
|
20-30 |
14 |
|
30-40 |
10 |
|
40-50 |
8 |
|
50-60 |
6 |
Here, the maximum class frequency is 14 and the class corresponding to frequency is 20-30.
So, the modal class is 20-30.
Thus, we have
Modal class = 20-30
l (lower limit of modal class) = 20
h (class size) = 10
f1 = (frequency of the modal class) = 14
f0 = (frequency of class preceding the modal class) = 12
f2 = (frequency of the class succeeding the modal class) = 10
Now, substituting these values in the formula of mode, we get
The marks distribution of 30 students in a mathematics are given in the table. Find the mode of this data. Also compare and interpret the mode and mean.
|
Class interval |
Number of students (f) |
|
10-25 |
2 |
|
25-40 |
3 |
|
40-55 |
7 |
|
55-70 |
6 |
|
70-85 |
6 |
|
85-100 |
6 |
Case I: Finding Mode :
Here, the maximum class frequency is 7 and the class corresponding to frequency is 40 - 55. So the modal class is 40 - 55.
Thuse we have
Model class = 40.55
l = 40
f1 = 7
f0 =3
f2 = 6
and h = 15
Npw substituting these values in the formula of mode, we get
Case II: Finding Mean
|
C.I. |
fi |
xi |
di = x2- 62.5 |
fiui = di/h |
fiui |
|
10-25 |
2 |
17.5 |
- 45 |
- 3 |
- 6 |
|
25-40 |
3 |
32.5 |
- 30 |
- 2 |
- 6 |
|
40-55 |
7 |
47.5 |
- 15 |
- 1 |
- 7 |
|
55-70 |
6 |
62.5 = A |
0 |
0 |
0 |
|
70-85 |
6 |
77.5 |
15 |
1 |
6 |
|
85-100 |
6 |
92.5 |
30 |
2 |
12 |
|
Σfi = 30 |
Σfiui= -1 |
Hence, the mode of the given data is 52 while median is 62.
Interpretation : Maximum number of students obtained 52 marks, while on an average student obtained 62 marks.
The following table shows the marks obtained by 100 students of class X in a school during a particular academic session. Find the mode of this distribution.
|
Marks |
No. of students |
|
Less than 10 |
7 |
|
Less than 20 |
21 |
|
Less than 30 |
34 |
|
Less than 40 |
46 |
|
Less than 50 |
66 |
|
Less than 60 |
77 |
|
Less than 70 |
92 |
|
Less than 80 |
100 |
|
Class Marks (C.I.) |
No. of students (c.f.) |
f |
|
Less than 10 |
7 |
7 |
|
Less than 20 |
21 |
14 |
|
Less than 30 |
34 |
13 |
|
Less than 40 |
46 |
12 |
|
Less than 50 |
66 |
20 |
|
Less than 60 |
77 |
11 |
|
Less than 70 |
92 |
15 |
|
Less than 80 |
100 |
8 |
The class (40-50) has maximum frequency i.e. 20 therefore, modal class is 40-50.
f = 20
Now, l = 40, h = 10m, f1 = 20 and f2 = 11
f0 = 12 and modal class = 40 - 50 
Calculate the median of the following distribution of incomes of employees of a company.
|
Income |
No. of Persons |
|
400-500 |
25 |
|
500-600 |
69 |
|
600-700 |
107 |
|
700-800 |
170 |
|
800-900 |
201 |
|
900-1000 |
142 |
|
1000-1100 |
64 |
Forming the cumulative frquency table, we get
|
Income |
No. of Persons |
c.f. |
|
400-500 |
25 |
25 |
|
500-600 |
69 |
94 |
|
600-700 |
107 |
201 |
|
700-800 |
170 |
371 |
|
800-900 |
201 |
572 |
|
900-1000 |
142 |
714 |
|
1000-1100 |
64 |
778 |
Find the mean, mode and median for the following data :
|
Class |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
Total |
|
Frequency |
8 |
16 |
36 |
34 |
6 |
100 |
III. Finding Mode
We have,
Modal class = 20-30 l = 20, h = 10, f1 = 36, f0 = 16 and f2 = 34
Now,
The median of the following data is 52.5. Find the values of x and y if the total frequency is 100.
|
Class Interval |
Frequency |
|
0-10 |
2 |
|
10-20 |
5 |
|
20-30 |
x |
|
30-40 |
12 |
|
40-50 |
17 |
|
50-60 |
20 |
|
60-70 |
y |
|
70-80 |
9 |
|
80-90 |
7 |
|
90-100 |
4 |
|
Total 100 |
|
C.l. |
fi |
c.f. |
|
0-10 |
2 |
2 |
|
10-20 |
5 |
7 |
|
20-30 |
x |
7 + x |
|
30-40 |
12 |
19 + |
|
40-50 |
17 |
36 + x |
|
50-60 |
20 |
56 + x |
|
60-70 |
y |
56 + x + y |
|
70-80 |
9 |
65 + x + y |
|
80-90 |
7 |
72 + x + y |
|
90-100 |
4 |
76 + x + y |
It is given that n = 100
∴ 76 + x + y 100 ⇒ x + y = 24 ...(i)
The median is 52.5 which lies in the class 50-60.
∴ l = 50,f = 20, c.f. = 36 + x, h = 10
Using the formula.
52.5 - 50 = (14 - x) x 0.5
2.5 = 7 -0.5x 
25 = 70 - 5x
5x = 70 - 25 = 45
x = 9
Putting this value of x in (i), we get
9 + y = 24
y = 24 - 9 = 15
Hence, x = 9 and y= 15
Calculate the median from the following data:
|
Value |
Frequency |
|
Less than 10 |
4 |
|
Less than 20 |
16 |
|
Less than 30 |
40 |
|
Less than 40 |
76 |
|
Less than 50 |
96 |
|
Lc9s than 60 |
112 |
|
Less than 70 |
120 |
|
Less than 80 |
125 |
Forming the Cumulative frequency table we have.
|
Value |
Frequency |
C.f |
|
0-10 |
4 |
4 |
|
10-20 |
12 |
16 |
|
20-30 |
24 |
40 |
|
30-40 |
36 |
76 |
|
40-50 |
20 |
96 |
|
50-60 |
16 |
112 |
|
60-70 |
8 |
120 |
|
70-80 |
5 |
125 |
or 62.5th obser vation, which lies in class 30-40.
Calculate the median from the following data:
|
Size |
Frequency |
|
More than 50 |
0 |
|
More than 40 |
40 |
|
More than 30 |
98 |
|
More than 20 |
123 |
|
More than 10 |
165 |
Forming the cumulatiue frequency table, we have
|
C.I. |
f |
C.f |
|
10-20 |
42 |
42 |
|
20-30 |
25 |
67 |
|
30-40 |
58 |
125 |
|
40-50 |
40 |
165 |
or 82.5th obser vation, which lies in class 30-40.
The median of the following data is 20.75. Find the missing frequencies x and y, if the total frequency is 100.
|
Class Interval |
0 - 5 |
5 - 10 |
10 - 15 |
15 - 20 |
20 - 25 |
25 - 30 |
30 - 35 |
35 - 40 |
|
Frequency |
7 |
10 |
x |
13 |
y |
10 |
14 |
9 |
Here, the missing frequencies are x and y.
|
Class intervals |
Frequency (f) |
Cumulative frequency |
|
0-5 |
7 |
7 |
|
5-10 |
10 |
17 |
|
10-15 |
x |
17 + x |
|
15-20 |
13 |
30 + x |
|
20-25 |
y |
30 + x + y |
|
25-30 |
10 |
40 + x + y |
|
30-35 |
14 |
54 + x + y |
|
35-40 |
9 |
63 + x + y |
|
Total |
100 |
It is given that N = 100 = Total frequency
∴ 63 + x + y = 100
⇒ x + y = 100 - 63
⇒ x + y = 37
⇒ y = 37-x ...(i)
The median is 20.75 (given), which lies in the class 20-25 ⇒ Median class = 20-25
Here, we have l = lower limit of median class = 20, Median class = 20-25
f = frequency of median class = y
cf = cumulative frequency of class preceding the median class = 30 + x
and h = class size = 5
Substituting these values in the formula of median, we get
3y = 400 - 20 x
3(37 - x) = 400 - 20x [From (i)]
111 - 3x = 400 - 20x
17x = 289
x = 17
Substituting x = 17 in (i), we get
y = 37 -17 = 20
Hence, the missing frequencies are x = 17 and y = 20.
A survey regarding the heights (in cm) of 50 girls of class X of a school was conducted and the following data was obtained:
|
Height in cm |
120 - 130 |
130 - 140 |
140 - 150 |
150 - 160 |
160 - 170 |
Total |
|
Number of girls |
2 |
8 |
12 |
20 |
8 |
50 |
Finding Median:
|
Class Interval |
Frequency (f1) |
Cf |
|
120 - 130 |
2 |
2 |
|
130 - 140 |
8 |
10 |
|
140 - 150 |
12 |
22 |
|
150 - 160 |
20 |
42 |
|
160 - 170 |
8 |
50 |
observationwhich lies in the class 150-160
Thus we have
Median class = 150 - 160
l = 150, 
, h = 10, cf = 22 and f = 20
Putting these values in the formula of Median, we get
III. Finding Mode
|
Class Interval |
No. of Students (f) |
|
120 - 130 |
2 |
|
130 - 140 |
8 |
|
140 - 150 |
12 |
|
150 - 160 |
20 |
|
160 - 170 |
8 |
Here, the man class frequency is 20 and the class corresponding to frequency is 150 - 160, So, the modal class is 150 - 160.
Thus, we have
Modal class = 150-160
l = 150,f1 = 20
f0 = 12, f2 = 8, n = 10
Now,
Substituting these values in the formula of made, we get
Find mean, median and mode of the following data:
|
Classes |
Frequency |
|
0-20 |
6 |
|
20-40 |
8 |
|
40-60 |
10 |
|
60-80 |
12 |
|
80-100 |
6 |
|
100-120 |
5 |
|
120-140 |
3 |
Therefore, 60 - 80 is the median class. Thus, the lower limit (l) of the median class is 60.
Using the formula for calculating the median.
II. Finding median: 
= 60 + 5 = 65
Find mean, median and mode of the following data:
|
Classes |
Frequency |
|
0 - 50 |
2 |
|
50 - 100 |
3 |
|
100 - 150 |
5 |
|
150 - 200 |
6 |
|
200 - 250 |
5 |
|
250 - 300 |
3 |
|
300 - 350 |
1 |
From the table, 
|
I. Finding Mean :
Now, 150 - 200 is the class whose cumulative frequency 16 is greater than 
Therefore, 150 - 200 is the median class. Thus, the lower limit (l) of the median class is 150.
II. Finding median:
III. Finding mode:
= 150 + 25 = 175
The following table gives the daily income of 50 workers of a factory :
|
Daily income (in Rs.) |
No. of Workers |
|
100-120 |
12 |
|
120-140 |
14 |
|
140-160 |
8 |
|
160-180 |
6 |
|
180-200 |
10 |
Find the Mean, Mode and Median of the above data.
|
C.I. |
xi |
fi |
fiixi |
c.f |
|
100-120 |
110 |
12 |
1320 |
12 |
|
120-140 |
130 |
14 |
1820 |
26 |
|
140-160 |
150 |
8 |
1200 |
34 |
|
160-180 |
170 |
6 |
1020 |
40 |
|
180-200 |
190 |
10 |
1900 |
50 |
|
Σfi = 50 |
Σfixi = 7260 |
Model class is 120 - 140, 
Class size h = 20
which lies in the class 120 - 140
l = 120. c.f. = 12, f = 14, h = 20
Find the mode, median and mean for the following data :
|
Marks Obtained |
Number of Students |
|
25-35 |
7 |
|
35-45 |
31 |
|
45-55 |
33 |
|
55-65 |
17 |
|
65-75 |
11 |
|
75-85 |
1 |
|
Marks obtained |
xi |
Frequency fi |
di |
c.f |
fidi |
|
25-35 |
30 |
7 |
-30 |
7 |
-210 |
|
35-45 |
40 |
31 |
-20 |
38 |
-620 |
|
45-55 |
50 |
33 |
-10 |
71 |
-330 |
|
55-65 |
60 |
17 |
0 |
88 |
0 |
|
65-75 |
70 |
11 |
10 |
99 |
110 |
|
75-85 |
80 |
1 |
20 |
100 |
20 |
|
Σf = 100 |
Σfidi = -1030 |
The table given below shows the frequency distribution of the scores obtained by 200 candidates in a BCA examination.
|
Score |
No. of candidates |
|
200-250 |
30 |
|
250-300 |
15 |
|
300-350 |
45 |
|
350-400 |
20 |
|
400-450 |
25 |
|
450-500 |
40 |
|
500-550 |
10 |
|
550-600 |
15 |
Draw cumulative frequency curves by using (i) 'less than series', (ii) 'more than series'.
e assume a class interval 150-200 prior to the first class interval 200-250 wtih zero frequency.
Cumulative frequency distribution [Less than Series]
|
Scroe |
c.f. |
|
Less than 200 |
0 |
|
Less than 250 |
30 |
|
Less than 300 |
45 |
|
Less than 350 |
90 |
|
Less than 400 |
110 |
|
Less than 450 |
135 |
|
Less than 500 |
175 |
|
Less than 550 |
185 |
|
Less than 600 |
200 |
More than Series
|
Scroe |
c.f. |
|
More than 200 |
200 |
|
More than 250 |
170 |
|
More than 300 |
155 |
|
More than 350 |
110 |
|
More than 400 |
90 |
|
More than 450 |
65 |
|
More than 500 |
25 |
|
More than 550 |
15 |
|
More than 600 |
0 |
Now, we plot the pts.: (200, 200), (250,170), (300,155), (350,110), (400, 90), (450, 65), (500, 25), (550, 15) and (600, 0).
Draw both types of cumulative freqneucy curve on the same graph paper and then determine the median.
|
Marks obtained |
No. of students |
|
50-60 |
4 |
|
60-70 |
8 |
|
70-80 |
12 |
|
80-90 |
6 |
|
90-100 |
6 |
c.f. distribution table :
|
Marks |
No. of students |
c.f. (Less than) |
c.f (More tlian) |
|
50-60 |
4 |
4 |
36 |
|
60-70 |
8 |
12 |
32 |
|
70-80 |
12 |
24 |
24 |
|
80-90 |
6 |
30 |
12 |
|
90-100 |
6 |
36 |
6 |
Now, we plot the points (60,4), (70,12), (80,24), (90,30), (100,36), for less than series.
And. (50,36), (60,32), (70,24), (80,12), (90,6) for more than series.
The two curves drawn intersect each other at point, say P. Through this point P, draw a vertical line, which meets x-axis at 76.
So, median = 76
Draw 'less than' and 'more than' ogive curve from the following and indicate the value of median.
|
Marks |
No. of students (f) |
|
0-5 |
7 |
|
5-10 |
10 |
|
10-15 |
20 |
|
15-20 |
13 |
|
20-25 |
12 |
|
25-30 |
10 |
|
30-35 |
14 |
|
35-40 |
9 |
c.f. distribution
|
Marks |
No. of students |
c.f. (Less than) |
c.f (More than) |
|
0-5 |
7 |
7 |
95 |
|
5-10 |
10 |
17 |
88 |
|
10-15 |
20 |
37 |
78 |
|
15-20 |
13 |
50 |
58 |
|
20-25 |
12 |
62 |
45 |
|
25-30 |
10 |
72 |
33 |
|
30-35 |
14 |
86 |
23 |
|
35-40 |
9 |
95 |
9 |
Now we plot the points (5,7), (10,17), (15,37), (20, 50), (25, 62), (30, 72), (35, 86), (40, 95) for less than series. And (0,95), (5,88), (10,78), (15,58), (20, 45), (25,33), (30,23), (35,9) for more than series.
The two curves drawn intersect each other at P. Through this P, draw a vertical line, which meets the x-axis at 20. So median = 20.
|
Daily Income (in Rs.) |
Number of Workers |
|
100-120 |
12 |
|
120-140 |
14 |
|
140-160 |
8 |
|
160-180 |
6 |
|
180-200 |
10 |
|
Marks |
No of students |
c.f. (Less than) |
c.f. (More than) |
|
100-120 |
12 |
12 |
50 |
|
120-140 |
14 |
26 |
38 |
|
140-160 |
8 |
34 |
24 |
|
160-180 |
6 |
40 |
16 |
|
180-200 |
10 |
50 |
10 |
No we plot the points (100, 0), (120, 12), (140, 26), (160, 34), (180, 40), (200, 50) for less than series. And (100, 50), (120, 38), (140, 24), (160,16), (180,10) for more than series.
During the medical check-up of 35 students of a class their weights were recorded as follows:
|
Weight (in kg.) |
No. of Students |
|
38—40 |
3 |
|
40—42 |
2 |
|
42—44 |
4 |
|
44—46 |
5 |
|
46—48 |
14 |
|
48—50 |
4 |
|
50—52 |
3 |
Draw a less than type and a more than type ogive from the given data. Hence obtain the median weight from the graph.
|
Classes |
f |
c.f. |
c.f. |
||
|
38-40 |
3 |
3 |
(40, 3) |
35 |
(38, 35) |
|
40-42 |
2 |
5 |
(42, 5) |
32 |
(40, 32) |
|
42-44 |
4 |
9 |
(44, 9) |
30 |
(42, 30) |
|
44-46 |
5 |
14 |
(46, 14) |
26 |
(44. 26) |
|
46-48 |
14 |
28 |
(48, 28) |
21 |
(46, 21) |
|
48-50 |
4 |
32 |
(50, 32) |
7 |
(48, 7) |
|
50-52 |
3 |
35 |
(52, 35) |
3 |
(50, 3) |
We plot the points (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35). We join these points with a smooth curve to get the 'less than' ogive as shown in Fig. 14.20.
Then, we plot the points (38, 35), (40, 32), (42, 30), (44, 26), (46, 21), (48, 7), (50, 3) on the same axes. By joining these points with a smooth curve to get 'more than' ogive. Since, the two curves intersect at the point, whose abscissa is 47 (approx). Hence, the required median weight is 47 kg (approx.).
The marks obtained by 40 students of class-X of a certain school in Mathematics unit test consisting of 10 marks are presented in table below. Find the mean of the marks obtained by the students.
|
Marks obtained (xi) |
5 |
6 |
7 |
8 |
9 |
|
No. of students (fi) |
4 |
8 |
14 |
11 |
3 |
Solution not provided.
Ans. 7.025
A survey was conducted by a group of persons as a part of energy consumption awareness in which they collected the following data regarding the number of electric summer cooler in 15 houses in a locality in North Delhi. Find the mean number of coolers per house.
|
No. of coolers |
0-2 |
2-4 |
4-6 |
6-8 |
8-10 |
|
No. of houses |
5 |
3 |
4 |
2 |
1 |
Which method do you use for finding the mean and why?
Solution not provided.
Ans. 3.8
Consider the following distribution of daily wages of 100 workers in a jute Mill.
|
Daily wages (in Rs.) |
80-100 |
100-120 |
120-140 |
140-160 |
160-180 |
|
No. of workers |
40 |
20 |
15 |
15 |
10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution not provided.
Ans. 117
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 7.5. Find the missing frequency f.
|
Daily Pocket allowance (in Rs.) |
2-4 |
4-6 |
6-8 |
8-10 |
10-12 |
12-14 |
|
No. of children |
6 |
8 |
15 |
f |
8 |
4 |
Solution not provided.
Ans. 3
If the mean of the following frequency distribution is 8. Find the value of p.
|
x |
3 |
5 |
7 |
9 |
11 |
13 |
|
f |
6 |
8 |
15 |
P |
8 |
4 |
Solution not provided.
Ans. 25
If the mean of the following distribution is 6. Find the value of p.
|
x |
2 |
4 |
6 |
10 |
p + 5 |
|
f |
3 |
2 |
3 |
1 |
2 |
Solution not provided.
Ans. 7
The data of number of patients attending a hospital in a month are given below. Find the average number of patients attending the hospital in a day.
|
No. of patients |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
|
No. of days attending hospitals |
2 |
6 |
9 |
7 |
4 |
2 |
Solution not provided.
Ans. 28.67
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.
|
No. of cars |
Frequency |
|
0-10 |
7 |
|
10-20 |
14 |
|
20-30 |
13 |
|
30-40 |
12 |
|
40-50 |
20 |
|
50-60 |
11 |
|
60-70 |
15 |
|
70-80 |
8 |
Solution not provided.
Ans. 44.7
The given distribution shows the number of wickets taken by some top bowlers of the world in one day cricket matches.
|
Wickets taken |
No. of bowlers |
|
50-80 |
50 |
|
80-110 |
40 |
|
110-140 |
30 |
|
140-170 |
60 |
|
170-200 |
15 |
|
200-230 |
10 |
|
230-260 |
8 |
Find the mode of the data.
Solution not provided.
Ans. 152
The following is the frequency distribution of marks obtained by class-X students of a school in CBSE BOARD EXAMINATION in Mathematics. Calculate the modal Marks.
|
Marks |
No. of Students |
|
40-50 |
20 |
|
50-60 |
15 |
|
60-70 |
30 |
|
70-80 |
12 |
|
80-90 |
15 |
|
90-100 |
8 |
Solution not provided.
Ans. 64.54
Find the modal age in years from the frequency distribution given below:
|
Age (in yrs.) |
Frequency |
|
25-29 |
4 |
|
30-34 |
14 |
|
35-39 |
22 |
|
40-44 |
16 |
|
45-49 |
6 |
|
50-54 |
5 |
|
55-59 |
3 |
Solution not provided.
Ans. 37.38
Calculate the modal income for the following data :
|
Income (in rupees per month) |
No. of employees |
|
5000-5500 |
8 |
|
5500-6000 |
15 |
|
6000-6500 |
25 |
|
6500-7500 |
20 |
|
7500-8000 |
10 |
|
8000-8500 |
5 |
Solution not provided.
Ans. 6200
Calculate mode for the following data :
|
Rent (in Rs/month) |
No. of houses |
|
150-250 |
8 |
|
250-350 |
10 |
|
350-450 |
15 |
|
450-550 |
25 |
|
550-650 |
40 |
|
650-750 |
20 |
|
750-850 |
15 |
|
850-950 |
7 |
Solution not provided.
Ans. Rs. 571.40
The following table gives the length of life of 150 electric bulbs:
|
Life (in hours) |
No. of bulbs |
|
0-400 |
4 |
|
400-800 |
12 |
|
800-1200 |
40 |
|
1200-1600 |
41 |
|
1600-2000 |
27 |
|
2000-2400 |
13 |
|
2400-2800 |
9 |
|
2800-3200 |
4 |
Calculate the modal life.
Solution not provided.
Ans. 1226.67
Find the mode from the following data :
|
Marks |
No. of students |
|
Above 0 |
80 |
|
Above 10 |
77 |
|
Above 20 |
72 |
|
Above 30 |
65 |
|
Above 40 |
55 |
|
Above 50 |
43 |
|
Above 60 |
28 |
|
Above 70 |
16 |
|
Above 80 |
10 |
|
Above 90 |
8 |
|
Above 100 |
0 |
Solution not provided.
Ans. 55
Find mode for the following distribution:
|
Class |
Frequency |
|
65-7.5 |
5 |
|
7.5-8.5 |
12 |
|
8.5-9.5 |
25 |
|
9.5-10.5 |
48 |
|
10.5-11.5 |
32 |
|
11.5-12.5 |
6 |
|
12.5-13.5 |
1 |
Solution not provided.
Ans. 10.2
Find the mode for the following data concerning the employees of a factory:
|
Daily income (in rupees) |
No. of employees |
|
|
0-50 |
90 |
|
|
50-100 |
150 |
|
|
100-150 |
100 |
|
|
150-200 |
80 |
|
|
200-250 |
70 |
|
|
250-300 |
10 |
|
Solution not provided.
Ans. 77.27
For the following grouped frequency distribution, find the mode:
|
Class |
Frequency |
|
3-6 |
2 |
|
6-9 |
5 |
|
9-12 |
10 |
|
12-15 |
23 |
|
15-18 |
21 |
|
18-21 |
12 |
|
21-24 |
3 |
Solution not provided.
Ans. 14.6
Find mode for the following distribution:
|
Score |
No. of Pupil |
|
80-90 |
18 |
|
90-100 |
27 |
|
100-110 |
48 |
|
110-120 |
39 |
|
120-130 |
12 |
|
130-140 |
6 |
|
140-150 |
16 |
Solution not provided.
Ans. 107
Find mode for the following distribution:
|
Wages (in Rs.) |
No. of workers |
|
51-56 |
12 |
|
57-62 |
24 |
|
63-68 |
40 |
|
69-74 |
30 |
|
75-80 |
18 |
|
81-86 |
8 |
|
87-92 |
20 |
Solution not provided.
Ans. 68.4
Find mode for the following distribution:
|
Marks |
Frequency |
|
10-19 |
19 |
|
20-29 |
21 |
|
30-39 |
27 |
|
40-49 |
21 |
|
50-59 |
22 |
|
60-69 |
20 |
Solution not provided.
Ans. 34.5
Find the median from the following data.
|
Marks |
No. of Students |
|
Below 10 |
12 |
|
Below 20 |
32 |
|
Below 30 |
57 |
|
Below 40 |
80 |
|
Below 50 |
92 |
|
Below 60 |
116 |
|
Below 70 |
164 |
|
Below 80 |
200 |
Solution not provided.
Ans. 53.33
Compute the median from the following distribution of monthly income (in Rs.) of a locality.
|
No. of families |
Income |
|
Below 100 |
50 |
|
100-200 |
50 |
|
200-300 |
555 |
|
300-400 |
100 |
|
400-500 |
3 |
|
500 and above |
2 |
Solution not provided.
Ans. 250.45
The following table gives Intelligent Quotient (I.Q.) of 100 students:
|
I.Q. |
No. of students |
|
55-65 |
1 |
|
65-74 |
2 |
|
75-84 |
9 |
|
85-94 |
22 |
|
95-104 |
33 |
|
105-114 |
22 |
|
115-124 |
8 |
|
125-134 |
2 |
|
135-144 |
1 |
Find the median.
Solution not provided.
Ans. 99.35
The following table gives heights (in cms) of 420 persons:
|
Heights (in cms) |
No. of persons |
|
160-162 |
15 |
|
163-165 |
118 |
|
166-168 |
142 |
|
169-171 |
127 |
|
172-174 |
18 |
Find the median height.
Solution not provided.
Ans. 167.13
Find the median for the following distribution:
|
Marks |
No. of students |
|
Below 10 |
3 |
|
Below 20 |
10 |
|
Below 30 |
23 |
|
Below 40 |
55 |
|
Below 50 |
70 |
|
Below 60 |
75 |
|
Below 70 |
80 |
Solution not provided.
Ans. 35.3
Find the median wage of the workers from the following frequency distribution table:
|
Wages (in Rs.) |
No. of workers |
|
More than 150 |
0 |
|
More than 140 |
10 |
|
More than 130 |
29 |
|
More than 120 |
60 |
|
More than 110 |
104 |
|
More than 100 |
134 |
|
More than 90 |
151 |
|
More than 80 |
160 |
Solution not provided.
Ans. 115.45
Frequency table of the marks obtained by 50 students is given below :
|
Marks obtained |
No. of students |
|
0-10 |
3 |
|
10-20 |
f1 |
|
20-30 |
20 |
|
30-40 |
10 |
|
40-50 |
5 |
|
50-60 |
f2 |
Given that the median marks are 28.5, find the missing frequencies f1 and f2.
Solution not provided.
Ans. 5.7
Find the missing frequencies in the following distribution. It is given that median of the distribution is 41 and the total number of observations is 82.
|
Class |
Frequency |
|
10-20 |
10 |
|
20-30 |
f1 |
|
30-40 |
15 |
|
40-50 |
20 |
|
50-60 |
f2 |
|
60-70 |
11 |
Solution not provided.
Ans. 14, 12
In the following frequency distribution, locate the median:
|
Monthly consumption of electricity |
No. of consumers |
|
65-85 |
4 |
|
85-105 |
5 |
|
105-125 |
13 |
|
125-145 |
20 |
|
145-165 |
14 |
|
165-185 |
7 |
|
185-205 |
4 |
Problems Based on Graphical Representation of Cumulative Frequency Distribuition
Solution not provided.
Ans. 136.5
In the following distribution frequency of the class-interval 40-50 and 70-80 are missing. If the mean of the distribution is 52. Find the missing frequencies.
|
Wages (Rs.) |
Number of workers |
|
10-20 |
5 |
|
20-30 |
3 |
|
30-40 |
4 |
|
40-50 |
- |
|
50-60 |
2 |
|
60-70 |
6 |
|
70-80 |
- |
|
Total |
4 |
Solution not provided.
Ans. 7, 13
The mean of the following data is 38.7. Find the missing frequencies f1 and f2.
|
Classes |
Frequencies |
|
0-10 |
5 |
|
10-20 |
7 |
|
20-30 |
f1 |
|
30-40 |
3 |
|
40-50 |
f2 |
|
50-60 |
9 |
|
60-70 |
6 |
|
Total |
100 |
Solution not provided.
Ans. 20, 50
Find the mean marks of the following data :
|
Marks |
Number of students |
|
Below 20 |
15 |
|
Below 40 |
31 |
|
Below 60 |
55 |
|
Below 80 |
70 |
|
Below 100 |
80 |
Solution not provided.
Ans. 47.25
Find the mean marks from the following data :
|
Marks |
Number of students |
|
Below 10 |
4 |
|
Below 20 |
12 |
|
Below 30 |
22 |
|
Below 40 |
34 |
|
Below 50 |
48 |
|
Below 60 |
64 |
|
Below 70 |
78 |
|
Below 80 |
87 |
|
Below 90 |
92 |
|
Below 100 |
100 |
Solution not provided.
Ans. 50.9
Find the mean of following cumulative frequency distribution:
|
Marks |
Number of observations |
|
less than 10 |
0 |
|
less than 15 |
5 |
|
less than 20 |
11 |
|
less than 25 |
19 |
|
less than 30 |
31 |
|
less than 35 |
37 |
|
less than 40 |
40 |
Solution not provided.
Ans. 26.6
Find the mean marks of students from the following frequency table.
|
Marks |
Number of students |
|
0 and above |
100 |
|
10 and above |
88 |
|
20 and above |
80 |
|
30 and above |
68 |
|
40 and above |
54 |
|
50 and above |
44 |
|
60 and above |
36 |
|
70 and above |
34 |
|
80 and above |
21 |
|
90 and above |
9 |
|
100 and above |
0 |
Solution not provided.
Ans. 48.4
Find the mean of following cumulative frequency distribution.
|
Marks |
Number of observations |
|
10 and above |
100 |
|
20 and above |
89 |
|
30 and above |
74 |
|
40 and above |
54 |
|
50 and above |
24 |
|
60 and above |
10 |
|
70 and above |
0 |
Solution not provided.
Ans. 40.1
Find the mean marks of students from the following cumulative frequency distribution.
|
Marks |
Number of students |
|
0 and above |
80 |
|
10 and above |
77 |
|
20 and above |
72 |
|
30 and above |
65 |
|
40 and above |
55 |
|
50 and above |
43 |
|
60 and above |
28 |
|
70 and above |
16 |
|
80 and above |
10 |
|
90 and above |
8 |
|
100 and above |
0 |
Solution not provided.
Ans. 51.75
The follwowing table shown the weekly wages drawn by number of workers in a factory.
|
Weekly wages (Rs) |
Number of workers |
|
0-100 |
40 |
|
100-200 |
39 |
|
200-300 |
34 |
|
300-400 |
30 |
|
400-500 |
45 |
Find the median wage income of the workers.
Solution not provided.
Ans. 244.12
Complete the median for the following cumulative frequency distribution.
|
Less than 10 |
0 |
|
Less than 15 |
4 |
|
Less than 20 |
16 |
|
Less than 25 |
30 |
|
Less than 30 |
46 |
|
Less than 35 |
66 |
|
Less than 40 |
82 |
|
Less than 90 |
92 |
|
Less than 100 |
100 |
Solution not provided.
Ans. 62
If the median of following frequency distribution is 46, find the missing frequencies.
|
Variable |
Frequency |
|
10-20 |
12 |
|
20-30 |
30 |
|
30-40 |
x |
|
40-50 |
65 |
|
50-60 |
y |
|
60-70 |
25 |
|
70-80 |
18 |
|
Total |
229 |
Solution not provided.
Ans. 34, 45
Find the median marks of the following data :
|
Marks |
No. of students |
|
More than 150 |
0 |
|
More than 140 |
12 |
|
More than 130 |
27 |
|
More than 120 |
60 |
|
More than 110 |
105 |
|
More than 100 |
124 |
|
More than 90 |
141 |
|
More than 80 |
150 |
Solution not provided.
Ans. 116.67
Calculate the mode of the following table. (i)
|
Weight (in kg.) |
No. of students |
|
40-44 |
4 |
|
45-49 |
16 |
|
50-54 |
24 |
|
55-59 |
20 |
|
60-64 |
12 |
|
65-69 |
8 |
(ii)
|
Weight (in cm) |
No. of students |
|
160-162 |
15 |
|
163-165 |
118 |
|
166-168 |
142 |
|
169-171 |
127 |
|
172-174 |
18 |
Solution not provided.
Ans. (i) 52.83 (ii) 167.35
Find the mean of the following frequency distributions:
|
Class Interval |
Frequency |
|
10-30 |
90 |
|
30-50 |
20 |
|
50-70 |
30 |
|
70-90 |
20 |
|
90-110 |
40 |
Solution not provided.
Ans. 50
Find the mean of the following data :
|
Class Interval |
Frequency |
|
50-60 |
8 |
|
60-70 |
6 |
|
70-80 |
12 |
|
80-90 |
11 |
|
90-100 |
13 |
Solution not provided.
Ans. 78
The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f1 and f2 :
|
Class Interval |
Frequency |
|
0-20 |
5 |
|
20-40 |
f1 |
|
40-60 |
10 |
|
60-80 |
f2 |
|
80-100 |
7 |
|
100-120 |
8 |
|
Total |
50 |
Solution not provided.
Ans. 8, 12
The mean of the following frequency distribution is 57.6 and the sum of the observation is 50. Find the missing frequency f1 and f2.
|
Class Interval |
Frequency |
|
0-20 |
7 |
|
20-40 |
f1 |
|
40-60 |
12 |
|
60-80 |
f2 |
|
80-100 |
8 |
|
100-120 |
5 |
Solution not provided.
Ans. 8, 10
For what value of x the mode of the following data is 17.
15, 16, 17, 13, 17, 16, 14, x, 17, 16, 15, 15.
Solution not provided.
Ans. 17
For what value of x the mode of the following data is 6.
13, 15, 16, 14, 16, 14, 13, 14, 13, x, 16, 17, 18, 17.
Solution not provided.
Ans. 16
Solution not provided.
Ans. 9
For what value of x, is the mode of the following data is 11?
8, 9, 11, 10, 12, 14, 11, 14, 15, 14, x, 11.
Solution not provided.
Ans. 11
Find the median of the following data :
31, 15, 47, 91, 12, 56, 29, 61, 65, 55.
Solution not provided.
Ans. 51
Find the median of the following data: 29, 31, 18, 41, 57, 45, 23, 35.
If 29 is replaced by 21 and 31 replaced by 33, find the new median.
Solution not provided.
Ans. 34, 33
Find the mode of the following:
3, 5, 6, 6, 5, 3, 5, 3, 6, 5, 3, 5, 7, 6, 5, 7, 5.
Solution not provided.
Ans. 5
If the mean of the following data is 21.5, find the value of k.
|
x |
5 |
15 |
25 |
35 |
45 |
|
y |
6 |
4 |
3 |
K |
2 |
Solution not provided.
Ans. 5
Solution not provided.
Ans. 5
Find the median of the following data: 18, 37, 24, 59, 41, 26, 63, 45, 57, 29.
In the above data, if 37 replaced by 73, find the new median.
Solution not provided.
Ans. 43
If the mean of the following data is 25, find the value of k.
|
x |
5 |
15 |
25 |
35 |
45 |
|
f |
3 |
k |
3 |
6 |
2 |
Solution not provided.
Ans. 4
Solution not provided.
Ans. 40
For what value of k, is the mode of the following data 7 ? 3, 5, 7, 4, 7, 8, 3, 6, 7, 4, k, 3.
If 5 is added to each element of the above data, find the new mode.
Solution not provided.
Ans. 7, 12
Solution not provided.
Ans. 153.5
Tips: -
Calculate the arithmetic mean of the following frequency distribution:
|
Class Interval |
Frequency |
|
0-30 |
12 |
|
30-60 |
21 |
|
60-90 |
34 |
|
90-120 |
52 |
|
120-150 |
20 |
|
150-180 |
11 |
Solution not provided.
Ans. 91
Calculate the arithmetic mean of the following distribution:
|
Class Interval |
Frequency |
|
10-20 |
11 |
|
20-30 |
15 |
|
30-40 |
20 |
|
40-50 |
30 |
|
50-60 |
14 |
Solution not provided.
Ans. 40.1
Calculate of arithmetic mean of the following frequency distribution.
|
Class Interval |
Frequency |
|
0-20 |
12 |
|
20-40 |
18 |
|
40-60 |
15 |
|
60-80 |
25 |
|
80-100 |
26 |
|
100-120 |
15 |
|
120-140 |
9 |
Solution not provided.
Ans. 69.34
Solution not provided.
Ans. 23, 35
Solution not provided.
Ans. 8, 11
Find the mean of the following data :
|
Classes |
Frequency |
|
0-100 |
12 |
|
100-200 |
18 |
|
200-300 |
40 |
|
300-400 |
17 |
|
400-500 |
13 |
Solution not provided.
Ans. 251
Find the mean of the following data :
|
Classes |
Frequency |
|
0-50 |
5 |
|
50-100 |
10 |
|
100-150 |
15 |
|
150-200 |
12 |
|
200-250 |
8 |
Solution not provided.
Ans. 133
The arithmetic mean of the following frequency distribution is 50. Find the value of P.
|
Classes |
Frequency |
|
0-20 |
17 |
|
20-40 |
P |
|
40-60 |
32 |
|
60-80 |
24 |
|
80-100 |
19 |
Solution not provided.
Ans. 28
The arithmetic mean of the following frequency distribution is 53. Find the value of P.
|
Classes |
Frequency |
|
0-20 |
12 |
|
20-40 |
15 |
|
40-60 |
32 |
|
60-80 |
P |
|
80-100 |
13 |
Solution not provided.
Ans. 28
The arithmetic mean of the following frequency distribution is 52.5. Find the value of P.
|
Classes |
Frequency |
|
0-20 |
15 |
|
20-40 |
22 |
|
40-60 |
37 |
|
60-80 |
P |
|
80-100 |
21 |
Solution not provided.
Ans. 25
The arithmetic mean of the following frequency distribution is 47. Determine the value of P.
|
Classes |
Frequency |
|
0-20 |
8 |
|
20-40 |
15 |
|
40-60 |
20 |
|
60-80 |
P |
|
80-100 |
5 |
Solution not provided.
Ans. 12
The arithmetic mean of the following frequency distribution is 50. Determine the value of P.
|
Classes |
Frequency |
|
0-20 |
16 |
|
20-40 |
P |
|
40-60 |
30 |
|
60-80 |
32 |
|
80-100 |
14 |
Solution not provided.
Ans. 28
The arithmetic mean of the following frequency distribution is 25. Determine the value of P.
|
Classes |
Frequency |
|
0-10 |
5 |
|
10-20 |
18 |
|
20-30 |
15 |
|
30-40 |
P |
|
40-50 |
6 |
Solution not provided.
Ans. 16
If the mean of the following distribution is 27. Find the value of P.
|
Classes |
Frequency |
|
0-10 |
8 |
|
10-20 |
P |
|
20-30 |
12 |
|
30-40 |
13 |
|
40-50 |
10 |
Solution not provided.
Ans. 7
If the mean of the following distribution is 54. Find the value of P:
|
Classes |
Frequency |
|
0-20 |
7 |
|
20-40 |
P |
|
40-60 |
10 |
|
60-80 |
9 |
|
80-100 |
13 |
Solution not provided.
Ans. 11
If the mean of the following distribution is 50. Find the value of f1.
|
Class |
Frequency |
|
0-20 |
17 |
|
20-40 |
28 |
|
40-60 |
32 |
|
60-80 |
f1 |
|
80-100 |
19 |
Solution not provided.
Ans. 24
Find the mean of the following frequency distribution:
|
Class |
Frequency |
|
25-29 |
14 |
|
30-34 |
22 |
|
35-39 |
16 |
|
40-44 |
6 |
|
45-49 |
5 |
|
50-54 |
3 |
|
55-59 |
4 |
Solution not provided.
Ans. 36.36
The mean of the following frequency distribution is 62.8. Find the missing frequency x.
|
Class |
10 - 20 |
20 - 40 |
40 - 60 |
60 - 80 |
80 - 100 |
100 - 120 |
|
Frequency |
5 |
8 |
x |
12 |
7 |
8 |
Solution not provided.
Ans. 10
Find the mean of the following distribution:
|
Class |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
|
Frequency |
8 |
12 |
10 |
11 |
9 |
Solution not provided.
Ans. 25.2
A survey regarding the heights (in cm) of 50 girls of class X of a school was conducted and the following data was obtained.
|
Heights in cm |
120 - 130 |
130 - 140 |
140 - 150 |
150 - 160 |
160 - 170 |
Total |
|
Number of girls |
2 |
8 |
12 |
20 |
8 |
50 |
Find the mean, median and mode of the above data.
Solution not provided.
Ans. Mean : 149.8, Median : 151.15, Mode = 154
100 surnames were randomly picked up from a local telephone directory and the distribution of the number of letters of the English alphabet in the surnames was obtained as follows:
|
Heights of letters |
1 - 4 |
4 - 7 |
7 - 10 |
10 - 13 |
13 - 16 |
16 - 19 |
|
Number of surnames |
6 |
30 |
40 |
16 |
4 |
4 |
Determine the median and mean number of letters in the surnames. Also find the modal size of surnames.
Find the mean, mode and median of the following data :
|
Classes |
Frequency |
|
0-10 |
5 |
|
10-20 |
10 |
|
20-30 |
18 |
|
30-40 |
30 |
|
40-50 |
20 |
|
50-60 |
12 |
|
60-70 |
5 |
Find the mean, mode and median of the following data :
|
Classes |
Frequency |
|
0-10 |
3 |
|
10-20 |
8 |
|
20-30 |
10 |
|
30-40 |
15 |
|
40-50 |
7 |
|
50-60 |
4 |
|
60-70 |
3 |
Find the mean, mode and median of the following data:
|
Classes |
Frequency |
|
0-10 |
3 |
|
10-20 |
4 |
|
20-30 |
7 |
|
30-40 |
15 |
|
40-50 |
10 |
|
50-60 |
7 |
|
60-70 |
4 |
Find the mean, mode and median of the given data:
|
Classes |
Frequency |
|
10-20 |
4 |
|
20-30 |
8 |
|
30-40 |
10 |
|
40-50 |
12 |
|
50-60 |
10 |
|
60-70 |
4 |
|
70-80 |
2 |
Find the mean, mode and median of the following data:
|
Classes |
Frequency |
|
0-10 |
5 |
|
10-20 |
8 |
|
20-30 |
15 |
|
30-40 |
20 |
|
40-50 |
14 |
|
50-60 |
8 |
|
60-70 |
5 |
Find the mean, mode and median of the following data :
|
Classes |
Frequency |
|
0-20 |
6 |
|
20-40 |
8 |
|
40-60 |
10 |
|
60-80 |
12 |
|
80-100 |
6 |
|
100-120 |
5 |
|
120-140 |
3 |
Find the mean, mode and median of the following data:
|
Classes |
Frequency |
|
0-10 |
6 |
|
10-20 |
8 |
|
20-30 |
10 |
|
30-40 |
15 |
|
40-50 |
5 |
|
50-60 |
4 |
|
60-70 |
2 |
Solution not provided.
Ans. Mean : 30, Median : 30.67, Mode :33.33
If mean of the following data is 18.75, then the value of P is.
|
xi |
10 |
15 |
P |
25 |
30 |
|
fi |
5 |
10 |
7 |
8 |
2 |
20
25
23
24
A.
20
Calculate the median for the following :
|
Heights (cm) |
120 |
121 |
122 |
123 |
124 |
125 |
|
No. of students |
8 |
12 |
17 |
14 |
13 |
6 |
120
121
123
122
D.
122
C.
A trial is made to answer a true-false question. The answer is right or wrong.D.
A baby is born. It is a boy or a girl.It is given that :
P(E) = 0.05, then Probability of ‘not E’ = 1 – P(E)
⇒ P(E) = 1 – 0.05 = 0.95.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out :
(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
(i) 0, because the bag contains the lemon flavoured candies only.
(ii) 1, because the bag contains lemon flavoured candies only.
Probability that the 2 students have the same birthday
= 1 – Probability that the 2 students have not the same birthday
= 1 – 0.992 = 0.008.
Number of red balls in the bag = 3
Number of black balls in the bag = 5
Total number of balls in the bag = 3 + 5 = 8
i.e. n(s) = 8
(i) Let A be the favourable outcomes of getting red balls, then
n(A) = 3
Therefore, P(A) 
Let Be be the favourable outcomes of getting ‘not red'balls. Then,
n(B) = 5
Therefore, P(B) = 
Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
Total number of marbles = 5 + 8 + 4 = 17
i.e. n(s) = 17
(i) Let A be the favourable outcomes of getting red marble, then
n(A) = 5
Therefore, P(A) = 
(ii) Let B be the favourable outcomes of getting white marbles, then n(B) = 8
Therefore, P(B) = 
(iii) Let C be the favourable outcomes of getting ‘not green’ marble, then
n(C) = 5 + 8 = 13
Therefore, P(C) = 
Number of 50 p coins = 100
Number of Re. 1 coins = 50
Number of Rs. 2 coins = 20
Number of Rs. 5 coins = 10
Total number of coins = 180
i.e. n(s) = 180
(i) Let A be the favourable outcomes of getting a 50 p coin, then
n(A) = 100
Therefore, P(A) = 
(ii) Let B be the favourable outcomes of getting ‘not a Rs. 5 coin’. Then
n(B) = 170
Therefore, P(B) = 
Gopibuys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish.
What is the probability that the fish taken out is a male fish?
Number of male fish = 5
Number of female fish = 8
Total number of fish = 13
i.e. n(s) = 13
Let A be the favourable outcomes of getting a male fish. Then
N(a) = 5
Therefore, P(A) = 
(i) 8?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
Total number of possible outcomes in the game = 8
i.e. n(s) = 8
(i) Let A be the favourable outcomes of getting 8. Then
n(A) = 1
Therefore, P(A) = 
(ii) Let B be the favourable outcomes of getting an odd number. Then
B = {1, 3, 5, 7}
i.e. n(B) = 4
Therefore, P(B) = 
(iii) Let C be the favourable outcomes of getting a number greater than 2. Then,
C = {3, 4, 5, 6, 7, 8}
i.e. n(C) = 6
Therefore, P(C) = 
(iv) Let D be the favourable outcomes of getting a number less than 9. Then,
D = {1, 2, 3, 4, 5, 6, 7, 8}
i.e. n(D) = 8
Therefore, P(D) = 
A die is thrown once, find the probability of getting
(i) a prime number.
(ii) a number lying between 2 and 6.
(iii) an odd number.
If we throw a die once, then possible outcomes (s) are
S = {1, 2, 3, 4, 5, 6}
⇒ n( S) = 6
(i) Let E be the favourable outcomes of getting a prime number then
E = {2, 3, 5}
n(E) = 3
Therefore, P(E) = 
(ii) Let F be the favourable outcomes of getting a number lying between 2 and 6, then
F = {3, 4, 5}
n(F) = 3
Therefore, P(F) = 
(iii) Let G be the favourable outcomes of getting an odd number, then
G = {1, 3, 5}
n(G) = 3
Therefore, P(G) = 
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour.
Let E be the favourable outcomes of getting a king of red colour, then n(E) = 2
Therefore,
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
a face card.
Let F be the favourable outcomes of getting a face card, then
n(F) = 12
Therefore,
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
a red face card.
Let G be the favourable outcomes of a red face card, then
n(G) = 6
Therefore,
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
the Jack of hearts.
Let H be the favourable outcomes of getting the jacks from the hearts, then n(H) = 1
Therefore,
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
a spade
Let I be the favourable outcomes of getting a spade, then
n(I) = 13
Therefore : 
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting
the queen of diamonds.
Let J be the favourable outcomes of getting thez queen of diamonds, then n(J) = 1
Therefore,
Five cards - the ten, jack, queen, king and ace of diamonds are well-shuffled with their face down wards. One card is then picked up at random.
(i) What is the probability that the card is a queen?
Let E be the favourable outcomes of getting a queen, then
E = {1}
⇒ n(E) = 1
Therefore,
Five cards - the ten, jack, queen, king and ace of diamonds are well-shuffled with their face down wards. One card is then picked up at random.
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace (b) a queen.
If the Queen is drawn and put aside then possible outcomes become 4
i.e., n(S) = 4
(a) Let A be the favourable outcomes that the picked up card is an ace, then
n(A) = {1}
Therefore,
(b) Let B be the favourable outcomes that the picked up card is a queen, then n(B) = 0
Therefore,
Number of defective pens = 12
Number of non-defective pens = 132
Total number of pens = 12 + 132 = 144
Let ‘A’ be the favourable outcomes of getting the pen taken out is a good one. Then, n(A) = 132
(i) A lot of 20 bulbs contain 4 defectuve ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective ?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced.
Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
(i) Total number of bulbs = 20
i.e. n(S) = 20
Let A be the favourable outcomes of getting defective bulbs. Then, n(A) = 4
Therefore, P(A) = 
(ii) Total number of bulbs = 19 Let B be the favourable outcomes of getting non defective bulbs. Then n(B) = 19 - 4 = 15
Therefore, P(B) = 
Total number of discs in the box = 90 i.e. n(S) = 90
(i) Let A be favourable outcomes of getting a two digit number. Then n(A) = 81
Therefore, P(A) = 
(ii) Let B be the favourable outcomes of getting a perfect square number. Then
B = {1, 4, 9, 16, 25, 36, 49, 64, 81} i.e. n(B) = 9
Therefore, P(B) = 
(iii) Let C be the favourable outcomes of setting a number divisible by 5.
Then, n(C) = 18
Therefore, P(C) = 
Total number of faces in a die = 6 i.e. n(S) = 6
(i) Let E be the favourable outcomes of setting A. Then,
n(E) = 2
Therefore, P(E) = 
(ii) Let A be the favourable outcomes of setting D. Then
n(D) = 1
Therefore, P(D) = 
Suppose you drop a die at random on the rectangular region show in Fig. 15.6. What is the probability that it will tend inside the circle with diameter 1 m?
Radius = 
A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
(i) She will buy it?
(ii) She will not buy it?
Total number of ball pens = 144 i.e. n( S) = 144
(i) Let A be the favourable outcomes of buying a ball pen by her. Then
n( A) = 144 – 20 = 124
Therefore, P(A) = 
(ii) Let B be the favourable outcomes of not buying a ball pen. Then,
P(B) = 1 – P(A)
(i) Complete the following table :
(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Therefore, each of them has a probability 
. Do you agree with this argument ? Justify your answer.
Total number of possible outcomes = 36 i.e. n( S) = 36
(i) Let A be the favourable outcomes of getting the sum as 2. Then
A = (1,1) i.e. n(A) = 1
Therefore, P(A) = 
(ii) Let B be the favourable outcomes of getting the sum as 3. Then
B = (1,2), (2,1) i.e. n(B) = 2
Therefore, P(B) = 
(iii) Let C be the favourable outcomes of getting the sum as 4. Then
C = (2,2), (1,3), (3,1) i.e. n(C) = 3
Therefore, P(C) = 
(iv) Let D be the favourable outcomes of getting the sum as 5. Then
D = {(1,4), (4,1), (2,3), (3,2)}.
(v) Let E be the favourable outcomes of getting the sum as 6. Then
E = {(1,5), (5,1), (2,4), (4,2), (3,3)} i.e. n(E) = 5
Therefore, P(E) = 
(iv) Let D be the favourable outcomes of getting the sum as 5. Then
D = {(1,4), (4,1), (2,3), (3,2)}.
(v) Let E be the favourable outcomes of getting the sum as 6. Then
E = {(1,5), (5,1), (2,4), (4,2), (3,3)} i.e. n(E) = 5
Therefore, P(F) = 
(vii) Let G be the favourable outcomes of getting the sum as 8. Then
G = (2,6), (6,2), (3,5), (5,3), (4,4) i.e. n(G) = 5
Therefore, P(G) = 
(viii) Let H be the favourable outcomes of getting the sum as 9. Then
H = (3,6), (6,3), (4,5), (5,4) i.e. n(H) = 4
Therefore , P(H) = 
(ix) Let I be the favourable outcomes of getting the sum as 10. Then
1 = (4,6), (6,4), (5,5)
Le. n(1) = 3
Therefore, P(I) = 
(x) Let J be the favourable outcomes of getting the sum as 11. Then
J = (6,5), (5,6) i.e. n(J) = 2
Therefore, P(J) = 
(xi) Let K be the favourable outcomes of getting the sum as 12. Then
K = (6,6) i.e. n(K) = 1
Therefore, P(K) = 
If we toss a one rupee coin then possible outcomes are
S= {HHH,TTT, HHT, HTH, HTT,THH,THT,TTH}
i.e. n(S) = 8
Let A be the favourable outcomes of losing the game. Then
A = {HHT, HTH, HTT,THH,THT,TTH}
i.e., (A) = 6
Therefore, P(A) = 
A die is thrown twice. What is the probability that
(i) 5 will not Come up either time?
(ii) 5 will come up at least once?
(i) If two coins are tossed at the same time the possible outcomes are :
S = (H,H), (H,T), (T,H), (T,T) i.e. n(S) = 4
So, the probability of each occurrence = 
Thus, the given statement is wrong.
A number is chosen at random from the number –3, –2, –1, 0, 1, 2, 3. What will be the probability that square of this number is less then or equal to 1?
S = {−3, −2, −1, 0, 1, 2, 3}
Let E be the event of getting a number whose square is less than or equal to 1.
So, E = {−1, 1, 0}
P(E)=3/7.
Hence, the probability of getting a number whose square is less than or equal to is 3/7.
Two different dice are thrown together. Find the probability that the numbers obtained
(i) have a sum less than 7
(ii) have a product less than 16
(iii) is a doublet of odd numbers.
The outcomes when two dice are thrown together are
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Total number of outcomes = 36
(i) Let A be the event of getting the numbers whose sum is less than 7.
The outcomes in favour of event A are (1, 1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2) and (5,1).
Number of favourable outcomes = 15
(ii) Let B be the event of getting the numbers whose product is less than 16.
The outcomes in favour of event B are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1) and (6,2).
Number of favourable outcomes = 25
iii) Let C be the event of getting the numbers which are doublets of odd numbers.
The outcomes in favour of event C are (1,1), (3,3) and (5,5).
Number of favourable outcomes = 3
Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die and squares the number obtained. Who has the better chance to get the number 25.
Let us first write the all possible oucomes when Peter throws two different dice together.
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
∴ Total number of outcomes = 36
The favorable outcome for getting the product of numbers on the dice equal to 25 is (5, 5).
Favourable number of outcomes = 1
∴ Probability that Peter gets the product of numbers as 25
= 
The outcomes when Rina throws a die are 1, 2, 3, 4, 5, 6.
∴ Total number of outcomes = 6
Rina throws a die and squares the number, so to get the number 25, the favourable outcome is 5.
Favourable number of outcomes = 1
∴ Probability that Rina gets the square of the number as 25
As, 1/6>1/36, so Rina has better chance to get the number 25.
Two different dice are tossed together. Find the probability:
(i) of getting a doublet
(ii) of getting a sum 10, of the numbers on the two dice.
The outcomes when two dice are thrown together are
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Total number of outcomes = 36
n (s) = 36
i) A = getting a doublet
A = {(1,1), (2,2), (3,3),(4,4), (5,5), (6,6)}
n(A) = 6
B = getting sum of numbers as 10.
B = {(6, 4), (4, 6), (5, 5)}
n(B) =3
An integer is chosen at random between 1 and 100. Find the probability that it is :
(i) divisible by 8
(ii) not divisible by 8
An integer is chosen at random from 1 to 100
Therefore n(S) = 100
(i) Let A be the event that number chosen is divisible by 8
∴ A = { 8,16,24,32,40,48,56,64,72,80,88,96}
∴ n (A) = 12
Now, P (that number is divisible by 8)
(ii) Let ‘A’ be the event that number is not divisible by 8.
The probability of getting an even number, when a die is thrown once, is
A.
S = { 1, 2, 3, 4, 5, 6 }
let event E be defined as 'getting an even number'.
n(E) = { 2, 4, 6 }
A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime-number less than 23,is
C.
S = { 1, 2, 3,........90 }
n(s) = 90
The prime number less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19
Let event E be defined as 'getting a prime number less than 23'.
n(E) = 8
A card is drawn at random from a well shuffled pack of 52 playing cards.Find the probability that the drawn card is neither a king nor a queen
Let E be the event that the drwan card is neither a king nor a queen.
Total number of possible outcomes = 52
total number of kings and queens = 4+4 = 8
Therefore, there are 52 - 8 = 44 cards that are neither king nor queen.
total number of favourable outcomes = 44
A Group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is
(i)extremely patient
(ii) extremely kind or honest. Which of the above values you prefer more?
The group consist of 12 persions.
In a family of 3 children, the probability of having at least one boy is
A.
There are in all 23 = 8 combinations or outcomes for the gender of the 3 children
The 8 combinations are as follows
BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG
Thus the probability of having at least one boy in a family is
Two different dice are tossed together. Find the probability
(i) That the number on each die is even.
(ii) That the sum of numbers appearing on the two dice is 5.
The total number of outcomes when two dice are tossed together is 36.
The sample space is as follows
(i). Favourable outcomes = { (2,2) (2,4) (2,6) (4,2) (4,4)
(6,4) (6,2) (6,4) (6,6) }
Porbability that the number on each dice is even
=
(ii) Favourable outcomes = { (1,4) (2,3) (3,2) (4,1) }
Probability that the sum of the number appearing on the two dice is 5
=
All the red face cards are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the drawn card is
(i) of red colour
(ii) a queen
(iii) an ace
(iv) a face card
(i) Face card are removed from a pack of 52 playing card = 6
Total favourable outcomes = 52 - 6 = 46
Number of all possible outcomes = 26 - 6 = 20
(ii) Number of all possible outcomes a queen = 2
(iii) Number of all possible outcomes an ace = 2
(iv) Number of all possible outcomes = 6
The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?
Let the total number of rotten apples in a heap = n
Total number of apples in a heap = 900
probability of selecting a rotten apple from a heap = 0.18
Now.
A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.
Let the number of black balls in the bag be x.
Number of white balls = 15
Hence, total number of balls in the bag = x + 15
Given, P(black ball) = 3 x P(white ball)
Thus, the number of black balls in the bag is 45.
Two different dice are thrown together. Find the probability that the numbers obtained have
(i) even sum, and
(ii) even product
Elementary events associated to the random experiment of throwing two dice are:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6),
Total number of elementary events = 6 x 6 = 36.
(i) Let A be the event of getting an even number as the sum.
i.e. 2, 4, 6, 8, 10, 12.
Elementary events favourable to event A are:
(1,1), (1,3), (1,5), (2,2), (2,4), (2,6),
(3,1), (3,3), (3,5), (4,2), (4,4), (4,6),
(5,1), (5,3), (5,5), (6,2), (6,4), (6,6).
Total number of favourable events = 18.
Hence, required probability =
(ii) Let B be the event of getting an even number as the product.
i.e. 2, 4, 6, 8, 10, 12, 16, 18, 20, 24, 30, 36.
Elementary events favourable to event B are:
(1,2), (1,4), (1,6), (2,1), (2,2), (2,3),
(2,4), (2,5), (2,6), (3,2), (3,4), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,2), (5,4), (5,6), (6,1), (6,2), (6,3),
(6,4), (6,5), (6,6).
Total number of favourable events = 27.
Hence, required probability = .
Two dice are thrown together. The probability of getting the same number on both dice is:
C.
When two dice are thrown together, the total number of outcomes is 36.
favourable outcomes = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }
A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.
Total mumber of outcomes is 50.
Favourable outcomes = { 12, 24, 36, 48 }
A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting
(i) a king of red colour
(ii) a face card
(iii) the queen of diamonds.
Total number of outcomes = 52
(i) Probability of getting a red king
Here the number of favourable outcomes = 2
(ii) probability of getting a face card
Total number of face cards = 12
(iii) Probability of queen of diamonds
Number of queens of diamond = 1
Sponsor Area
observation
th observation and 
observation
