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Which average would be suitable in the following cases:
(i) Average size of readymade garments.
(ii) Average intelligence of students in a class.
(iii) Average production in a factory per shift.
(iv) Average wages in an industrial concern.
(v) When the sum of absolute deviations from average is least.
(vi) When quantities of the variable are in ratios.
(vii) In case of open ended frequency distribution.Classroom Activity. Try yourself with the help of your teacher.
Comment on whether the following statements are True or False.
A.
The sum of deviation of items from median is zero.
B.
An average alone is not enough to compare series.
C.
Arithmetic mean is a positional value.
D.
Upper quartile is the lowest value of top 25% of items.
E.
Median is unduly affected by extreme observations.
Following information pertains to the daily income of 150 families. Calculate the arithmetic means.
|
Income (in Rs.) |
No. of Families |
|
More than 75 More than 85 More than 95 More than 105 More than 115 More than 125 More than 135 More than 145 |
150 140 115 95 70 60 40 25 |
|
Income |
M.V. |
f |
d |
d' (d/10) |
fd' |
|
|
75-85 85-95 95-105 105-115 115-125 125-135 135-145 145-155 |
80 90 100 110 120 130 140 150 |
10 25 20 25 10 20 15 25 |
-40 -30 -20 -10 0 10 20 30 |
-4 -3 -2 -1 0 +1 +2 +3 |
-40 -75 -40 -25 0 20 30 75 |
-180 +125 |
|
Σ f = 150 |
Σ fd' = 55 |
|||||
If the arithmetic mean of the data given below is 28, find (a) the missing frequency and (b) the median of the series.
|
Profit per retail shop (in Rs.) |
Number of retail shops |
|
0-10 10-20 20-30 30-40 40-50 50-60 |
12 18 27 – 17 6 |
|
Profit per retail shop (in Rs.) |
Mid-Value M |
Number of Retail Shops |
d |
d' |
fd' |
|
|
0-10 10-20 20-30 30-40 40-50 50-60 |
5 15 25 35 45 55 |
12 18 27 (X) suppose 17 6 |
-30 -20 -1 0 +1 +2 |
-3 -2 -1 0 +1 +2 |
-36 -36 -27 0 +17 +12 |
-99 +29 |
|
Σ f = X+80 |
Σ fd' = 70 |
Calculation of Median :
|
Profit of Retail Shops |
Frequency (f) |
cf |
|
0-10 10-20 20-30 30-40 40-50 50-60 |
12 18 27 20 17 6 |
12 20 57 77 94 100 |
|
Σ f= 100 |
The following series relates to the daily income of workers employed in a firm. Compute the (a) highest income of lowest 50% workers, (6) minimum income earned by the top 25% and (c) maximum income earned by lowest 25% workers.
|
Daily Income (in Rs.) |
Number of Workers |
|
10-14 15-19 20-24 25-29 30-34 35-39 |
5 10 15 20 10 5 |
Value of 32.5 item lies in 24.5 - 29.5 class interval.
Value of 16.25 item lies in 19.5 - 24.5 class interval
Value of 47.75 lies in 24.5 - 29.5 class interval.
The following table gives production yield in kg. per hectare of wheat of 150 arms in a village. Calculate the mean, median and mode production yield.
|
Production yield (kg. per hectare) |
50-53 |
53-56 |
56-59 |
59-62 |
62-65 |
65-68 |
68-71 |
71-74 |
74-77 |
|
Number of farms |
3 |
8 |
14 |
30 |
36 |
28 |
16 |
10 |
5 |
Calculation of Mean and Medians
|
Production |
M.V. |
d |
d' |
f |
fd' |
cf |
|
50-53 53-56 56-59 59-62 62-65 65-68 68-71 71-74 74-77 |
51.5 54.5 57.5 60.5 63.5 66.5 69.5 72.5 75.5 |
-12 -9 -6 -3 0 + 3 + 6 + 9 +12 |
-4 -3 -2 -1 0 +1 + 2 + 3 + 4 |
3 8 14 30 36 28 16 10 5 |
-12 -24 -28 -30 -94 0 28 32 30 20 +110 |
3 11 25 55 91 119 135 145 150 |
|
Σ f = 150 |
Σ fd' = 16 |
(Answer 63.29 kg. per hectare)
The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.
|
Size of Land Holdings (in acres) |
Number of families |
|
less than 100 100-200 200-300 300-400 400 and above |
40 89 148 64 39 |
Calculation of median size of land holdings :
Value of 190th item lies in class interval of 200-300.
What do you mean by 'Average'?
Average value or central tendency is the critical value which represents all the items in a series.
Pocket expenditures of 6 students are (Rs.) 6,12,18,24,30 and 36 respectively. Find out arithmetic mean.
Arithmetic Mean
= 6+12+18+24+30+36
6
= 126
6
= Rs. 21
Name various kinds of statistical averages?
(i) Mathematical Averages : (a) Arithmetic Mean, (b) Geometric Mean, (c) Harmonic Means.
(ii) Positional Averages : (a) Median, (b) Partition value, (c) Mode.
What is arithmetic mean?
Arithmetic mean is a simple average of all items in a series. It is simply mean value which is obtained by adding the values of all the items and dividing the total by the number of items.
Define median.
Median is a centrally located value of a series such that half of the values of the series are above it and other half below it.
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Give formula for estimating mode in frequency distribution series.
(i) First we find out the model class
Z = Value of the mode
I1 = Lower limit of the model class.
i = Size of the modal group.
Give formula for finding out median of a continuous series.
l1 = Lower limit of the median class.
c.f = Cumulative frequency of the class.
f = Preceding the median class.
i = Size of the median class interval.
Define quartile.
If a statistical series is divided into four equal parts, then the end value of each part is called a quartile.
What does Central Tendency refer to?
Central Tendency refers to a central value or a representative value of statistical data.
Enumerate any three essentials of a good average.
(i) It should be simple, (ii) It must be certain on character, (iii) It should be capable of further mathematical or algebric treatement.
Name the three statistical measures of Central Tendency.
Three statistical measure of central tendency are : (i) Arithmetic Mean, (ii) Median and (iii) Mode.
Write down the formula for calculating mean for ungrouped data.
Calculate correct average marks.
stands for arithmetic mean.
Σ X stands for summation of all the observations in a series.
N stands for number of observations.
Write down an interesting property of A.M.
The sums of deviations of items about arithmetic mean is always equal to zero. Symbolically
How is arithmetic mean affected, if two is added to all the observations in the series?
Arithmetic mean will increase by 2.
What is the arithmetic mean of 5, 6, 7, 8 and 9 ?
Arithmetic mean of 5, 6, 7, 8 and 9 is = 35
5
= 7
What is the basic difference between simple arithmetic mean and weighted arithmetic mean?
In simple arithmetic mean, all items of the series are taken as of equal importance. On the hand, in the weighted arithmetic mean, different items are taken as of different importance.
What does the following symbol indicate:
The symbol indicates that the total of frequencies multiplied by Arithmetic Mean is always equal to sum of the products of variables and their respective frequencies.
When can the direct method of computation of arithmetic mean be used?
The direct method of computation of arithmetic mean can be used when the items in the series are less.
When will the weighted arithmetic mean be less than the simple arithmetic mean?
The weighted arithmetic mean will be less than the simple arithmetic mean when items of small value are given greater weights and items of big values are given less weights.
When will the weighted mean be greater than the simple arithmetic mean?
The weighted arithmetic mean will be greater than the simple arithmetic mean when items of small values are given less weights and items of big values are given more weights.
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What is Median?
Median is that positional value of the variable which divides the distribution into two equal parts, one part comprises all values greater than or equal to median value and the other comprises all values less than or equal to it.
What is the first and most important rule for calculating the median?
The first and most important rule for calculating median is that data should be arranges in ascending or descending order.
Calculate the median of 5, 7, 6, 6, 1, 8, 10,12, 4 and 3.
By arranging the data in ascending order we get 1, 3, 4, 5, 6, 7, 8, 10, 12.
Median = Value of N + 1 item = 9 + 1 = 5th
2 2
Calculate the median of 3, 5, 7 and 4.
Median = size of N+1 item
2
= 4+1 = 2.5 item = 5+7 = 6
2 2
Write down the formula for locating the median from median group.
l1 = lower limit of median group,
c.f. = cumulative frequency of the class preceeding the median class.
f = frequency of the median group,
i = the class interval of the median group.
You are given the data in the form of cumulative series for calculating median. What would be your step for calculating?
Our first step will be to convert the cumulative series with simple series.
Give two merits of median.
1. It is easy to calculate and understand median.
2. It is well defined as an ideal average should be.
Give two demerits of median.
1. Median is not based on all observations.
2. It cannot be given further algebric treatment.
What is partition value?
The value that divides the series into more than two parts is called partition value. Q1, Q2, Q3 etc. are partition value.
What are quartiles?
Quartiles are those values which divide the series into four equal parts.
What is the most important rule for getting partition values?
The most important rule for getting partitional values is that the values must be arranged in ascending order only
Write down the difference between averages and partition values.
An average is representative of whole series while quartiles are average of parts of series.
What are percentiles?
Percentiles are the measures which divide the distribution in to hundred equal parts. There are 99 percentiles denoted by P1, P2, P3, ...P99.
Suppose a person has secured 82 percentiles in a mangement entrance examination. What does it mean?
It means that his position is below 18 percent of total candidates appeared in the examination.
Find the mode from the following data :
1,2, 3, 4,4, 5.
The mode for this data is 4 because 4 occures most frequently twice in the data.
Calculate mode from the following discrete series:
|
Variable |
10 |
20 |
30 |
40 |
50 |
|
Frequency |
2 |
8 |
20 |
10 |
5 |
The value of mode is 30, because it is the most frequently observed data.
What do you mean by unimodal data?
Unimodal data is that data which has only one modal. For example, there is one modal (4) in the following data:
1, 2, 3, 4, 4, and 5.
Which type of modal in the following series:
|
Variable |
10 |
20 |
30 |
40 |
50 |
|
Frequency |
2 |
8 |
20 |
10 |
5 |
It is unimodal because it has one mode i.e. 30.
What is the relative position of Arithmetic Mean, Median and Mode?
(i) A.M. > Median > Mode
or
(ii) A.M. < Median < Mode
In other words, the median is always between the arithmetic mean and the mode.
A shoe company, making shoes for adults only, wants to know the most popular size of shoes. Which average will be the most appropriate for it?
Mode will be the most appropriate for knowing the popular size of shoes.
What are deciles?
Deciles are measures which divide the total set of values into ten equal parts.
You are given inclusive series for the calculation of mode. What will be your step?
The first step will be to convert the inclusive series into exclusive series.
Which formula is applied for calculating Mode from median and mean in case of normal distribution?
_
Z=3M-2 X
= 3 x 18.8-2 x 20.2
= 56.4 - 40.4 = 16
Give any two Mertis of Mode.
Merits : (i) Mode is not affected by extreme values, (ii) It can be determined in open-end distribution.
Give two Demerits of Mode.
(i) Mode cannot be divided in biomodal and multi-modal distribution, (ii) Mode is unsuitable to further mathematical treatement.
Write any four Merits of Mean.
Merits of Mean are as follows:
1. Certainity : Arithemetic Mean is a certain value. It has no scope for estimated values.
2. Based on all items : Arithmetic Mean is based on all the items in a series. It is therefore, represehtative value of the different items.
3. Simplity : From the view point of calculation and use, arithmetic mean is the simplest of all the measures of central tendency.
4. Stability : Arithmetic Mean is a stable measure of central tendency. This is because, changes in the sample of a series having minimum effect or the arithmetic average series.
Write any four Demerits of Mean.
Demerits of arithmetic average are as follows :
1. Effect of extreme value: The main effect of it is that it gets distorted by extreme values of the series.
2. Unsuitability : Arithmetic mean is a suitable measures in case of percentage or proportionate values.
3. Misleading conclusion : Arithmetic mean sometimes offer misleading conclusions.
4. Laughable conclusion : Arithmetic mean sometimes offer laughable conclusion.
Compare the arithmetic mean, median and mode as measure of Central Tendency. Describe situations where one is more suitable than the others.
As compared to Mean and Median, Mode is less suitable. Mean is simple to calculate, its value is definite, it can be given algebraic treatment and is not affected by fluctuations of sampling. Median is even more simple to calculate but is affected by fluctuations and cannot be given algebraic treatment. Mode is the most popular item of a series but it is not suitable for most elementary studies because it is not based on all the observations of the series and is unrepresentative. i
In case of arithmetic mean it is the numerical magnitude of the deviations that balances. In case of median it is the number of values greater than the Median which balances against the number of values less than the Median. The median is always between the arithmetic mean and the mode.
The arithmetic mean is described as the centre of gravity of the distribution of values of the variable. Explain.
his is rightly said that arithmetic mean is the centre of gravity of the distribution of values of the variable because it is the number obtained by dividing the total values of different items by their number. We use this method in our every day life. It is a popular and most widely used measure.
Find mean and median for all four values of the series and answer the questions given at the end of table :
|
Series |
X (Variable Values) |
Mean |
Median |
|
A B C D |
1,2,3 1, 2, 30 1, 2, 300 1, 2, 3000 |
? ? ? ? |
? ? ? ? |
1. Is median affected by extreme values?
2. Is median a better method than mean?
|
Series |
X (Variable Values) |
Mean |
Median |
|
A B C D |
1,2,3 1, 2, 30 1, 2, 300 1, 2, 3000 |
2 11 101 1001 |
2 2 2 2 |
1. No, median is not affected by extreme values.
2. Yes, median is a better method than mean.
Prove that the sum of the squares of the deviations from arithmetic mean is the least i.e. less than of the squares of the deviations of observation taken from any value.
For proving the statement given in the question we take the following observations:
1,2,3, 4,5
|
X |
X-3 |
X-4 (x') |
X'2 |
|
|
(x) |
(x2) |
|||
|
1 2 3 4 5 |
-2 -1 0 +1 +2 |
4 1 0 1 4 |
-3 -2 -1 0 +1 |
9 4 1 0 1 |
|
Σ x = 15 |
Σ x2=10 |
Σ x2= 15 |
||
We take another value i.e. 4. We observe that the sum of squares of deviations taken from mean (3) is less than that taken from other value (4) is less.
Prove with an example that the weighted arithmetic mean will be less than the simple mean, when items of small values are given greater weights and items of big values are given less weight.
In order to prove the statement given in the example, we take the following table :
|
Type |
Weight (Rs.) X |
Workers W |
WX |
|
Man Woman Children |
8 6 4 |
50 20 10 |
400 120 40 |
|
Σ x=18 |
Σ W=80 |
ΣWX=560 |
With the help of following data, prove that the sum of the deviations of items about arithmetic mean is always equal to zero.
X : 4, 6, 8, 10 and 12.
|
X |
D (X- |
|
4 6 8 10 12 |
-4 -2 0 +2 +4 |
|
Σ X = 40 |
Σ D = 0 |
From the given data it has been proved that the sum of the deviations of items about the arithmetic mean is equal to zero. Symbolically
Calculate median from the following series:
|
X |
10 |
20 |
30 |
80 |
90 |
100 |
|
f |
3 |
7 |
6 |
2 |
8 |
4 |
|
X |
f |
c.f. |
|
10 20 30 80 90 100 |
3 7 6 2 8 4 |
3 10 16 18 26 30 |
Value of 15.5 th item = 30.
Hence, median is equal to 30.
Calculate median from the following:
|
Sr. No. |
No. of Items |
|
1 2 3 4 5 6 7 8 9 |
3 4 4 5 6 8 8 8 10 |
Median = value of N+1 th item
2
= 9+1 = Value of 5th item
2
= 6
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Write down the mertis of median.
Merits of Median : Following are the merits of median :
1. Median is definite.
2. It is easy to calculate and understand.
3. It can also be determined graphically.
4. It is not affected by extreme values.
5. It can be calculated in the absence of any one of the item.
6. It is helpful in qualitative facts such as ability, stability etWrite down the demerits of median.
Demerits of Median : Following are the demerits of median:
1. In it, all the items of a series are not given equal importance.
2. If the number of items are given the correct value of median cannot be calculated.
3. It is affected by flqctuations of sampling.
4. It cannot be given further algebraic treatment.
5. Data needs to be arranged in ascending or descending order.
Under what circumstances are median mostly used to measure central tendency?
1. In open ended series, median and mode can be easily calculated, while it is difficult to compute arithmetic mean, because it is not possible to calculate mid-pqjnts of these class intervals.
2. Median and Mode are not affected by extreme values while arithmetic mean is affected by extreme values.
3. Median and Mode can be calculated graphically, while there is no graphical method for calculating arithmetic mean.
Why is arithmetic mean the most popular measure of the central tendency?
Arithmetic mean is the most popular measure of the central tendency due to following reasons :
1. It is easy to calculate and understand arithmetic mean.
2. Arithmetic mean is based on all the values of the series.
3. Arithmetic mean is stable measure of central tendency. It is because change in the sample of a series have minimum effect on the arithmetic average.
Give the comparative study of the measures of Central Tendency.
Comparative study of the measures of Central Tendency : Arithmetic is the most commonly used average. It is simple to calculate and is based on all the observations. But it is unduly affected by the presence of extreme items. Median is a better summary for such data. Mode is generally used to describe the qualitative data. Median and mode can be easily computed graphically. In case of open ended distribution, they can also be easily computed.
Give the special features of Arithmetic Mean.
Special features of Arithmetic Mean:
1. If we replace each item of observations by the calculated mean, then the total of these replaced values will be equal to the sum of the given observations.
2. The sum of deviations of items from arithmetic mean is always equal to zero.
3. The total of frequencies multiplied by Arithmetic Mean is always equal to the sum of the product of mid-points of various classes and their respective frequencies,
N
= Σ fm.
4. The sum of the deviation of mid-points from arithmetic mean being multipled by their frequencies is always equal to zero.
Daily incomes of eight families are given below. Calculate average daily family income.
170, 500, 250, 700, 300, 400, 200 and 350.
|
Sr. No. |
Daily Income (Rs.) X |
|
1 2 3 4 5 6 7 8 |
170 500 250 700 300 400 200 350 |
|
N = 8 |
Σ X = 2870 |
Calculate mean wages from the following table :
|
Wages (in Rs.) |
10 |
20 |
30 |
40 |
|
No. of Workers |
4 |
5 |
3 |
2 |
|
Wages in (Rs.) (X) |
No. of Workers (f) |
fX |
|
10 20 30 40 |
4 5 3 2 |
40 100 90 80 |
|
N = 14 Σ fX = 310 |
||
A student secured 60%, 75% and 63% marks in English, Hindi and Maths respectively in Senior Secondary Examination. These subject have been given weights of 1,1 and 2 respectively. Calculate weighted arithmetic mean.
|
Subjects |
Marks Secured (X) |
Weight (W) |
XW |
|
English Hindi Maths |
60 75 63 |
1 1 2 |
60 75 126 |
|
Σ W = 4 |
Σ XW = 261 |
||
Hence, weighted arithmetic mean = 65.25 marks.
alculate median from the following data:
|
Marks Obtained |
No. of Students |
|
20 40 60 80 100 |
5 10 20 15 10 |
Marks Obtained (X)
|
No. of Students(f) |
cf |
|
|
20 40 60 80 100 |
5 10 20 15 10 |
5 15 35 50 60 |
i.e. Value of 30.5 th item
Value of 30.5 is 60
Hence, Median = 6.
Monthly incomes of five families are given below. Calculate mean by Short Cut Method.
6550, 7550,9550, 4550 and 8000.
Calculation of A. M. by Short Cut Method:
|
S. No. |
Montly Income X |
Deviation from |
|
1 2 3 4 5 |
6550 7550 9550 4550 800 |
1000 2000 4000 -1000 2450 |
|
Σd = 8450 |
Calculate arithmetic mean of the following distribution by Direct Method:
|
X |
Y |
|
1 2 3 4 5 6 7 8 |
5 15 25 35 10 5 3 2 |
|
Total |
100 |
Calcuation of Arithmetic Mean:
|
X |
f |
fx |
|
1 2 3 4 5 6 7 8 |
5 15 25 35 10 5 3 2 |
5 30 75 140 50 30 21 16 |
|
Σ f = 100 |
Σ fx = 367 |
How will you calculate correct arithmetic mean from incorrect arithmetic mean due to writing incorrect value.
Steps involved in calculating correct A.M. from incorrect A.M. due to writing incorrect value:
1. Multiply incorrect arithmetic mean by X i.e. 
x N.
2. Subtract incorrect value and add correct value to 
x N.
3. Divide the resultant by N and get the correct arithmetic mean.
The average marks obtained by 100 students is 50. Later, it was found that 63 marks obtained by a student were written as 93. Calculate correct average marks.
Given incorrect marks 
= 50,
N = 100.
Hence, incorrect total marks
Σx = 50 × 100 = 5000
Correct total marks
= 5000 - 93 + 63 = 4970.Correct 
= 4970/100 = 49.70.
The average marks of 100 students in a class are 48. But while calculating it the marks of a student were written 73 instead of 53. Find out the correct arithmetic mean.
Incorrect ΣX
= N × 
= 100 × 48 = 4800
Correct ΣX = 4800 - 73 + 53 = 4780
4780
Hence Correct 
= 4780/100 = 47.8.
The arithmetic mean of 5 items is 7, but on checking it was found that two items were written as 4 and 8 in place of 5 and 9 respectively. Calculate correct arithmetic mean.
Incorrect ΣX
= 7 × 5 = 35
Correct ΣX = 35- 4- 8 + 5 + 9 = 37
Correct 
= 37/5 = 7.4
Values of median and mode are 26 and 25 respectively. Calculate mode.
Mode = 3 Median-2Arithmetic Mean
= 3x26-2x 25
= 78-50 = 28
Arithmetic mean is affected by a large value or a small value. Prove it by taking an example.
For example, we take daily wages of 5 workers:
Rs. 45, 55, 55, 65, 70
Now, we suppose the daily wages of the 6 workers is Rs. 90. In this case
From the above example, it is proved that arithmetic mean is sufficiently affected by taking a large value. Earlier A.M. was 58 and now it is 80.
Calculate weighted arithmetic mean from the following data:
|
X |
10 |
20 |
30 |
40 |
50 |
|
Weight |
2 |
3 |
4 |
6 |
5 |
|
X |
W |
WX |
|
10 20 30 40 50 |
2 3 4 6 5 |
20 60 120 240 250 |
|
Σ W = 20 |
Σ WX = 690 |
Suppose mean of a series of 5 items is 30. Four values are respectively 10, 15,30 and 35. Estimate the missing 5th value of the series.
Suppose missing 5th item is 5 X.
__
Hence, X = 10+15+30+35+5X
5
= 90 + X
5
90 + X = 150
N = 150-90 = 60
Hence, missing value is 60.
The following table shows prices per 100 gram of tea of different brands using quantity as weight. Find out weighted arithmetic mean of prices.
|
Price per 100 gram in Rs. |
Quantity |
|
2.50 3.00 3.50 4.00 4.25 5.00 |
10 8 8 4 4 2 |
|
X |
W |
WX |
|
2.50 3.00 3.50 4.00 4.25 5.00 |
10 8 8 4 4 2 |
25.00 24.00 28.00 16.00 17.00 10.00 |
|
Σ W = 36 |
Σ WX = 120.00 |
Calculate weighted mean of the following data:
|
Item (X) |
Weight |
|
5 10 25 20 25 20 |
8 4 5 10 7 6 |
|
X |
W |
WX |
|
5 10 25 20 25 20 |
8 4 5 10 7 6 |
40 40 125 200 175 120 |
|
Σ W = 40 |
Σ WX = 700 |
Calculate weighted mean of the following data:
|
Marks |
81 |
76 |
74 |
58 |
70 |
73 |
|
Weight |
2 |
3 |
6 |
7 |
3 |
7 |
|
X |
W |
WX |
|
81 76 74 58 70 73 |
2 3 6 7 3 7 |
162 228 444 406 210 511 |
|
Σ W = 28 |
Σ WX= 1,961 |
A candidate obtained 46% in English, 67% in Mathematics, 53% in Hindi, 72% in History and 58% in Economics. It is agreed to give single weights to marks in English and double weights to marks in Mathematics as compared to other subjects. Calculate weighted mean.
|
Subject |
X |
W |
WX |
|
English Mathematics Hindi History Economics |
46 67 53 72 58 |
3 2 1 1 1 |
138 134 53 72 58 |
|
Σ W = 8 |
Σ WX = 455 |
||
Calculate weighted mean by weighing each price by the quantity consumed:
|
Articles of Food |
Quantity consumed in kgs. |
Price per kg. in Rs. |
|
Flour Ghee Sugar Potato Oil |
11.50 5.60 0.28 0.16 0.35 |
5.8 58.4 8.2 2.5 20.0 |
|
Price per kg. in Rs. (X) |
Quantity consumed in kgs. (W) |
WX |
|
5.8 58.4 8.2 2.5 20.0 |
11.50 5.60 0.28 0.16 0.35 Σ W = 17.89 |
66.700 327.040 2.296 0.400 7.000 Σ WX = 403.436 |
Write down the uses of Weighted Mean.
Uses of Weighted Mean:
1. Weighted mean is used for calculating of wage bill by factory authorities or contractors.
2. Weighted mean is used for comparison of the results of two or more universities or boards.
3. It is used in the construction of index numbers.
4. It is used to calculate standardised birth rate and death rate.
A candidate obtains the following percentage of marks:
|
Subjects |
Marks |
|
History Geography Sanskrit Maths Economics English Politics |
54 47 75 84 56 78 57 |
It is agreed to give double weights to marks in English, Maths and Sanskrit. What is the weighted mean?
|
Subject |
Marks (X) |
Weightage (W) |
WX |
|
History Geography Sanskrit Maths Economics English Politics |
54 47 75 84 56 78 57 |
1 1 2 2 1 2 1 |
54 47 150 168 56 156 57 |
|
Σ W = 100 |
Σ WX = 688 |
Calculate weighted mean by weighting each price by the quantity consumed:
|
Food Items |
Quantity consumed (in kgs.) |
Price in Rs. (per kg.) |
|
Flour Ghee Sugar Potato Oil |
500 200 30 15 40 |
1.25 20.00 4.50 0.50 5.50 |
|
Food Items |
Quantity consumed (W) |
Price in Rs. (per kg.) (X) |
WX |
|
Flour Ghee Sugar Potato Oil |
500 200 30 15 40 |
1.25 20.00 4.50 0.50 5.50 |
625.00 4000.00 135.00 7.50 220.00 |
|
Σ W = 785 |
Σ WX = 4987.50 |
The mean weight of 25 boys in a group A of a class is 61 kg. and the mean weight of 35 boys in group B of the same class is 58 kg. Find the mean weight of 60 bovs.
Total weight of 25 boys in group A
_
=X XN = 61 x 25 = 1525 kgs.
Total weight of 35 boys in group B
= 58 x 35 = 2030 kgs./
Total wight of 60 boys
=1525 + 2030 = 3555 kgs.
Mean wight of 60 boys
3555/60 = 59.25 kg.
The average marks for students in a class of 30 were 52. The six students has an average of 31 marks. What were the average marks of the other students.
_
Total marks of 30 students = X x N = 52 x 30 = 1560
Total marks of six students
= 31 x 6 = 186
Total marks of 24 students
= 1560 - 186 = 1374 marks
Average marks of students = 1374/24 = 57.25 marks.
The mean marks obtained in an examination by a group of 100 students were found to be 49.6. The mean marks obtained in the same examination by another group of 200 students were 53.32. Find out the mean of marks obtained by both the groups of students taken together.
Total marks of 300 students
= (49.4 x 100) + (52.32 x 200)
= 4946+10464
=15410
There are two branches of an examination employing 100 and 80 person respectively. If the arithmetic mean of the monthly salaries by two branches are Rs. 275 and Rs. 225 respectively. Find out the arithmetic mean of the salaries of the employees of the establishment as a whole.
Combined Mean
= (275 X 100) + (225 X 80)
100 + 80
= 27500 + 18000 = 45500
180 1800
= Rs. 252.8.
Eight workers earn the following income:
30,36,34,40,42,46, 54,62 Find the arithmetic mean.
The mean marks of 100 students of combined sections of A and B are 38 marks. The mean marks of section A are 40 and that of section B are 35. Find out the number of students in sections A and B.
Hence, students in Sections A and B are 60 and 40 respectively.
Calculate median from the following distribution:
|
Marks |
No. of Students |
|
10 20 30 40 50 60 70 80 |
2 8 16 26 20 16 7 4 |
By arranging the data we get the following table:
|
Daily Wages (in Rs.) |
No. of Workers |
Cumulative Frequencies |
|
50 70 80 100 110 |
20 15 12 15 18 |
20 35 47 62 80 |
Sponsor Area
Calculate Q3 from the following data:
|
Heights in inches |
No. of Persons |
|
58 59 60 61 62 63 64 65 66 |
2 3 6 15 10 5 4 3 1 |
|
Heights in inches (X) |
No. of Persons (f) |
cf |
|
58 59 60 61 62 63 64 65 |
2 3 6 15 10 5 4 3 1 |
2 5 11 26 36 41 45 48 49 |
|
Σ f =49 |
Calculate the median of following data :
145, 130, 200, 210, 198, 234, 159, 160, 178, 257, 260, 300, 345, 360 and 390.
By arranging data we get the following data:
130, 145, 159, 160, 178, 198, 200, 210, 234.1 257, 260, 300, 345, 360 and 390.
Median = Value of N + 1 item
2
= 15 + 1
2
= Value of 7th item
Value of 8th item = 210
Calculate Q1 from the following data:
|
Heights in inches |
No. of Persons |
|
58 59 60 61 62 63 64 65 66 |
2 3 6 15 10 5 4 3 1 |
|
Heights in inches (X) |
No. of Persons (f) |
cf. |
|
58 59 60 61 62 63 64 65 66 |
2 3 6 15 10 5 4 3 1 |
2 5 11 26 36 41 45 48 49 |
|
Σ f =49 |
Calculate the median of following data:
145, 130, 200, 210, 198, 234, 159, 160, 178, 257, 260, 300, 345, 360 and 390.
By arranging data we get the following data:
130, 145, 159, 160, 178, 198, 200, 210, 234.1 257, 260, 300, 345, 360 and 390.
Median = Value of N + 1 item
2
= 15 + 1
2
= Value of 8th item
Value if 8th item = 210
We have the following frequency distribution of the size of 51 households. Calculate the arithmetic median:
|
Size : |
2 |
3 |
4 |
5 |
6 |
7 |
|
No. of households |
2 |
3 |
9 |
21 |
11 |
5 |
|
Size |
No. of |
Cumulative Frequencies (cf) |
|
2 3 4 5 6 7 |
2 3 9 21 11 5 |
2 5 14 35 46 51 |
|
Σ f =51 |
Calculate mode for the following data :
|
No. of Persons |
Families |
|
1 2 3 4 5 6 7 8 9 10 |
26 113 120 95 60 42 21 14 5 4 |
Mode is 3 because this item has maximum frequencies.
Daily expenditure on vegetables (in Rs. of 20 households) in a certain locality are given below:
|
25.00 |
26.50 |
30.25 |
28.00 |
23.00 |
|
31.40 |
34.00 |
33.00 |
30.50 |
27.20 |
|
28.00 |
35.00 |
38.60 |
34.00 |
22.50 |
|
24.00 |
23.70 |
28.00 |
29.00 |
32.20 |
Show that the value of the median lies between the arithmetic mean and the mode.
(i) Mode = 28.00 items with the highest frequency.
(iii) For calculating median we will arrange the items in ascending order as given below :
22.50,23.00,23.70,24.00,25.00,26.50,27.20, 28.00, 28.00, 28.00, 29.00, 30.25, 30.50, 31.40, 32.20, 33.00, 34.00, 34.00, 35.00, 38.60.
i.e. The average of the values of 10th and 11th items
Now, Median = 28.50
Mean = 29.19
Mode = 28.00
28.50 is between 28 and 29.19. Hence, it is proved that median lies between mean and median.
In a certain examination there were 100 candidates of whom 21 failed, 6 secured distinction, 43 were placed in the third division and 18 in second division. It is know that atleast 75% marks are required for distinction, at least 40% for passing, at least 50% for second division and at least 60% for first division. Calculate the median of the distribution of marks.
No. of candidate = 100
Failed = 21
Passed =100-21 = 79
Secured distinction = 6
3rd divisioner = 43
2nd divisioner = 18
1st divisioner = 18 (79 - 43 - 18)
(including distinction holders)
|
Class Interval |
f |
cf |
|
0-40 40-50 50-60 60-70 75 and above |
21 43 18 12 6 |
21 64 82 94 100 |
|
Σ f = 100 |
The value of 50th item lies in 40 - 50 class interval.
Calculate the upper and lower quartiles for the following frequency distribution:
|
Class Interval |
Frequency |
|
13-25 25-37 37-49 49-61 61-73 |
6 11 23 7 3 |
|
50 |
The daily expenditure in rupees of 50 households is given as follows:
|
Daily Expenditure |
No. of Households |
|
100-150 150-200 200-300 300-500 500-1000 Above 1000 |
3 9 21 10 5 2 |
|
Total |
50 |
|
C.I. |
Frequency |
c.f. |
|
100-150 150-200 200-300 300-500 500-1000 1000 and above |
3 9 21 10 5 2 |
3 12 33 43 48 50 |
|
50 |
Calculate arithmetic mean by (i) Direct Method, (ii) Short Cut Method and Step Deviation Method.
|
Daily Expenditure (in Rs.) |
No. of Persons |
|
0-10 10-20 20-30 30-40 40-50 50-60 60-70 |
3 2 5 8 4 6 2 |
(i) Calculation of Arithmetic Mean by Direct Method:
|
Daily Expenditure (Rs.) |
M.V. |
f. |
fx |
|
0-10 10-20 20-30 30-40 40-50 50-60 60-70 |
5 15 25 35 45 55 65 |
3 2 5 8 4 6 2 |
15 30 125 280 180 330 130 |
|
N = 30 |
Σ fx =1090 |
(ii) Calculation of Arithmetic Mean by Short Cut Method :
|
Daily |
X M.V |
f |
fx |
fdx |
|
0-10 10-20 20-30 30-40 40-50 50-60 60-70 |
5 15 25 35 45 55 65 |
3 2 5 8 4 6 2 |
-30 -20 -10 0 + 10 + 20 + 30 |
-90 -40 -50 0 + 40 + 120 + 60 |
|
N=30 |
Σ fdx = 40 |
(iii) Calculation of Arithmetic Mean by Step Deviation Method :
|
Daily |
X M.V |
f |
dx |
dx' |
fdx |
|
0-10 10-20 20-30 30-40 40-50 50-60 60-70 |
5 15 25 35 45 55 65 |
3 2 5 8 4 6 2 |
-30 -20 - 10 0 + 10 + 20 + 30 |
-3 -2 - 1 0 + 1 + 2 + 3 |
-9 -4 -5 0 + 4 + 12 + 6 |
|
N |
Σ fdx' = 4 |
||||
Following is the height of 10 students:
|
Students |
A |
B |
C |
D |
E |
F |
G |
H |
I |
J |
|
Height (cm) |
155 |
153 |
168 |
160 |
162 |
166 |
164 |
180 |
157 |
165 |
Calculate arithmetic mean by using direct and short cut method.
(a) Calculation of arithmetic mean by direct method:
|
Students |
Height (c.m.) |
|
A B C D E F G H I J |
155 153 168 160 162 166 164 180 157 165 |
|
N= 10 |
Σ = 1630 |
(b) Calculation of arithmetic mean by indirect method :
|
Students |
Height (c.m.) |
d |
|
A B C D E F G H I J N= 10 |
155 153 168 160 162 166 164 180 157 165 |
-5 -7 + 8 + 0 + 2 + 6 + 4 + 20 -3 + 5 Σ d = 30 |
Average age of the people of a country is shown in the following table:
|
Age (years) |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
|
People (000) |
30 |
32 |
15 |
12 |
9 |
|
Age |
People (000) f |
M.V. (X) |
fx |
|
10-20 20-30 30-40 40-50 50-60 |
30 32 15 12 9 |
15 25 35 45 55 |
450 800 525 540 495 |
|
Σ f =98 |
Σ fx = 2810 |
Find out the arithmetic mean from the following distribution by short cut method:
|
Items |
10-8 |
8-6 |
6 – 4 |
4 —2 |
2-0 |
|
Frequency |
10 |
8 |
6 |
4 |
2 |
|
Items |
X (MV) |
f |
(X-AM) d |
fd |
|
10-8 8-6 6-4 4-2 2-0 |
9 7 5 3 1 |
10 8 6 4 2 Σ f=30 |
4 2 0 -2 -4 |
40 16 0 -8 -8 Σ fd = 40 |
Mean marks obtained by a student in five subjects are 15. In English, he secures 8 marks, in Economics 12, in Mathematics 18 and in Commerce 9. Find out the marks secured in 5th subject i.e. Statistics.
1. Total marks obtained in 5 subjects = 15 x 5 = 75
2. Total marks obtained in 4 subjects = 8 + 12 + 18 + 9 = 47
3. Marks obtained in 5th subject i.e. Statistics = 75 - 47 = 28
Calculate weighted mean of the following data:
|
Item |
81 |
76 |
74 |
58 |
70 |
73 |
|
Weightes |
2 |
3 |
6 |
7 |
3 |
7 |
|
X |
W |
|
|
81 76 74 58 70 73 |
2 3 6 7 3 7 Σ W = 28 |
162 228 444 406 2 1 0 5 1 1 Σ Wx= 1961 |
In the following frequency distribution if the arithmetic mean is 45.6. Find out the missing frequency.
|
Wage |
No. of Workers |
|
10-20 20-30 30-40 40-50 50-60 60-70 70-80 |
5 6 7 ? 4 3 9 |
|
Wage (Rs.) |
X (M.V.) |
f |
fx |
|
10-20 20-30 30-40 40-50 50-60 60-70 70-80 |
15 25 35 45 55 65 75 |
5 6 7 X (suppose) 4 3 9 |
75 150 245 45X 220 195 675 |
|
Σ f= X + 34 |
Σ fx = 1560 + 45X |
Hence missing frequency is 16
The mean of 10 items was 70. Later on it was found that an item 92 was misread as 29. What was the correct mean?
Since,
Mean of 200 items = 50
No.ofitems
sum of 200 items = mean
sum of 200 items = mean of x no of items
50×200=10,000
wrong sum of 200 items = 10,000
Correct sum = 10,000+192+88-92-8
= 10,280-100
=10,180∴ correct mean = 200
10,180 = 50.9
Given below is the age of 9 children of a street. Find the median.
5, 8, 7, 3, 4, 6, 2, 9, 1
By arranging the age we get
1, 2, 3, 4, 5, 6, 7, 8, 9
Find out the median of the series of the following table:
|
Items |
3 |
4 |
5 |
6 |
7 |
8 |
|
Frequency |
6 |
9 |
11 |
14 |
23 |
10 |
|
Item |
Frequency |
Cumulative Frequency |
|
3 4 5
7 8 |
3 9 11 14 23 10 |
6 15 26
63 73 |
Hence median is 6.
Calculate mode of following series using the graphic technique. Counter check the model value with the frequency.
|
Wage |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
|
No. of Workers |
28 |
46 |
54 |
42 |
30 |
Checking the model value with the formula
From observation we come to know that model lies in 20-30 class interval as it has highest frequency.
Calculate median by the following data :
|
Mid value |
20 |
30 |
40 |
50 |
60 |
70 |
|
Cummulative Frequency |
12 |
25 |
42 |
46 |
48 |
50 |
|
Class Interval |
Frequency |
Commutative Frequency |
|
15—25 25—35 35—45 45—55 55—65 65—75 |
12 13 17 4 2 2 Σ f= 50 |
12 25 42 46 48 50 |
Size of 25th item lies in 25-35 class interval.
Find out median from the following data.
|
Age |
20-25 |
25-30 |
30-35 |
35-40 |
40-45 |
45-50 |
50-55 |
55-60 |
|
No. of Students |
50 |
70 |
100 |
180 |
150 |
120 |
70 |
60 |
|
Age |
No. of Students |
cf |
|
20-25 25-30 30-35 35-40 40-45 45-50 50-55 55-60 |
50 70 100 180 150 120 70 60 |
50 120 220 400 550 670 740 800 |
|
Σ f=800 |
Calculate mode from the following data:
|
X |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
|
f |
5 |
10 |
40 |
20 |
25 |
|
X |
f |
|
|
0-10 10-20 (l1) 20-30 30-40 40-50 |
5 10 (f0) 40 (f1) 20 (f2) 25 |
|
Ans.
Locate mode graphically for the following table:
|
X |
0-20 |
20-40 |
40-60 |
60-80 |
80-100 |
|
Frequency |
5 |
10 |
20 |
10 |
5 |
|
X |
f |
|
0-20 20-40 (l1) 40-60 60-80 80-100 |
5 10 (f0) 20 (f1) 10 5 |
From the diagram we come to know that mode lies in class-interval 40—60.
Calculate median in a asymmetrical distribution, if mode is 83 and arithmetic mean is 92.
Mode = 3 median - 2 mean
83 = 3 median -2 x 92
83 + 184 = 3 median
Hence Median = 83 + 184 =267 = 89
3 3
Calculate mode when arithmetic mean is 146 and median is 130.
Mode = 3 median - 2 mean
Mode = 3 x 130 - 2 x 146
Mode =390-292 = 98
If mode is 63, and median is 77, calculate arithmetic mean:
Mode = 3 median - 2 mean
63 = 3 x 77 - 2mean
or 63-231 = - 2 mean
or-168 =-2 mean
Hence, Mean
= 168 = 84
2
Calculate Median from the following data:
|
Daily Wages (In Rs.) |
No. of Workers |
|
55-60 50-55 45-50 40-45 35-40 30-35 25-30 20-25 |
7 13 15 20 30 33 28 14 |
By arranging the data in ascending order, we get as follows and we calculate median with the help of following data :
|
Daily Wages (in Rs.) |
No. of Workers (f) |
Cumulative Frequencies (cf) |
|
20-25 25-30 30-35 35-40 40-45 45-50 50-55 55-60 |
14 28 33 30 20 15 13 7 |
14 42 75 105 125 140 153 160 |
Value of 80th item lies in 35 - 40 class interval.
Hence median class is 35 - 40.
Calculate the median of the following distribution of salaries of a employees of a firm.
|
Income (in Rs.) |
No. of Persons |
|
400-500 500-600 600-700 700-800 800-900 900-1000 1000-1100 |
25 69 107 170 201 142 64 |
|
ncome (Rs.) |
No. of Persons (f) |
c.f |
|
400-500 500-600 600-700 700-800 800-900 900-1000 1000-1100 |
25 69 107 170 201 142 64 |
25 94 201 371 572 714 778 |
11 = Lower limit of median group.
c.f. = Cumulative frequency of the class preceding the median class.
f = Frequency of the median group,
c = The class interval of the median group
Calculation of Median :
Median lies in the group 800 - 900. Applying the formula we get,
Hence Median Income = 800.89.
Calculate Median of the following series:
|
Wages Rate (in Rs.) |
No. of Workers |
|
Less than 10 Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 |
15 35 60 84 96 127 198 250 |
Here, we have been given cumulative frequency of "less than". This frequency has been converted into simple frequency as under for calculating median.
Calculation of Median:
|
Wages (Rs.) |
Cumulative frequency |
Frequency |
|
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 |
15 35 60 84 96 127 198 250 |
15-0= 15 35-15 = 20 60-35 = 25 84 - 60= 24 96 - 84 = 12 127-96 = 31 198-127 = 71 250-198 = 52 |
Calculate median of the following:
|
Marks |
No. of Students |
|
46-50 41-45 36-40 31-35 26-30 21-25 16-20 11-15 |
5 11 22 35 26 13 10 7 |
In the question the inclusive series has been given in the descending order. For calculating, it would be converted into an exclusives series in the ascending order.
Calculation of Median
|
Class Interval |
No. of Students |
Cumulative frequency (c.f.) |
|
10.5-15.5 15.5-20.5 20.5-25.5 25.5-30.5 30.5-35.5 35.5-40.5 40.5-45.5 45.5-50.5 |
7 10 13 26 35 22 11 5 |
7 17 30 50 91 113 124 129 |
|
N=129 |
Hence Median = 31.7 marks.
From the following distribution, calculate the missing frequency if N = 100 and Median = 30.
|
Marks |
No. of Students |
|
0-10 10-20 20-30 30-40 40-50 50-60 |
10 ? 25 30 ? 10 |
In the question, two frequencies are missing f1 and f2. We will get one frequency from summation of frequencies and the other from median formula :
|
Marks |
f |
c.f. |
|
0-10 10-20 20-30 30-40 40-50 50-60 |
10 f1 25 30 f2 10 |
10 10 + f1 35 + f1 65 + f1 65 + f1 + f2 75 +f1 + f2 |
|
Σ f = 100 |
We two that the sum of frequencies is equal to the cumulative frequency of the last class interval. Hence
75 + f1 + f2 = 100
or f1 + f2 = 100-75 = 25 .... (i)
We have been given median (30) which must be in 30 - 40 class interval.
Hence class interval = 30-40
Median = 30 (given)
Substituting the value of f, in (i) equation we get,
15 + f2 = 25
Hence f2 = 25 - 15 = 10
Calculate the mode of the following frequency distribution:
|
Size of Items : |
|||||||||||
|
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
17 |
|
|
No. of Items : |
|||||||||||
|
2 |
3 |
6 |
12 |
20 |
24 |
25 |
7 |
5 |
3 |
1 |
|
On inspection we find that 12 is the mode but it does not appear to be correct since maximum items are not concentrated around. This value let us use the grouping method to determine mode.
Grouping Table
|
Size Item |
Col. 1 |
CoL 2 |
CoL 3 |
CoL 4 |
CoL 5 |
CoL 6 |
|
7 |
2 |
|||||
|
8 |
3 |
5 |
9 |
11 |
21 |
|
|
9 |
6 |
|||||
|
10 |
12 |
18 |
32 |
|||
|
11 |
20 |
56 |
38 |
|||
|
12 |
24 |
44 |
69 |
|||
|
13 |
25 |
49 |
56 |
|||
|
14 |
7 |
32 |
12 |
37 |
||
|
15 |
5 |
8 |
15 |
|||
|
16 |
3 |
4 |
||||
|
17 |
1 |
9 |
Analysis Table
|
Col. No. |
Size of Items |
||||||||||
|
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
17 |
|
|
1 |
1 |
||||||||||
|
2 |
1 |
1 |
1 |
||||||||
|
3 |
1 |
||||||||||
|
4 |
1 |
1 |
1 |
1 |
|||||||
|
5 |
1 |
1 |
|||||||||
|
6 |
1 |
1 |
1 |
||||||||
|
Total |
1 |
3 |
5 |
4 |
1 |
||||||
Since size 12 occur the maximum number of times i.e. 5 times the mode is 12.
Calculate mode from the following data:
|
Class Interval : |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
|
Frequency : |
2 |
5 |
7 |
5 |
2 |
By inspection we come to know that modal class is 20 - 30 as it has highest frequencies.
Here 1, = 20
D1 = 7-5 = 2
D2 = 7-5 = 2
h = 10
Calculate the value of modal worker family's monthly income from the following data:
|
Income per month Below (in '000Rs.) |
50 |
45 |
40 |
35 |
30 |
25 |
20 |
15 |
|
No. of families |
97 |
95 |
90 |
80 |
60 |
30 |
12 |
4 |
We have given cumulative frequencys. We will convert it into an exclusive series in order to calculate mode :
Grouping Table
|
Income (in'000Rs.) |
I |
n |
III |
IV |
V |
VI |
|
45-50 |
2 |
|||||
|
40-45 |
5 |
7 |
17 |
|||
|
35-40 |
10 |
15 |
||||
|
30-35 |
20 |
30 |
35 |
|||
|
25-30 |
30 |
50 |
60 |
|||
|
20-25 |
18 |
48 |
68 |
|||
|
15-20 |
8 |
26 |
56 |
|||
|
10-15 |
4 |
12 |
30 |
|
Columns |
45-50 |
40 -45 |
35-40 |
30-35 |
25-30 |
20-25 |
15-20 |
10-15 |
|
I |
✓ |
|||||||
|
n |
✓ |
✓ |
||||||
|
ni |
✓ |
✓ |
||||||
|
IV |
✓ |
✓ |
✓ |
|||||
|
V |
✓ |
✓ |
✓ |
|||||
|
VI |
✓ |
✓ |
✓ |
|||||
|
Total |
1 |
3 |
6 |
3 |
1 |
The value.of the mode lies in 25 - 30 class interval. By inspection also, it can be seen that this is a modal class.
Using the formula we can obtain the value of mode as follows :
We have given cumulative frequencys. We will convert it into an exclusive series in order to calculate mode :
Grouping Table
|
Income (in'000Rs.) |
I |
n |
III |
IV |
V |
VI |
|
45-50 |
2 |
|||||
|
40-45 |
5 |
7 |
17 |
|||
|
35-40 |
10 |
15 |
||||
|
30-35 |
20 |
30 |
35 |
|||
|
25-30 |
30 |
50 |
60 |
|||
|
20-25 |
18 |
48 |
68 |
|||
|
15-20 |
8 |
26 |
56 |
|||
|
10-15 |
4 |
12 |
30 |
|
Columns |
45-50 |
40 -45 |
35-40 |
30-35 |
25-30 |
20-25 |
15-20 |
10-15 |
|
I |
✓ |
|||||||
|
n |
✓ |
✓ |
||||||
|
ni |
✓ |
✓ |
||||||
|
IV |
✓ |
✓ |
✓ |
|||||
|
V |
✓ |
✓ |
✓ |
|||||
|
VI |
✓ |
✓ |
✓ |
|||||
|
Total |
1 |
3 |
6 |
3 |
1 |
The value.of the mode lies in 25 - 30 class interval. By inspection also, it can be seen that this is a modal class.
Using the formula we can obtain the value of mode as follows :
Prove with an example that if we add, subtract, multiply and divide all the items by any constant value (say 2). We get the arithmetic mean add by 2, subtracted by 2, multiplied by and divided by 2 repectively.
We take by following values 10, 20, 30, 40 and 50.
|
X |
X + 2 |
X-2 |
1x2 |
X + 2 |
|
|
10 20 30 40 50 |
12 22 32 42 52 |
8 18 28 38 48 |
20 40 60 80 100 |
5 10 15 20 25 |
|
|
Total |
150 |
160 |
140 |
300 |
75 |
Determine the median value of the following series using graphic method (less than ogive approach)
|
Marks: |
0-5 |
5-10 |
10-15 |
15-20 |
20-25 |
25-30 |
30-35 |
35-40 |
|
Number of Students |
4 |
6 |
10 |
10 |
25 |
22 |
18 |
5 |
|
Marks : |
less than 5 |
10 |
15 |
20 |
25 |
30 |
35 |
40 |
|
Number of Students |
4 |
10 |
20 |
30 |
55 |
77 |
95 |
100 |
Find out N/2 th item and mark it on y axis i.e. 50.
Using more than ogive approach. Calculate median value of the following series
|
Marks |
0-5 |
5-10 |
10-15 |
15-20 |
20-25 |
25-30 |
30-35 |
35-40 |
|
Number of Students |
4 |
6 |
10 |
10 |
25 |
22 |
18 |
5 |
Calculation of the Median by more than ogive approach:
|
Marks . More than |
0 |
5 |
10 |
15 |
20 |
25 |
30 |
35 |
40 |
|
Cumulative frequency |
100 |
96 |
90 |
80 |
70 |
45 |
23 |
5 |
0 |
Calculate the median by using less than and more than ogive:
|
Marks |
0-5 |
5-10 |
10-15 |
15-20 |
20-25 |
25-30 |
30-35 |
35-40 |
|
Number of Students |
4 |
6 |
10 |
10 |
25 |
22 |
18 |
5 |
Calculation of the Median less than and more than ogive :
|
Marks |
cf. |
Marks |
c.f. |
||
|
less than 5 less than 10 less than 15 less than 20 less than 25 less than 30 less than 35 less than 40 |
4 10 20 30 55 77 95 100 |
more than 0 more than 5 more than 10 more than 15 more than 20 more than 25 more than 30 more than 35 more than 40 |
100 96 90 80 70 45 23 5 0 |
||
Less than and more than ogive :
Determine the value of mode of the following distribution graphically and verify the results.
|
Marks |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
|
Number of Students |
5 |
12 |
14 |
10 |
8 |
6 |
Verification : Mode lies in the class 20 - 30.
With the help of following data calculate mode:
|
Class Interval |
0-5 |
5-10 |
10-15 |
15-20 |
20-25 |
25-30 |
30-35 |
35-40 |
40-45 |
|
Frequency |
6 |
20 |
32 |
10 |
25 |
30 |
28 |
20 |
15 |
By inspection we come to know that 10 -15 class interval is Modal class, because it has the highest frequency (32). But the frequencies immediately after this class interval the frequency is very low. Hence the modal class will be determined with the help of grouping table.
Grouping Table
Analysis Table
|
Col. No. |
0-5 |
5-10 |
10-15 |
15-20 |
20-25 |
25-30 |
30-35 |
35-40 |
40-45 |
|
I II III IV V VI |
✓ |
||||||||
|
✓ |
✓ |
||||||||
|
✓ |
|||||||||
|
✓ |
✓ |
✓ |
✓ |
||||||
|
✓ |
✓ |
✓ |
|||||||
|
✓ |
✓ |
||||||||
|
Total |
— |
— |
1 |
1 |
3 |
5 |
3 |
— |
— |
From the table we come to know that modal class is 25-30. Now we shall calculate the mode with the help of following formula :
(a) What will you do under following situation to calculate mode when:
(i) class intervals are inclusive,
(ii) when class intervals are unequal.
(b) Calculate mode from the following frequency distribution:
|
Class Interval |
20-24 |
25-29 |
30-34 |
35-39 |
40-44 |
45-49 |
50-54 |
55-59 |
|
Frequency |
3 |
5 |
10 |
20 |
12 |
6 |
3 |
1 |
(a) (i) If class interval is inclusive, convert it in exclusive class interval for calculating mode.
(ii) When class intervals are unequal then make them equal for calculating mode.
(b) Converted Frequency Distribution:
|
Class Interval |
19.5-24.5 |
24.5-29.5 |
29.5-34.5 |
34.5-39.5 |
39.5-44.5 |
44.5-49.5 |
49.5-54 |
554.5-59.5 |
|
Frequency |
3 |
5 |
10 |
20 |
12 |
6 |
3 |
1 |
By inspection we come to know that Modal Class is 34.5-39.5 as it has highest frequency.
Calculate mode with the help of graph and verify the answer by calculation:
calculation :
|
X : |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
|
f: |
5 |
10 |
20 |
25 |
20 |
10 |
5 |
Calculation of Mode Graphically:
Verification of result by calculation :
By observation we come to know that Modal class is 30 - 40.
Calculate mode with the help of graph and verify the answer by calculation:
|
X : |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
|
f: |
5 |
10 |
20 |
25 |
20 |
10 |
5 |
Calculation of Mode Graphically :
Verification of result by algebrical method :
By inspection we come to know that modal class is 20 - 30.
Calculate mode of the following using graphic technique. Check the result calculating mode through its standard formula:
|
Expenditure |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
|
No. of families |
14 |
23 |
27 |
21 |
15 |
Calculation of Mode Graphically:
Verification : By inspection we come to know that 20 - 30 is modal class.
Prove that the arithmetic mean of a series after addition will be equal to the addition by the same constant to it their mean.
We take the following observations to prove the statement
X : 1,2,3,4 and 5
A. M. = 1+2+3+4+5 = 15 =3
5 5
Now we add 2 to all observations and get the following observations :
3,4,5,6 and 7
New A. M. = 3+4+5+6+7 = 25 = 5
5 5
Thus, we see that the A.M. of a series after addition is equal to the addition by the same constant to their mean.
With an example that if we replace each item of observation by the calculated mean, then the total of these replaced values will be equal to the sum of the given observations.
We take following three observations calculate their mean and replace the observations by the calculated mean:
|
Worker Wages |
A |
B |
C |
|
Daily wages (in Rs) |
100 |
150 |
200 |
|
Workers |
Wages (Rs.) X |
Mean X |
|
A B C |
100 150 200 |
150 150 150 |
|
N = 3 |
Σ X = 450 |
450 |
Thus we see that the total of replaced values (Rs. 450/-) is equal to the sum of the given observation.
Prove that the sum of the deviations of the items about the median ignoring ± sign will be less than any other point.
Example :
|
X |
Deviation from Median d |
Deviation from 10 d1 |
|
10 11 12 13 14 |
2 1 0 1 2 |
0 1 2 3 4 |
|
N = 5 |
Σ d=6 |
Σ d1 = 10 |
Suppose other point is 10.
From the above example we come to know that the sum of the deviations taken from median (12) ignoring ± in 6 which is less than the sum of deviations taken from 10.
You are given the values 5 items as 4, 6, 8,10,12.
(i) In the above example if mean is increased by 2, then what happens to the individual observations if all are equally affected.
(ii) If first three items increase by 2, then what would be the value of the last two items so that mean remains the same.
(iii) Replace the value 12 by 9 to what happen to the arithmetic mean comment.
(i) All the observation will be increase by 2 as it is clear from below :
Suppose increased in all the individual observations = X
Hence value of each individual observation will be increased by two
Suppose decrease in the value of last two observation = X
Hence New values will be =
Hence the value of last two items will be increased by 3.
(ii) 
of new values of observation
It shows that arithmetic mean is affected by extreme values.
With example that if there are unequal class intervals in a series, then the median will be same without matching the equal class intervals.
We can prove it by taking the following table:
|
Marks |
No. of Students |
|
0-10 10-30 30-60 60-80 80-90 |
5 12 28 10 5 |
Calculation of median without adjusting the class-intervals.
|
X |
f |
cf |
|
0-10 10-30 30-60 60-80 80-90 |
5 12 28 10 5 |
5 17 45 55 60 |
|
Total |
Σ P = 60 |
The value of 30th item lies in 30-60 class interval.
Now we convert the unequal class intervals into equal intervals and calculate the median
|
Marks |
Quantity |
Cumulative Frequency (cf) |
|
0-30 30-60 69-90 |
5+12 = 17 28 10+5 =15 |
17 45 60 |
|
60 |
The value of 30th item lies in 30-60 class interval
Then we see that the median is same by the both method.
Calculate the value of Median and Q1 from the following data:
|
Marks: |
5-15 |
15-25 |
25-35 |
35 – 5 |
45-55 |
|
Students : |
8 |
12 |
15 |
9 |
6 |
|
Class interval |
Frequency |
cf |
|
5-15 15-25 25-35 35-45 45-55 |
8 12 15 9 6 |
8 20 35 44 50 |
|
Class interval |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
|
Frequency |
5 |
8 |
10 |
14 |
3 |
|
Class interval |
Frequency |
cf |
|
0-10 10-20 20-30 30-40 40-50 |
5 8 10 14 3 |
5 13 23 37 40 |
Calculate the median. Also derive the median geometrically by drawing two ogives and locating the point of intersection.
|
Class Interval |
Frequency |
|
13-25 25-37 37-49 49-61 61-75 |
6 11 23 7 3 |
|
Total |
50 |
Median = Value of 50/2th item = 25th item
Hence median lies in 37-49 class interval.
|
Less than |
Frequency |
|
25 37 49 61 75 |
6 17 40 47 50 |
|
More than |
Frequency |
|
13 25 37 49 61 |
50 44 33 10 3 |
From the following table, find the value of median by making use of both the ogive curves (less than and more than).
|
Marks (X) |
Number of Students |
|
0-10 10-20 20-30 30-40 40-50 |
10 20 30 20 10 |
|
Marks (X) (less than) |
No. of Students (ft) |
Marks (X) |
No. of |
|
10 20 30 40 50 |
10 30 60 80 90 |
0 10 20 30 40 |
90 80 60 30 10 |
Calculate the number of shops under class-interval of 30-40 from the following frequency distribution:
|
Profit |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
|
No. of Shop |
12 |
18 |
27 |
9 |
17 |
6 |
What is a central tendency?
A central tendency is that technique of statistical analysis which represents the whole mass of data.
What are main types of statistical average?
Main types of statistical averages : 1. Arithmetic mean, 2. Median and 3. Mode.
Name the three methods by which the arithmetic mean can be computed.
Methods of computation of arithmetic average are (i) Direct method, (ii) Short cut method and (iii) Step deviation method.
How is arithmetic mean calculated by direct method?
By direct method the arithmetic mean of series of item is calculated by dividing the summation of values by number of items. Following formula is used :
Write down the ssteps involved in the calculation of median in continuous series.
Steps involved : (i) Arrange the series either in ascending or descending order.
(ii) If inclusive series is given convert into exclusive series.
(iii) Convert the simple frequencies into cumulative frequencies.
(iv) After that find the size of (N/2) th item.
(v) Determine the median class in which median lies.
(vi) Apply the following formula :
M = Median.
L1 = Lower limit of median group.
i = Difference between lower and upper limits of median group.
f = Frequency of median group.
m = Median number.
c = Cumulative frequency of the group preceding the group.
If the series is arranged in descending order, then the following formula will be applied :
How is median determined graphically when you are given ‘less than ogive’ or ‘more than ogive’?
Determination of median graphically : Following steps are involved in this case :
1. Convert the frequency distribution series into ‘less than’ or ‘more than’ cumulative series.
2. Present the data graphically to make a ‘less than’ or ‘more than’ ogive.
3. Determine N/2 th item of series.
4. From this point (on the Y axis of the graph), draw a perpendicular to the right to cut cumulative frequency curve.
5. From the point where perpendicular from Y-axis cuts cumulative frequency draw a perpendicular on OX-axis.
6. That point will be median where this perpendicular cuts X-axis.
In which case, should the inspection method be used for the calculation of mode in discrete series?
Inspection method should be used when different items show different frequencies.
What is the partition value?
The value that divides the series into more than two parts is called partition value.
What is positional average?
Positional average is that average whose value is worked out on the basis of his position.
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