Statistics

More Topic from Mathematics

Question 1

A value of C for which the conclusion of Mean Value Theorem holds for the function f(x) = logex on the interval [1, 3] is

  • 2log3e

  • loge3/2

  • log3e

  • loge3

Solution

A.

2log3e

Using mean value theorem
straight f apostrophe left parenthesis straight c right parenthesis space equals space fraction numerator straight f left parenthesis 3 right parenthesis minus straight f left parenthesis 1 right parenthesis over denominator 3 minus 1 end fraction
rightwards double arrow 1 over straight c space equals space fraction numerator log space 3 minus log 1 over denominator 2 end fraction
rightwards double arrow space straight c space equals space fraction numerator 2 over denominator 1 og subscript straight e 3 end fraction space equals space 2 space log subscript 3 straight e

Question 2

All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given?

  • Mean

  • Median

  • Mode

  • Variance

Solution

D.

Variance

If initially, marks of students were xi, then
straight sigma subscript 1 superscript 2 space equals fraction numerator straight capital sigma left square bracket left parenthesis straight x subscript straight i space minus straight x with bar on top right parenthesis squared over denominator straight N end fraction
Now comma space each space is space increased space by space 10
therefore space straight sigma subscript 2 superscript 2 space equals space fraction numerator straight capital sigma left square bracket left parenthesis straight x subscript straight i space plus 10 right parenthesis minus left parenthesis straight x with bar on top space plus 10 right parenthesis squared right square bracket over denominator straight N end fraction space equals space straight sigma subscript 1 superscript 2
So, the variance will not change whereas mean, median and mode will increase by 10.

Question 3

For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is

  • 5/2

  • 11/2

  • 6

  • 13/2

Solution

B.

11/2

because space straight sigma subscript straight x superscript 2 space equals space 4 space and space straight sigma subscript straight y superscript 2 space equals 5
Also comma space straight x with minus on top space equals space 2 space and space straight y with minus on top space equals space 4 space
Now comma space Σx subscript straight i over 5 space equals space 2
rightwards double arrow space straight capital sigma space straight x subscript straight i space equals 10 space semicolon space space Σy subscript straight i space equals space 20
and space straight sigma subscript straight x superscript 2 space equals space open parentheses 1 half Σx subscript 1 superscript 2 close parentheses minus left parenthesis straight x with minus on top space right parenthesis squared space equals space 1 fifth space left parenthesis straight capital sigma space straight y subscript straight i superscript 2 right parenthesis minus 16
rightwards double arrow space Σy subscript straight i superscript 2 space equals space 105
space straight sigma subscript straight x superscript 2 space space equals space 1 over 10 left parenthesis straight capital sigma space straight x subscript straight i superscript 2 space plus space straight capital sigma space straight y subscript straight i superscript 2 right parenthesis space minus open parentheses fraction numerator begin display style straight x with minus on top end style plus begin display style straight y with minus on top end style over denominator 2 end fraction close parentheses squared
space equals space 1 over 10 left parenthesis 40 plus 105 right parenthesis minus 9 space equals space fraction numerator 145 minus 90 over denominator 10 end fraction space equals space 55 over 10 space equals space 11 over 2
Question 4

If in a frequently distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

  • 22.0

  • 20.5

  • 25.5

  • 24.0

Solution

D.

24.0

Mode + 2Mean = 3 Median
⇒ Mode = 3 × 22 – 2 × 21= 66 – 42= 24.

Question 5

If the mean deviation about the median of the numbers a, 2a, ....., 50a is 50, then |a| equals 

  • 2

  • 3

  • 4

  • 5

Solution

C.

4

Median = 25.5 a
Mean deviation about median = 50
rightwards double arrow space fraction numerator straight capital sigma space vertical line straight x subscript straight i space minus 25.5 straight a vertical line over denominator 50 end fraction space equals space 50
⇒24.5 a + 23.5a + ..... + 0.5a + 0.5a + .... + 24.5a = 2500
⇒ a + 3a + 5a + ..... + 49a = 2500
⇒ 25/2 (50a) = 2500 ⇒ a = 4