Answer to Question #311392 in Statistics and Probability for Star

Question

Heights of the students in a class are given in the distribution below:

5' - 5'2"      10

5'2" - 5'4" 40

    5'4" - 5'6"   25 .

Find:

1. Central Tendency

2. Dispersion

3. Skewness and kurtosis

Answer 1

Solution

1. Central tendency is the measurement of mean, median and the mode.

(a) Mean.

$\bar X=\dfrac{\sum fX}{\sum f}$

$=\dfrac{400.5}{75}=5.34$

(b) median

$\dfrac{N}{2}=\dfrac{75}{2}=37.5$

Hence the median lies in the class $5.2-5.4$

$Median=L+(\dfrac{\frac{N}{2}-pcf}{f})i$

$=5.2+(\dfrac {37.5-11}{40})0 .2$

$=5.3325$

Hence the median height is $5'3$

(c) Mode

The class with the highest frequency $(40)$ is $5.2-5.4$ and hence is the modal class.

$Mode=L(\dfrac{\Delta_1}{\Delta_1+\Delta_2})i$

$M_0=5.2(\dfrac{(40-10)}{(40-10)+(40-25)})0.2$

$=5.333$

Modal height is "5'3""

2. Dispersion is the measurement of variance and standard deviation.

(a) Variance

$\sigma^2=\dfrac {\sum f(X-mean)^2}{n-1}$

$\sigma^2=\dfrac{1.28}{74}=0.017$

(b) Standard deviation

$\sigma=\sqrt {variance}=\sqrt{0.017}$

$\sigma=0.132$

3. (a) Skewness

$=\dfrac{\sum (X-mean)^3}{n.\sigma^3}$

$=\dfrac{-0.009792}{75\times0.132^3}=-0.057$

$=-0.057$

(b) Kurtosis

$=\dfrac{\sum(X-mean)^4}{n.\sigma^4}$

$=\dfrac{0.00397568}{75\times0.132^4}=0.175$

$=0.175$