Answer to Question #295771 in Statistics and Probability for Ali

Question

The number of inquiries received per day by the Office of

Admissions in a certain university is shown below. Find the variance

and standard deviation.

Number of inquiries: 22, 23,24,25,26,27

Probability: 0.08, 0.19, 0.36, 0.25, 0.07, 0.05

Answer 1

Given the probability distribution,

inquiries: 22, 23,24,25,26,27

Probability: 0.08, 0.19, 0.36, 0.25, 0.07, 0.05

The expected value is

$E(x)=\sum xp(x)=(22\times0.08)+(23\times0.19)+(24\times0.36)+(25\times0.25)+(26\times0.07)+(27\times0.05)=24.19$

The variance is given as,

$var(x)=\sum (x^2)-(\sum(x))^2$ 

We need to find $E(x^2)=\sum x^2p(x)=(484\times0.08)+(529\times0.19)+(576\times0.36)+(625\times0.25)+(676\times0.07)+(729\times0.05)=586.61$

Now,

$var(x)=586.61-(24.19)^2=586.61-585.1561=1.4539$

The standard deviation is

$sd(x)=\sqrt{var(x)}=\sqrt{1.4539}=1.2058$