Comment
Share
Q)

A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X.

$\begin{array}{1 1}Mean = 17.53, Variance = 4.89, Standard Deviation = 2.21 \\Mean = 18.53, Variance = 4.89, Standard Deviation = 2.14 \\Mean = 17.53, Variance = 5.89, Standard Deviation = 2.21 \\ Mean = 17.53, Variance = 5.89, Standard Deviation = 2.14 \end{array}$

Comment
A)
• Mean of the probability distribution = $\sum (X_i \times P(X_i))$ The Expected value of X is nothing but the mean of X.
• Standard Deviation = $\sqrt{\text{Variance}}$, where Variance $= E (X^2) - E(X)^2$
Mean of the probability distribution = $\sum (X_i \times P(X_i))$
Mean or $E(X) = \sum (X_i \times P(X_i)) = $$14 \times \large\frac{2}{15} +$$ 15\times\large\frac{1}{15} + $$16\times\large\frac{2}{15} +$$17\times\large\frac{3}{15}+ $$18\times\large\frac{1}{15} +$$19\times \large\frac{2}{15} + $$20\times\large\frac{3}{15} +$$21\times\large\frac{1}{15}$
$E (X) = \large \frac{28+15+32+51+18+28+60+21}{15} = \frac{263}{15} =$$17.53 E(X^2) = \sum ((X_i)^2\times P(X_i)) =$$14^2 \times \large\frac{2}{15} + $$15^2\times\large\frac{1}{15} +$$16^2\times\large\frac{2}{15} + $$17^2\times\large\frac{3}{15}+$$18^2\times\large\frac{1}{15} + $$19^2\times \large\frac{2}{15} +$$ 20^2\times\large\frac{3}{15} + $$21^2\times\large\frac{1}{15} E (X^2) = \large \frac{392+225+512+876+324+722+1200+441}{15} = \frac{4683}{15} =$$312.2$
Therefore Variance = $312.2 - 17.53^2 = 4.89$
Therefore Standard Deviation = $\sqrt{4,89} = 2.21$