Question

In Exercise 5, let \({{\bf{X}}_{\bf{3}}}\) denote the number of juniors in the random sample of 15 students and the number of seniors in the sample. Find the value of \({\bf{E}}\left( {{{\bf{X}}_{\bf{3}}} - {{\bf{X}}_{\bf{4}}}} \right)\) and the value of \({\bf{Var}}\left( {{{\bf{X}}_{\bf{3}}} - {{\bf{X}}_{\bf{4}}}} \right)\)

Short answer

\(\begin{array}{c}E\left( {{X_3} - {X_4}} \right) = 0.9\\Var\left( {{X_3} - {X_4}} \right) = 10.4460\end{array}\)

Step-by-step solution

Given information

It \({X_3}\)denotes the number of juniors in a random sample of 15 students.

Similarly, \({X_3}\)denotes the number of seniors in a random sample of 15 students.

Adding previous data

It is given that

\(\begin{array}{l}{p_1} = 0.16\\{p_2} = 0.14\\{p_3} = 0.38\\{p_4} = 0.32\end{array}\)

Which denotes the percent of students in freshman, sophomores, juniors, and seniors, respectively. Let the respective number of students be denoted by \({x_1},{x_2},{x_3}\,and\,{x_4}\).

In this case, n=15

Defining the pdf

This is a case of a multinomial distribution.

The pdf of a standard multinomial distribution is:

\(\)\(f\left( {x|n,p} \right) = \left\{ \begin{array}{l}\frac{{n!}}{{{x_1}! \ldots {x_n}!}}{p_1}^{{x_1}} \ldots {p_k}^{{x_k}},if\,{x_1} + \ldots + {x_k} = n\\0,otherwise\end{array} \right.\)

\(\frac{{15!}}{{{x_1}!{x_2}!{x_3}!{x_4}!}}{\left( {0.16} \right)^{{x_1}}}{\left( {0.14} \right)^{{x_2}}}{\left( {0.38} \right)^{{x_3}}}{\left( {0.32} \right)^{{x_4}}}\)

By comparing with multinomial,

\({X_i} \sim Multinomial\left( {n = 15,{p_i}} \right),\,\,i = 1,2,3,4\)

Finding the required values

The expectation and variance of

\(\begin{aligned}{}{X_i}\,for\,i& = 1,2, \ldots ,k\\E\left( {{X_i}} \right) &= n{p_i} \ldots \left( 1 \right)\\Var\left( {{X_i}} \right) &= n{p_i}\left( {1 - {p_i}} \right) \ldots \left( 2 \right)\\Cov\left( {{X_i},{X_j}} \right)& = - n{p_i}{p_j} \ldots \left( 3 \right)\end{aligned}\)

Note that covariance between any pair of \(\left( {{X_i},{X_j}} \right)\) when \(i = 1,2, \ldots ,k\,and\,i \ne j\)

Now,

\(\begin{aligned}{}E\left( {{X_3} - {X_4}} \right) &= E\left( {{X_3}} \right) - E\left( {{X_4}} \right)\\ &= n \times {p_3} - n \times {p_4}\\& = 15 \times \left( {0.38 - 0.32} \right)\\ &= 0.9\end{aligned}\)

\(\begin{aligned}{}Var\left( {{X_3} - {X_4}} \right) &= Var\left( {{X_3}} \right) + Var\left( {{X_4}} \right) - 2{\mathop{\rm cov}} \left[ {{X_3},{X_4}} \right]\\& = n \times {p_3} \times \left( {1 - {p_3}} \right) - n \times {p_4} \times \left( {1 - {p_4}} \right) - \left( { - n \times {p_3} \times {p_4}} \right)\\ &= 15 \times 0.38 \times \left( {1 - 0.38} \right) + 15 \times 0.32 \times \left( {1 - 0.32} \right) - 2 \times \left( { - 15 \times 0.38 \times 0.32} \right)\\ &= 015 \times 0.38 \times 0.62 + 15 \times 0.32 \times 0.68 - 2 \times \left( { - 1.824} \right)\\ &= 3.5340 - 3.2640 - 2 \times \left( { - 1.824} \right)\\& = 10.4460\end{aligned}\)

Therefore,

\(\begin{aligned}{}E\left( {{X_3} - {X_4}} \right) &= 0.9\\Var\left( {{X_3} - {X_4}} \right) &= 10.4460\end{aligned}\)