Chapter 6.2, Problem 62E

To determine

To calculate; the probability that the scores differ by 5 or more points in either direction.

Answer to Problem 62E

  $\text{0}\text{.0768}$

Explanation of Solution

Given:

  $\begin{array}{l}{\mu }_X=24\\ {\text{σ}}_X\text{ = 2}\end{array}$

  $\begin{array}{l}{\text{μ}}_{\text{aX + bY}}{\text{ = a(μ}}_{\text{X}})+b({\text{μ}}_Y)\\ {\text{σ}}^{\text{2}}{}_{\text{aX + bY}}{\text{ = a}}^{\text{2}}{\text{(σ}}^2{}_{\text{X}})+{\text{b}}^{\text{2}}{\text{(σ}}^2{}_{\text{Y}})\end{array}$

  $\begin{array}{l}{\text{μ}}_{{\text{X}}_{\text{1}}-X_2}{\text{ = μ}}_{{\text{X}}_{\text{1}}}{\text{- μ}}_{\text{X2}}\text{= }24-24=0\\ {\text{σ}}_{{\text{X}}_{\text{1}}-X_2}\text{ = }\sqrt{{\sigma }^2{}_{X_1}+{\sigma }^2{}_{X_2}}=\sqrt{(2^2)+(2^2)}=2.8284\end{array}$

  $z=\frac{x-\mu }{\sigma }=\frac{(-5-0)}{2.8284}\approx -1.77$

  $\text{P( Z < 1}\text{.77 or Z > 1}\text{.77) = 2 }\times \text{ P (Z < - 1}\text{.77) = 2 }\times \text{ 0}\text{.0384 = 0}\text{.0768}$