Q. 8.3 - A First Course in Probability

Question

Use the central limit theorem to solve part $\left(c\right)$of the problem$8.2$.

Step-by-Step Solution

Answer

$n\ge \mathbf{3}$

Step 1 Given Information.

The test score of a student taking her final examination is a random

variable with a mean of$75$.

Step 2 Explanation.

Let $X_i$represents the test score of the $i$th student taking her final examination, and assume that $X_i$is a random variable with mean $\mu -75$and variance${\sigma }^2-25$.

let $n$represents the required number of students, and$\overline{X}$ represents the class average:

$\overline{X}-\frac{X_1+X_2+\cdots +X_n}{n}.$

Then, $E\left(\overline{X}\right)-\mu$and$Var\left(\overline{X}\right)-\frac{{\sigma }^2}{n}$.

The central limit theorem says that the average of a set of independent identically distributed random variables is approximately normally distributed, i.e. for each$a$,

$P\left\{\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\le a\right\}\to \Phi \left(a\right)$

$P\left\{\frac{\overline{X}-75}{\frac{5}{\sqrt{n}}}\le a\right\}\to \Phi \left(a\right). (*)$

We need to find a lower bound for $n$such that$P\left\{\right|\overline{X}-75|<5\}\ge .9$.

So,

$$\begin{gathered} .9\le P\left\{\right|\overline{X}-75|<5\}- \\ P\{-5\le \overline{X}-75\le 5\}-P\left\{-\sqrt{n}\le \frac{\overline{X}-75}{\frac{5}{\sqrt{n}}}\le \sqrt{n}\right\}\stackrel{(\backslash \mathrm{AA})}{\approx } \\ \Phi \left(\sqrt{n}\right)-\Phi (-\sqrt{n})-2\Phi \left(\sqrt{n}\right)-1\to \\ \Phi \left(\sqrt{n}\right)\ge \frac{.9+1}{2}-.95{}^{\text{Table }5.1(\text{ texthook, Chapter 5) }}\to \sqrt{n}\ge 1.64\to \\ n\ge 2.6896 \end{gathered}$$

But, since $n\in \mathrm{\mathbb{N}}$we take$n\ge \mathbf{3}$.