91Ó°ÊÓ

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

variable with a mean of$75$.

Let $X_i$represents the test score of the $i$th student taking her final examination, and assume that role="math" localid="1649757255175" $X_i$is a random variable with mean $\mathrm{μ}-75$and variance${\mathrm{σ}}^2-25$.

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

$\stackrel{¯}{X}-\frac{X_1+X_2+⋯+X_n}{n}.$

Then, $E\left(\stackrel{¯}{X}\right)-\mathrm{μ}$and$Var\left(\stackrel{¯}{X}\right)-\frac{{\mathrm{σ}}^2}{n}$.

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$a$,

$P\left\{\frac{\stackrel{¯}{X}-\mathrm{μ}}{\frac{\mathrm{σ}}{\sqrt{n}}}≤a\right\}→\mathrm{Φ}\left(a\right)$

$P\left\{\frac{\stackrel{¯}{X}-75}{\frac{5}{\sqrt{n}}}≤a\right\}→\mathrm{Φ}\left(a\right).(*)$

We need to find a lower bound for $n$such that$P\left\{\right|\stackrel{¯}{X}-75|<5\}â‰\yen .9$.

So,

$$\begin{gathered} .9≤P\left\{\right|\stackrel{¯}{X}-75|<5\}- \\ P\{-5≤\stackrel{¯}{X}-75≤5\}-P\left\{-\sqrt{n}≤\frac{\stackrel{¯}{X}-75}{\frac{5}{\sqrt{n}}}≤\sqrt{n}\right\}\stackrel{(\backslash \mathrm{AA})}{≈} \\ \mathrm{Φ}\left(\sqrt{n}\right)-\mathrm{Φ}(-\sqrt{n})-2\mathrm{Φ}\left(\sqrt{n}\right)-1→ \\ \mathrm{Φ}\left(\sqrt{n}\right)â‰\yen \frac{.9+1}{2}-.95{}^{\text{Table}5.1(\text{texthook, Chapter 5)}}→\sqrt{n}â‰\yen 1.64→ \\ nâ‰\yen 2.6896 \end{gathered}$$

But, since $n∈\mathrm{â„•}$we take$nâ‰\yen \mathbf{3}$.