91Ó°ÊÓ

The test score taking her final examination is a random variable with a mean of$75$.

Let $X$represents the test score of a student taking her final examination, assuming that $X$it is a random variable with a mean$\mathrm{μ}=75$.

By Markov's inequality, an upper bound for the probability that a student's test score will exceed $85$is

$P\{X>85\}≤\frac{E\left[X\right]}{85}=\frac{75}{85}=\frac{\mathbf{15}}{\mathbf{17}}.$

Additionally, assume that the professor knows the variance of a student's test scorelocalid="1649756112940" ${\mathrm{σ}}^2=25$. Then,$\underset{¯}{a}=10$we get:

$P\{X>\underset{=85}{\underset{ÁåŸ}{75+10}}\}≤\frac{{\mathrm{σ}}^2}{{\mathrm{σ}}^2+{10}^2}=\frac{25}{25+100}=\frac{\mathbf{1}}{\mathbf{5}}.$

By Chebyshev's inequality,

$P\left\{\right|X-75|â‰\yen 10\}≤\frac{{\mathrm{σ}}^2}{{10}^2}=\frac{25}{100}=\frac{1}{4}$

Therefore, since$P\left\{\right|X-75|â‰\yen 10\}=1-P\left\{\right|X-75|<10\}$, we have:

$P\left\{\right|X-75|<10\}â‰\yen 1-\frac{1}{4}=\frac{3}{4}.(*)$

On the other hand,

$$\begin{gathered} P\left\{\right|X-75|<10\}=P\{-10<X-75<10\}=P\{65<X<85\}\stackrel{(*)}{⇒} \\ P\{65<X<85\}â‰\yen \frac{3}{4}. \end{gathered}$$

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

$\stackrel{¯}{X}=\frac{X_1+X_2+⋯+X_n}{n},$

where $X_i$is the test score of $i$th student? Then, $E\left[X_i\right]=E\left[X\right]={\mathrm{μ}}_{r}Var\left(X_i\right)=Var\left(X\right)={\mathrm{σ}}^2$and therefore

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

By Chebyshev's inequalitylocalid="1649756135336" $P\left\{\right|\stackrel{¯}{X}-75|â‰\yen 5\}≤\frac{{\mathrm{σ}}^2}{n{5}^2}=\frac{25}{25n}=\frac{1}{n}$.

Therefore,

localid="1649755768696" $P\left\{\right|\stackrel{¯}{X}-75|<5\}â‰\yen 1-\frac{1}{n}.(**)$

On the other hand, it is given:$P\left\{\right|\stackrel{¯}{X}-75|<5\}â‰\yen .9\stackrel{(**)}{⇒}$

$1-\frac{1}{n}â‰\yen .9⇒nâ‰\yen \mathbf{10}$