Q. 8.2 - A First Course in Probability

Question

From past experience, a professor knows that the test score taking her final examination is a random variable with a mean of$75$.

$\left(a\right)$Give an upper bound for the probability that a student’s test score will exceed$85$.

$\left(b\right)$Suppose, in addition, that the professor knows that the variance of a student’s test score is equal$25$. What can be said about the probability that a student will score between $65$and$85$?

$\left(c\right)$How many students would have to take the examination to ensure a probability of at least $.9$that the class average would be within $5$of$75$? Do not use the central limit theorem.

Step-by-Step Solution

Answer

$X\to$the test score

$\left(a\right)$ $P\{X>85\}\le \frac{\mathbf{15}}{\mathbf{17}},{\sigma }^2=25\Rightarrow P\{X>85\}\le \frac{\mathbf{1}}{\mathbf{5}}$

$\left(b\right)$$P\{65<X<85\}\ge \frac{3}{4}$

$\left(c\right)n\ge 10$

Step 1 Given Information.

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

Step 2 Explanation.

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

Step 3 Part (a) Explanation.

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

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

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

$P\{X>\underset{=85}{\underbrace{75+10}}\}\le \frac{{\sigma }^2}{{\sigma }^2+{10}^2}=\frac{25}{25+100}=\frac{\mathbf{1}}{\mathbf{5}}.$

Step 4 Part (b) Explanation.

By Chebyshev's inequality,

$P\left\{\right|X-75|\ge 10\}\le \frac{{\sigma }^2}{{10}^2}=\frac{25}{100}=\frac{1}{4}$

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

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

On the other hand,

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

Step 5 Part (c) Explanation.

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

$\overline{X}=\frac{X_1+X_2+\cdots +X_n}{n},$

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

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

By Chebyshev's inequality$P\left\{\right|\overline{X}-75|\ge 5\}\le \frac{{\sigma }^2}{n{5}^2}=\frac{25}{25n}=\frac{1}{n}$.

Therefore,

$P\left\{\right|\overline{X}-75|<5\}\ge 1-\frac{1}{n}. (**)$

On the other hand, it is given:$P\left\{\right|\overline{X}-75|<5\}\ge .9\stackrel{(**)}{\Rightarrow }$

$1-\frac{1}{n}\ge .9\Rightarrow n\ge \mathbf{10}$