Question

Suppose in Problem $8.14$that the variance of the number of automobiles sold weekly is$9$.

$\left(a\right)$Give a lower bound to the probability that next week’s sales are between $10$and$22$, inclusively.

$\left(b\right)$Give an upper bound to the probability that next week’s sales exceed$18$.

Short answer

$\left(a\right)$the probability is at least $.75$

$\left(b\right)$$P\{X>18\}\le \frac{9}{13}=.69$.

Step-by-step solution

Given Information.

The variance of the number of automobiles sold weekly is$9$.

part (a) Explanation.

Let the random variable $X$represents the number of automobiles sold (in a week). It is given that $E\left[X\right]=16$and${\sigma }^2=Var\left(X\right)=9$.

$\left(a\right)$By Chebyshev's inequality,

$P\left\{\right|X-16|\ge 6\}\le \frac{{\sigma }^2}{6^2}=\frac{9}{36}=\frac{1}{4}$.

Since$P\{10\le X\le 22\}=P\left\{\right|X-16|\le 6\}$we get:

$P\{10\le X\le 22\}=1-P\left\{\right|X-16|\ge 6\}\ge \frac{\mathbf{3}}{\mathbf{4}}=.\mathbf{75}$.

Part (b) Explanation.

$\left(b\right)$Using Corollary $5.1$.(textbook) for$\underset{¯}{a=2}$we get:

$P\{X>\underset{=18}{\underbrace{16+2}}\}\le \frac{{\sigma }^2}{{\sigma }^2+2^2}=\frac{9}{9+4}=\frac{\mathbf{9}}{\mathbf{13}}=.\mathbf{69}$.