Chapter SH, Problem 11.12EP

a.

To determine

To find the probability that there will be more than $10$ students run in a given year.

Answer to Problem 11.12EP

  $0.03303$

Explanation of Solution

Given:

The number of students that run for students council is normally distributed with a mean of $16.8$ and a standard deviation of $3.7$ .

Calculation:

  $\mu =16.8$ , $\sigma =3.7$

The probability that there will be more than $10$ students run in a given year will be represented by. Compute the corresponding z-value for $X=10$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{10-16.8}{3.7}\\ =\frac{-6.8}{3.7}\\ =-1.838\end{array}$

Hence, the required probability will be given below:

  $\begin{array}{l}P(X>10)=P(z<-1.838)\\ =0.03303\end{array}$

Hence, the required probability is $0.03303$ .

b.

To determine

To find number of years that there were between $15$ and $20$ students run in a given year.

Answer to Problem 11.12EP

  $10$ Years.

Explanation of Solution

Given:

The number of students that run for students council is normally distributed with a mean of $16.8$ and a standard deviation of $3.7$ .

Calculation:

The probability that there were between $15$ and $20$ students run in a given year will be represented by $P(15\le X\le 20)$ . Compute the corresponding z-value for $X=15$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{15-16.8}{3.7}\\ =\frac{-1.8}{3.7}\\ =-0.486\end{array}$

Compute the corresponding z-value for $X=20$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{20-16.8}{3.7}\\ =\frac{3.2}{3.7}\\ =0.865\end{array}$

Hence, the required probability will be calculated below:

  $\begin{array}{l}P(15\le X\le 20)=P(-0.486\le z\le 0.865)\\ =P(z\le \le 0.865)-P(z\le -0.486)\\ =0.80648-0.31348\\ =0.49300\end{array}$

If the school is existed for $20$ years then the number of students in which between $15$ and $20$ students ran, is calculated below:

  $\begin{array}{l}20\times 0.49300\\ \approx 10\end{array}$

Hence, about $10$ years in which there were between $15$ and $20$ assemblies.