Chapter 11, Problem 21SGR

a.

To determine

To find the percent of games sell fewer than $75$ tickets.

Answer to Problem 21SGR

  $68.4\%$

Explanation of Solution

Given:

The mean $\mu =68.7$ and standard deviation $\sigma =13.1$ .

Calculation:

Consider the event with $\mu =68.7$ and standard deviation $\sigma =13.1$ .

Consider the random variable $X$ which is represent the number of tickets

Consider the events

  $\begin{array}{l}X<75\\ \frac{X-68.7}{13.1}<\frac{75-68.7}{13.1}\\ Z<0.48\end{array}$

Where $Z$ are the standard normal distributions.

Consider the probability:

  $P(X<75)=P(Z<0.48)$

Calculate this using TI-83 calculator.

Step 1: Press 2^nd, VARS and Selects the DRAW;

  

Step 2: Press $1$ to select the Shade Norm and enter the lower and upper bound to calculate the probability;

  

Therefore,

  $\begin{array}{l}P(X<75)=P(Z<0.48)\\ =0.684\end{array}$

Therefore $68.4\%$ of games sell fewer than $75$ tickets.

b.

To determine

To find the number of $200$ games that will sell more than $100$ tickets.

Answer to Problem 21SGR

  $2$

Explanation of Solution

Given:

The mean $\mu =68.7$ and standard deviation $\sigma =13.1$ .

Calculation:

Consider the events:

  $\begin{array}{l}X>100\\ \frac{X-68.7}{13.1}>\frac{100-68.7}{13.1}\\ Z>2.39\end{array}$

Where $Z$ are the standard normal distributions.

Consider the probability:

  $P(X>100)=P(Z>2.39)$

  

Therefore,

  $\begin{array}{l}P(X>100)=P(Z>2.39)\\ =0.01\end{array}$

Therefore, $1\%$ of games will sell more than $100$ tickets.

As total number of games in the sample is $200$ therefore the number of games more than $100$ tickets is;

  $1\%\text{ of 2}00=2$

Therefore, the numbers of games that will sell more than $100$ tickets is $2$ .