Chapter 11, Problem 22SGR

a.

To determine

To find the percent of time Phil take more than $20$ minutes to complete the work.

Answer to Problem 22SGR

  $34\%$

Explanation of Solution

Given:

The mean $\mu =18.6$ and standard deviation $\sigma =3.5$ .

Calculation:

Consider the event with the mean $\mu =18.6$ and standard deviation $\sigma =3.5$ .

Consider the random variable $X$ which is represent the number of minutes.

Consider the events

  $\begin{array}{l}X>20\\ \frac{X-18.6}{3.5}<\frac{20-18.6}{3.5}\\ Z>0.4\end{array}$

Where $Z$ are the standard normal distributions.

Consider the probability:

  $P(X>20)=P(Z>0.4)$

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,

  $\begin{array}{l}P(X>20)=P(Z>0.4)\\ =0.34\end{array}$

Therefore $34\%$ of time Phil will take more than $20$ minutes to complete the work

b.

To determine

To find the number of sample of $50$ days that will it take Phil less than $15$ minutes to work.

Answer to Problem 22SGR

  $7$

Explanation of Solution

Given:

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

Calculation:

Consider the events:

  $\begin{array}{l}X<15\\ \frac{X-18.6}{3.5}<\frac{15-18.6}{3.5}\\ Z<-1.03\end{array}$

Where $Z$ are the standard normal distributions.

Consider the probability:

  $P(X<15)=P(Z<-1.03)$

  

Therefore,

  $\begin{array}{l}P(X<15)=P(Z<-1.03)\\ =0.15\end{array}$

Therefore, $15\%$ of times it will take less than $15$ minutes to commute the work days.

As total number of days in the sample is $50$ therefore the number of days will it take Phil less than $15$ minutes is;

  $15\%\text{ of 50}=7.5$ Days.

Therefore, the required number of days is $7$ .