Chapter 11.5, Problem 4GP

a.

To determine

To find the probability $P(30,000<X<32,000)$

Answer to Problem 4GP

  $0.683$

Explanation of Solution

Given:

The lifetimes of a certain type of car tire are normally distributed. The mean life time is $\mu =31,066$ miles with a standard deviation of $\sigma =1644$ miles.

Calculation:

Find the percent of the data is between $30,000$ and $32,000$

  

The graph shows that $30,000$ − $32,000$ define an interval that can be represented by $\overline{X}\pm ts$

And the values of $30,000$ , $32,000$ can be written as

  $\begin{array}{l}934=32,000-31066\\ 1066=31066-30,000\end{array}$

Thus

  $\begin{array}{l}t\sigma =934\\ t(1644)=934\\ t=0.5\end{array}$

The probability that the score is in between $30,000$ and $32,000$ is $0.683$

b.

To determine

To find the probability $P(X>35,000)$

Answer to Problem 4GP

  $0.4775$

Explanation of Solution

Given:

The lifetimes of a certain type of car tire are normally distributed. The mean life time is $\mu =31,066$ miles with a standard deviation of $\sigma =1644$ miles.

Calculation:

Find the percent of the data is between $X>35,000$

  

The graph shows that $X>35,000$ define an interval that can be represented by $\overline{X}\pm ts$

And the values of $X>35,000$ can be written as

  $\begin{array}{l}35,000=40,000-5000\\ 40,000=40,000+0\end{array}$

Thus

  $\begin{array}{l}40,000-t\sigma =30,000\\ 40,000-t(5000)=30,000\\ t=2\end{array}$

The probability that the score is in between $35,000$ and $40,000$ is $0.4775$