Chapter 15.2, Problem 48E

a.

To determine

To estimate the value of $T(2010)$ and describe what it measures.

Answer to Problem 48E

  $6830$

Explanation of Solution

Given:

The following is the given graph-

  

Calculation:

Find the percent of the data is between $35,000$ and $45,000$

The graph shows that $35,000$ − $45,000$ define an interval that can be represented by $\overline{X}\pm ts$

And the values of $35,000$ , $45,000$ can be written as

  $\begin{array}{l}35,000=40,000-5000\\ 45,000=40,000+5000\end{array}$

Thus

  $\begin{array}{l}t\sigma =5000\\ t(5000)=5000\\ t=1\end{array}$

The probability that the score is in between $35,000$ and $45,000$ is $0.683$

Hence the total number of tires that expect to last between $35,000$ and $45,000$ is $0.683(10000)=6830$

b.

To determine

To estimate the value of $T\text{'}(2010)$ and describe what it measures.

Answer to Problem 48E

  $4775$

Explanation of Solution

Given:

The following is the given graph-

  

Calculation:

Find the percent of the data is between $35,000$ and $40,000$

The graph shows that $35,000$ − $40,000$ define an interval that can be represented by $\overline{X}\pm ts$

And the values of $35,000$ , $40,000$ 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 $0.4775$

Hence the total number of tires that expect to last between $35,000$ and $40,000$ is $0.4775(10000)=4775$