Chapter 6, Problem 44E

a)

To determine

To explain whether it would be reasonable to hope they’ll last 40,000 miles.

Answer to Problem 44E

It is not reasonable to hope that the tires will last 40,000 miles.

Explanation of Solution

Given:

  $\mu$ = 32000

  $\sigma$ = 2500

  $x$ =40000

Using z-score,

  $z=\frac{40000-32000}{2500}=3.2$

  $\begin{array}{l}P(Z>3.2)=1-P(Z<3.2)\\ P(Z>3.2)=1-0.9993=0.0007\end{array}$

Therefore, probability is very small or less than 0.05. Hence, it is not reasonable to hope that the tires will last 40,000 miles.

b)

To determine

To find approximate percent that tires expected to last less than 30,000 miles

Answer to Problem 44E

Approximately 21.19% of the tires are expected to be less than 30,000 miles.

Explanation of Solution

Given:

  $\mu$ = 32000

  $\sigma$ = 2500

  $x$ =30,000

Using z-score,

  $z=\frac{30000-32000}{2500}=-0.80$

  $P(Z<-0.80)=0.2119$

Hence, approximately 21.19% of the tires are expected to be less than 30,000 miles.

c)

To determine

To find approximate percent that tires expected to last between 30,000 and 35,000 miles.

Answer to Problem 44E

Approximately 67.30% of the tires are expected to be between 30,000 and 35,000 miles.

Explanation of Solution

Given:

  $\mu$ = 32000

  $\sigma$ = 2500

Using z-score,

  $\begin{array}{l}{z}_1=\frac{30000-32000}{2500}=-0.8\\ z_2=\frac{35000-32000}{2500}=1.2\end{array}$

  $\begin{array}{l}P(-0.8<Z<1.2)=P(Z<1.2)-P(Z<-0.8)\\ P(-0.8<Z<1.2)=0.8849-0.2119=0.6731\end{array}$

Hence, approximately 67.30% of the tires are expected to be between 30,000 and 35,000 miles.

d)

To determine

To find IQR.

Answer to Problem 44E

The IQR is 3350.

Explanation of Solution

Given:

  $\mu$ = 32000

  $\sigma$ = 2500

Formula:

  $IQR=Q_3-Q_1$

First, we need to find first quartile and third quartile. Therefore, we need to find z-score corresponding to a probability of 0.25 or 0.75.

  z = $\pm 0.67$

Using formula,

  $\begin{array}{l}{Q}_1=32000+(-0.67)\times 2500=30325\\ Q_3=32000+0.67\times 2500=33675\end{array}$

So, $IQR=33675-30325=3350$

e)

To determine

To find the mileage.

Answer to Problem 44E

the dealer can then guarantee that these tires will last 27,625 miles.

Explanation of Solution

Given:

  $\mu$ = 32000

  $\sigma$ = 2500

Formula:

  $z=\frac{x-\mu }{\sigma }$

Here, 1 in 24 corresponds to a probability of $\frac{1}{25}=0.04$

From given information, the dealer will have to give a refund to all customers whose tires fail in less than x miles.

So, P(Z< z) = 0.04

Z = -1.75

Therefore,

  $x=32000+(-1.75)\times 2500=27625$

Hence, the dealer can then guarantee that these tires will last 27,625 miles.