Chapter 11.5, Problem 17PPS

(a)

To determine

To find: The number of batteries that will last between $90,000$ and $110,000$ miles.

Answer to Problem 17PPS

The number of batteries will last between $90,000$ and $110,000$ miles are $7,000$ .

Explanation of Solution

Given:

The mean is $113,627$ .

The standard deviation is $14,266$ .

The total number of batteries made by company are $20,000$ .

Calculation:

Here find $P\left(90,000<X<110,000\right)$ .

Calculate the corresponding $z$ value for $X=90,000$ as follows.

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{90,000-113,627}{14,266}\\ =-1.656\end{array}$

Calculate the corresponding $z$ value for $X=110,000$ as follows.

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{110,000-113,627}{14,266}\\ =-0.2542\end{array}$

Calculate the required probability by using TI-83 calculator, and sketch the area between $z=-1.656$ and $z=-0.2542$ as follows.

1. Press “ $2^{\text{nd}}$ ” key and then “ $\text{DISTR}$” key.

  

Figure (1)

2. Go to “DRAW” and press “1.”

  

Figure (2)

3. Enter respective lower and upper $z$ values as shown below.

  

Figure (3)

4. Press “ENTER” to obtain the result as shown below.

  

Figure (4)

The area between $z=-1.656$ and $z=-0.2542$ s $0.35$ .

Therefore, about $35\%$ of batteries will last between $90,000$ and $110,000$ miles.

The total number of batteries are $20,000$ .

So, the number of batteries will last between $90,000$ and $110,000$ miles are,

  $0.35\times 20,000=7,000$

Therefore, the number of batteries will last between $90,000$ and $110,000$ miles are $7,000$ .

(b)

To determine

To find: The number of batteries that will last more than $125,000$ miles.

Answer to Problem 17PPS

The number of batteries that will last more than $125,000$ miles are $4200$ .

Explanation of Solution

Given:

The mean is $113,627$ .

The standard deviation is $14,266$ .

The total number of batteries made by company are $20,000$ .

Calculation:

Here, find $P\left(X>125,000\right)$

Calculate the corresponding $z$ value for $X=125,000$ as follows.

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{125,000-113,627}{14,266}\\ =0.797\end{array}$

Calculate the required probability by using TI-83 calculator, and sketch the area between $z=0.797$ and $z=4$ as follows.

5. Press “ $2^{\text{nd}}$ ” key and then “ $\text{DISTR}$” key.

  

Figure (5)

6. Go to “DRAW” and press “1.”

  

Figure (6)

7. Enter respective lower and upper $z$ values as shown below.

  

Figure (7)

8. Press “ENTER” to obtain the result as shown below.

  

Figure (8)

The area between $z=0.797$ and $z=4$ is $0.21$ .

Therefore, about $21\%$ of batteries will last more than $125,000$ miles.

The total number of batteries are $20,000$ .

So, the number of batteries that will last more than $125,000$ miles are,

  $0.21\times 20,000=4200$

Therefore, the number of batteries that will last more than $125,000$ miles are $4200$ .

(c)

To determine

To find: The probability that if a car battery is bought at random, then it will last less than $100,000$ miles.

Answer to Problem 17PPS

About $17\%$ batteries will last less than $100,000$ miles.

Explanation of Solution

Given:

The mean is $113,627$ .

The standard deviation is $14,266$ .

The total number of batteries made by company are $20,000$ .

Calculation:

Here, find $P\left(X<100,000\right)$

Calculate the corresponding $z$ value for $X=100,000$ as follows.

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{100,000-113,627}{14,266}\\ =-0.9552\end{array}$

Calculate the required probability by using TI-83 calculator, and sketch the area between $z=-4$ and $z=-0.9552$ as follows.

9. Press “ $2^{\text{nd}}$ ” key and then “ $\text{DISTR}$” key.

  

Figure (5)

10. Go to “DRAW” and press “1.”

  

Figure (6)

11. Enter respective lower and upper $z$ values as shown below.

  

Figure (7)

12. Press “ENTER” to obtain the result as shown below.

  

Figure (8)

The area between $z=-4$ and $z=-0.9552$ is $0.17$ .

Therefore, about $17\%$ of batteries will last more than $125,000$ miles.

Therefore, about $17\%$ batteries will last less than $100,000$ miles.