Chapter 11.5, Problem 18PPS

(a)

To determine

To find: The percent of the product lasts between $150$ , and $200$ days.

Answer to Problem 18PPS

About $71\%$ of the product lasts between 150 and 200 days.

Explanation of Solution

Given:

The mean is $173.3$ days.

The standard deviation is $23.6$ days.

Calculation:

Here find $P\left(150<X<200\right)$ .

Calculate the corresponding $z$ value for $X=150$ as follows.

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{150-173.3}{23.6}\\ =-987\end{array}$

Calculate the corresponding $z$ value for $X=200$ as follows.

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{200-173.3}{23.6}\\ =1.13\end{array}$

Calculate the required probability by using TI-83 calculator, and sketch the area between $z=-0.987$ and $z=1.13$ 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 $z=-0.987$ and $z=1.13$ is $0.71$ .

Therefore, about $71\%$ of the product lasts between 150 and 200 days.

(b)

To determine

To find: The percent of the product lasts more than $225$ days.

Answer to Problem 18PPS

About $1.4\%$ of the product lasts more than 225 days.

Explanation of Solution

Given:

The mean is $173.3$ days.

The standard deviation is $23.6$ days.

Calculation:

Here find $P\left(X>225\right)$ .

Calculate the corresponding $z$ value for $X=225$ as follows.

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{225-173.3}{23.6}\\ =2.19\end{array}$

Calculate the required probability by using TI-83 calculator, and sketch the area between $z=2.19$ 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 $z=2.19$ and $z=4$ is $0.014$ .

Therefore, about $1.4\%$ of the product lasts more than 225 days.

(c)

To determine

To find: The range of values represents the outside $5\%$ of the distribution.

Answer to Problem 18PPS

The range value that represents the outside $5\%$ of the distribution is $X<126.1$ , and $X>220.5$ .

Explanation of Solution

Given:

The mean is $173.3$ days.

The standard deviation is $23.6$ days.

Calculation:

The range will be $\mu +2\sigma$ to $\mu -2\sigma$ .

  $\begin{array}{c}\mu +2\sigma =173.3+2\left(23.6\right)\\ =220.5\\ \mu -2\sigma =173.3-2\left(23.6\right)\\ =126.1\end{array}$

Therefore, the range value that represents the outside $5\%$ of the distribution is $X<126.1$ , and $X>220.5$ .