Chapter 11.5, Problem 16PPS

(a)

To determine

To find: The probability for $P\left(X<8\right)$ , and then sketch the area under curve.

Answer to Problem 16PPS

The required probability is $0.3445$ , and graph is shown in figure (4).

Explanation of Solution

Given:

The mean is $8.2$ .

The standard deviation is $0.3$ .

Calculation:

Calculate the corresponding $z$ value for $X=8$ as follows.

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{8-8.2}{0.3}\\ =-0.667\end{array}$

Calculate the required probability by using TI-83 calculator, and sketch the area between $z=-4$ and $z=-0.667$ 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 probability is $0.2523$ .

Therefore, the required probability is $0.2523$ , and graph is shown in figure (4).

(b)

To determine

To find: The probability for $P\left(X>7.5\right)$ , and then sketch the area under curve.

Answer to Problem 16PPS

The required probability is $0.990129$ , and graph is shown in figure (8).

Explanation of Solution

Given:

The mean is $8.2$ .

The standard deviation is $0.3$ .

Calculation:

Calculate the corresponding $z$ value for $X=7.5$ as follows.

  $\begin{array}{c}z=\frac{X-\mu }{\sigma }\\ =\frac{7.5-8.2}{0.3}\\ =-2.33\end{array}$

Calculate the required probability by using TI-83 calculator, and sketch the area between $z=-4$ and $z=-2.33$ 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 the $z=-4$ and $z=-2.33$ is $0.0098$ .

Find the required probability.

  $\begin{array}{c}P\left(X>7.5\right)=1-P\left(X\le 7.5\right)\\ =1-0.009871\\ =0.990129\end{array}$

Therefore, the required probability is $0.990129$ , and graph is shown in figure (8).