Chapter 6.2, Problem 15E

Given that z is a standard normal random variable, find z for each situation.

1. a. The area to the left of z is .2119.
2. b. The area between –z and z is .9030.
3. c. The area between –z and z is .2052.
4. d. The area to the left of z is .9948.
5. e. The area to the right of z is .6915.

To determine

Find the value of z.

Answer to Problem 15E

The value of z is –0.80.

Explanation of Solution

Calculation:

The area to the left of z is given as 0.2119.

The area to the left of z, which is 0.2119 can be written as $P\left(Z\le z\right)=0.2119$.

z value:

Software Procedure:

Step-by-step procedure to obtain the z using the EXCEL:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =NORMSINV(0.2119) in the cell A1.
• Press Enter.

Output using the EXCEL software is given below:

Hence, $P\left(Z\le -0.80\right)=0.2119$.

Thus, the value of z is –0.80.

To determine

Obtain the value of z.

Answer to Problem 15E

The value of z is 1.66.

Explanation of Solution

Calculation:

The area between –z and z is given as 0.9030.

Thus, the area between –z and z, which is 0.9030 can be written as,

$\begin{array}{c}P\left(-z\le Z\le z\right)=0.9030\\ P\left(Z\le z\right)-P\left(Z\le -z\right)=0.9030\\ P\left(Z\le z\right)-\left[1-P\left(Z\le z\right)\right]=0.9030\\ 2P\left(Z\le z\right)-1=0.9030\\ P\left(Z\le z\right)=\frac{1+0.9030}{2}\\ P\left(Z\le z\right)=\frac{1.903}{2}\\ P\left(Z\le z\right)=0.9515\end{array}$

z value:

Software Procedure:

Step-by-step procedure to obtain the z using the EXCEL:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =NORMSINV(0.9515) in the cell A1.
• Press Enter.

Output using the EXCEL software is given below:

Hence, $P\left(Z\le 1.66\right)=0.9515$.

Thus, value of z is 1.66.

To determine

Compute the value of z.

Answer to Problem 15E

The value of z is 0.26.

Explanation of Solution

Calculation:

The area between –z and z is given as 0.2052.

Thus, the area between –z and z, which is 0.2052 can be written as,

$\begin{array}{c}P\left(-z\le Z\le z\right)=0.2052\\ P\left(Z\le z\right)-P\left(Z\le -z\right)=0.2052\\ P\left(Z\le z\right)-\left[1-P\left(Z\le z\right)\right]=0.2052\\ 2P\left(Z\le z\right)-1=0.2052\\ P\left(Z\le z\right)=\frac{1+0.2052}{2}\\ P\left(Z\le z\right)=\frac{1.2052}{2}\\ P\left(Z\le z\right)=0.6026\end{array}$

z value:

Software Procedure:

Step-by-step procedure to obtain the z using the EXCEL:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =NORMSINV(0.6026) in the cell A1.
• Press Enter.

Output using the EXCEL software is given below:

Hence, $P\left(Z\le 0.26\right)=0.6026$.

Thus, value of z is 0.26.

To determine

Calculate the value of z.

Answer to Problem 15E

The value of z is 2.56.

Explanation of Solution

Calculation:

The area to the left of z is given as 0.9948.

The area to the left of z, which is 0.9948 can be written as $P\left(Z\le z\right)=0.9948$.

z value:

Software Procedure:

Step-by-step procedure to obtain the z using the EXCEL:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =NORMSINV(0.9948) in the cell A1.
• Press Enter.

Output using the EXCEL software is given below:

Hence, $P\left(Z\le 2.56\right)=0.9948$.

Thus, the value of z is 2.56.

To determine

Compute the value of z.

Answer to Problem 15E

The value of z is –0.50.

Explanation of Solution

Calculation:

The area to the right of z is given as 0.6915.

Thus, the area to the right of z, which is 0.6915, can be written as,

$\begin{array}{c}P\left(Z\ge z\right)=0.6915\\ 1-P\left(Z\le z\right)=0.6915\\ P\left(Z\le z\right)=1-0.6915\\ P\left(Z\le z\right)=0.3085\end{array}$

z value:

Software Procedure:

Step-by-step procedure to obtain the z using the EXCEL:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =NORMSINV(0.3085) in the cell A1.
• Press Enter.

Output using the EXCEL software is given below:

Hence, $P\left(Z\le -0.50\right)=0.3085$.

Thus, value of z is –0.50.