Chapter 6, Problem 14CAP

1.

To determine

Find which one is the biggest area among the two areas or check whether the two areas or equal or not.

Answer to Problem 14CAP

The area left of z score 1.00 and the area right of z score –1.00 are same.

Explanation of Solution

Calculation:

The given situations are the area to the left of z score 1.00 and another one is the area to the right of z score –1.00.

Calculate the area left of z score 1.00 is, $P\left(z\le 1.00\right)=P\left(z<0\right)+P\left(0\le z\le 1.00\right)$.

Calculate the value of $P\left(0\le z\le 1.00\right)$.

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 1.00 in column A.
• The probability value located in the column B: Area Between mean and z corresponding to 1.00 is 0.3413.

Thus, the value of $P\left(0\le z\le 1.00\right)$ is 0.3413.

$\begin{array}{c}P\left(z\le 1.00\right)=P\left(z<0\right)+P\left(0\le z\le 1.00\right)\\ =0.5000+0.3413\\ =0.8413\end{array}$

Thus, the area left of z score 1.00 is 0.8413.

Calculate the area right of z score –1.00 is, $P\left(z\ge -1.00\right)=P\left(-1.00\le z<0\right)+P\left(z>0\right)$

Calculate the value of $P\left(-1.00\le z\le 0\right)$.

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 1.00 in column A.
• The probability value located in the column B: Area Between mean and z corresponding to 1.00 is 0.3413.

Thus, the value of $P\left(-1.00\le z\le 0\right)$ is 0.3413.

$\begin{array}{c}P\left(z\ge -1.00\right)=P\left(-1.00\le z\le 0\right)+P\left(z>0\right)\\ =0.3413+0.5000\\ =0.8413\end{array}$

Thus, the area right of z score –1.00 is 0.8413.

Thus, the area left of z score 1.00 and the area right of z score –1.00 are same.

2.

To determine

Find which one is the biggest area among the two areas or check whether the two areas or equal or not.

Answer to Problem 14CAP

The area to the right of z score –1.00 higher than the area to the right of z score 1.00.

Explanation of Solution

Calculation:

The given situations are the area to the right of z score 1.00 and another one is the area to the right of z score –1.00.

Calculate the area right of z score 1.00 is, $P\left(z\ge 1.00\right)$

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 1.00 in column A.
• The probability value locate in column C: Area Beyond z in tail corresponding to 1.00 is 0.1587.

The value of $P\left(z\ge 1.00\right)$ is 0.1587.

Thus, the area right of z score 1.00 is 0.1587.

Calculate the area right of z score –1.00 is, $P\left(z\ge -1.00\right)=P\left(-1.00\le z<0\right)+P\left(z>0\right)$

Calculate the value of $P\left(-1.00\le z\le 0\right)$.

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 1.00 in column A.
• The probability value locate in the column B: Area Between mean and z corresponding to 1.00 is 0.3413.

Since the normal distribution is symmetric,

$P\left(0\le z\le 1.00\right)=P\left(-1.00\le z\le 0\right)$.

From the table value $P\left(0\le z\le 1.00\right)$ is 0.3413.

Thus, the value of $P\left(-1.00\le z\le 0\right)$ is 0.3413.

$\begin{array}{c}P\left(z\ge -1.00\right)=P\left(-1.00\le z\le 0\right)+P\left(z>0\right)\\ =0.3413+0.5000\\ =0.8413\end{array}$

Thus, the area right of z score –1.00 is 0.8413.

Thus, the area right of z score 1.00 is 0.1587 and the area right of z score –1.00 is 0.8413.

Therefore, the area to the right of z score –1.00 higher than the area to the right of z score 1.00.

3.

To determine

Find which one is the biggest area among the two areas or check whether the two areas or equal or not.

Answer to Problem 14CAP

The area between the mean and z score 1.20 higher than the area to the right of z score 0.80.

Explanation of Solution

Calculation:

The given situations are the area between the mean and z score 1.20 and another one is the area right of the z score 0.80.

The mean of the standard normal distribution is 0.

Calculate the area between the mean and z score 1.20 is, $P\left(0<z<1.20\right)$.

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 1.20 in column A.
• The probability value located in column B: Area Between mean and z corresponding to 1.20 is 0.3849.

The value of $P\left(0<z<1.20\right)$ is 0.3849.

Thus, the area between the mean and z score 1.20 is 0.3849.

Calculate the area right of z score 0.80 is, $P\left(z\ge 0.80\right)$

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 0.80 in column A.
• The probability value located in column C: Area Beyond z in tail corresponding to 0.80 is 0.2119.

Thus, the value of $P\left(z\ge 0.80\right)$ is 0.2119.

Thus, the area right of z score 0.80 is 0.2119.

Thus, the area between the mean and z score 1.20 is 0.3849 and the area right of z 0.80 is 0.2119.

Therefore, the area between the mean and z score 1.20 higher than the area to the right of z score 0.80.

4.

To determine

Find which one is the biggest area among the two areas or check whether the two areas or equal or not.

Answer to Problem 14CAP

The area left the mean is lower than the area between the z scores –1.00 and 1.00.

Explanation of Solution

Calculation:

The given situations are the area left of the mean and another one is the area between the z scores –1.00 to 1.00.

The mean of the standard normal distribution is 0.

Calculate the area left the mean is, $P\left(z<0\right)$.

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 0.00 in column A.
• The probability value locate column C: Area Beyond z in tail corresponding to 0.00 is 0.5000.

The value of $P\left(z<0\right)$ is 0.5000.

Thus, the area left of the mean is 0.5000.

Calculate the area between of z scores –1.00 and 1.00 is, $P\left(-1.00\le z\le 1.00\right)=P\left(z\le 1.00\right)-P\left(z\le -1.00\right)$

Calculate the $P\left(z\le 1.00\right)=P\left(z<0\right)+P\left(0\le z\le 1.00\right)$.

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 1.00 in column A.
• The probability value locate in column B: Area Between mean and z corresponding to 1.00 is 0.3413.

The value of $P\left(z\le 1.00\right)$ is,

$\begin{array}{c}P\left(z\le 1.00\right)=P\left(z<0\right)+P\left(0<z<1.00\right)\\ =0.5000+0.3413\\ =0.8413\end{array}$

Calculate the $P\left(z\le -1.00\right)$.

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 1.00 in column A.
• The probability value locate in column C: Area Beyond z in tail corresponding to 1.00 is 0.1587.

Since the normal distribution is symmetric,

$P\left(z\ge 1.00\right)=P\left(z\le -1.00\right)$.

From the table value $P\left(z\ge 1.00\right)$ is 0.1587.

The value of $P\left(z\le -1.00\right)$ is 0.1587.

Thus, the area between of z scores –1.00 and 1.00 is,

$\begin{array}{c}P\left(-1.00\le z\le 1.00\right)=P\left(z\le 1.00\right)-P\left(z\le -1.00\right)\\ =0.8413-0.1587\\ =0.6826\end{array}$.

Thus, the area left of the mean is 0.5000 and the area between the z scores –1.00 and 1.00 is 0.6826.

Therefore, the area left the mean is lower than the area between the z scores –1.00 and 1.00.

5.

To determine

Find which one is the biggest area among the two areas or check whether the two areas or equal or not.

Answer to Problem 14CAP

The area right of z score 1.65 and the area left of z score –1.65 are same.

Explanation of Solution

Calculation:

The given situations are the area to the right of z score 1.65 and another one is the area to the left of z score –1.65.

Calculate the area right of z score 1.65 is, $P\left(z\ge 1.65\right)$

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 1.65 in column A.
• The probability value locate in column C: Area Beyond z in tail corresponding to 1.65 is 0.0495.

Thus, the value of $P\left(z\ge 1.65\right)$ is 0.0495.

Thus, the area right of z score 1.65 is 0.0495.

Calculate the area left of z score –1.65 is, $P\left(z\le -1.65\right)$

From Appendix C: Table C.1 “The Unit Normal Table”

• Locate the z value 1.65 in column A.
• The probability value locate in column C: Area Beyond z in tail corresponding to 1.65 is 0.0495.

Since the normal distribution is symmetric,

$P\left(z\le -1.65\right)=P\left(z\ge -1.65\right)$.

From the table value $P\left(z\ge 1.65\right)$ is 0.0495.

Thus, the value of $P\left(z\le -1.65\right)$ is 0.0495.

Thus, the area right of z score 1.65 and the area left of z score –1.65 are same.