Chapter 6, Problem 13CAP

(a)

To determine

Find the proportion value under the standard normal curve lies to the between the mean and the z score 1.96.

Answer to Problem 13CAP

The value of $P\left(0\le z\le 1.96\right)$ is 0.4750.

Explanation of Solution

Calculation:

The given z scores are 0 and 1.96.

Calculate the value of $P\left(0\le z\le 1.96\right)$.

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

• Locate the z value 1.96 in column A.
• The probability value locate in the column B: Area Between mean and z corresponding to 1.96 is 0.4750.

Thus, the value of $P\left(0\le z\le 1.96\right)$ is 0.4750.

(b)

To determine

Find the proportion value under the standard normal curve lies between the z score mean and 0.

Answer to Problem 13CAP

The value of $P\left(0<z<0\right)$ is 0.000.

Explanation of Solution

Calculation:

The given z scores are mean and 0. The mean value of the standard normal distribution is 0.

Thus, the area between the value mean and the value 0 is 0.000.

(c)

To determine

Find the proportion value under the standard normal curve lies to the between the z scores –1.50 and 1.50.

Answer to Problem 13CAP

The value of $P\left(-1.50\le z\le 1.50\right)$ is 0.8664.

Explanation of Solution

Calculation:

The given z scores are –1.50 and 1.50.

Calculate the value of $P\left(-1.50\le z\le 1.50\right)=P\left(z\le 1.50\right)-P\left(z\le -1.50\right)$.

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

• Locate the z value 1.50 in column A.
• The probability value locate in the column B: Area Between mean and z corresponding to 1.50 is 0.4750.

The value of $P\left(z\le 1.50\right)$ is,

$\begin{array}{c}P\left(z\le 1.50\right)=P\left(z<0\right)+P\left(0<z<1.50\right)\\ =0.5000+0.4332\\ =0.9332\end{array}$

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

• Locate the z value 1.50 in column A.
• The probability value locate in the column C: Area Beyond z in tail corresponding to 1.50 is 0.0668.

Since the normal distribution is symmetric,

$P\left(z\le -1.50\right)=P\left(z\ge 1.50\right)$.

From the table value $P\left(z\ge 1.50\right)$ is 0.0668.

The value of $P\left(z\le -1.50\right)$ is 0.0668.

$\begin{array}{c}P\left(-1.50\le z\le 1.50\right)=0.9332-0.0668\\ =0.8664\end{array}$

Thus, the value of $P\left(-1.50\le z\le 1.50\right)$ is 0.8664.

(d)

To determine

Find the proportion value under the standard normal curve lies to the between the z scores –0.30 and –0.10.

Answer to Problem 13CAP

The value of $P\left(-0.30\le z\le -0.10\right)$ is 0.0781.

Explanation of Solution

Calculation:

The given z scores are –0.30 and –0.10.

Calculate the value of $P\left(-0.30\le z\le -0.10\right)=P\left(z\le -0.10\right)-P\left(z\le -0.30\right)$.

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

• Locate the z value 0.10 in column A.
• The probability value locate in the column C: Area Beyond z in tail corresponding to 0.10 is 0.4602.

Since the normal distribution is symmetric,

$P\left(z\le -0.30\right)=P\left(z\ge 0.30\right)$.

From the table value $P\left(z\ge 0.30\right)$ is 0.4602.

The value of $P\left(z\le -0.10\right)$ is 0.4602.

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

• Locate the z value 0.30 in column A.
• The probability value locate in the column C: Area Beyond z in tail corresponding to 0.30 is 0.3821.

Since the normal distribution is symmetric,

$P\left(z\le -0.10\right)=P\left(z\ge 0.10\right)$.

From the table value $P\left(z\ge 0.10\right)$ is 0.3821.

The value of $P\left(z\le -0.30\right)$ is 0.3821.

$\begin{array}{c}P\left(-0.30\le z\le -0.10\right)=0.4602-0.3821\\ =0.0781\end{array}$

Thus, the value of $P\left(-0.30\le z\le -0.10\right)$ is 0.0781.

(e)

To determine

Find the proportion value under the standard normal curve lies to the between the z scores 1.00 and 2.00.

Answer to Problem 13CAP

The value of $P\left(1.00\le z\le 2.00\right)$ is 0.1359.

Explanation of Solution

Calculation:

The given z scores are 1.00 and 2.00.

Calculate the value of $P\left(1.00\le z\le 2.00\right)=P\left(z\le 2.00\right)-P\left(z\le 1.00\right)$.

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

• Locate the z value 2.00 in column A.
• The probability value locate in column B: Area Between mean and z corresponding to 2.00 is 0.4772.

The value of $P\left(z\le 2.00\right)$ is,

$\begin{array}{c}P\left(z\le 2.00\right)=P\left(z<0\right)+P\left(0<z<2.00\right)\\ =0.5000+0.4772\\ =0.9772\end{array}$

Thus, the value of $P\left(z\le 2.00\right)$ is 0.9772.

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,

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

Thus, the value of $P\left(z\le 1.00\right)$ is 0.8413.

$\begin{array}{c}P\left(1.00\le z\le 2.00\right)=0.9772-0.3413\\ =0.1359\end{array}$

Thus, the value of $P\left(1.00\le z\le 2.00\right)$ is 0.1359.