Find the value of z 0 such that P ( Z > z 0 ) = 0.5

Question

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Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 4.5, Problem 59E

a.

To determine

Find the value of z₀ such that P(Z>z0)=0.5

a.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z>z0)=0.5 is 0.0000.

Explanation of Solution

Calculation:

The value of z₀ such that P(Z>z0)= 0.5 is obtained below:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z>z0)=A(z0)A(z0)=A(0.50)

From Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas), the following is obtained:

Locate the area 0.5000 From Table 4: Appendix 3.
Move left until the first column and note the values as 0.0.
Move upward until the top row is reached and note the values as 0.00.

The value of z₀ such that P(Z>z0)= 0.5 is 0.000.

b.

To determine

Find the value of z₀ such that P(Z<z0)=0.8643

b.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z<z0)=0.86431−P(Z≥z0)=0.8643

Here, the following is obtained:

P(Z>z0)=A(z0)1−A(z0)=0.8643A(z0)=0.1357

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.8643 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.1
Move upward until the top row is reached and note the value as 0.0.

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

c.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.90.

c.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.90P(Z>z0)=A(z0)1−2A(z0)=0.90A(z0)=0.05

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.05 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.6
Move upward until the top row is reached and note the value as 0.04 and 0.05

The required value is as follows:

z0=1.64+1.652=1.645

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

d.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.99

d.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.99P(Z>z0)=A(z0)1−2A(z0)=0.99A(z0)=0.005

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.005 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 2.5
Move upward until the top row is reached and note the value as 0.07 and 0.08

The required value is as follows:

z0=2.57+2.582=2.575

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

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Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 4.5, Problem 59E

a.

To determine

Find the value of z₀ such that P(Z>z0)=0.5

a.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z>z0)=0.5 is 0.0000.

Explanation of Solution

Calculation:

The value of z₀ such that P(Z>z0)= 0.5 is obtained below:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z>z0)=A(z0)A(z0)=A(0.50)

From Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas), the following is obtained:

Locate the area 0.5000 From Table 4: Appendix 3.
Move left until the first column and note the values as 0.0.
Move upward until the top row is reached and note the values as 0.00.

The value of z₀ such that P(Z>z0)= 0.5 is 0.000.

b.

To determine

Find the value of z₀ such that P(Z<z0)=0.8643

b.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z<z0)=0.86431−P(Z≥z0)=0.8643

Here, the following is obtained:

P(Z>z0)=A(z0)1−A(z0)=0.8643A(z0)=0.1357

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.8643 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.1
Move upward until the top row is reached and note the value as 0.0.

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

c.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.90.

c.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.90P(Z>z0)=A(z0)1−2A(z0)=0.90A(z0)=0.05

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.05 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.6
Move upward until the top row is reached and note the value as 0.04 and 0.05

The required value is as follows:

z0=1.64+1.652=1.645

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

d.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.99

d.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.99P(Z>z0)=A(z0)1−2A(z0)=0.99A(z0)=0.005

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.005 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 2.5
Move upward until the top row is reached and note the value as 0.07 and 0.08

The required value is as follows:

z0=2.57+2.582=2.575

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

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Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 4.5, Problem 59E

a.

To determine

Find the value of z₀ such that P(Z>z0)=0.5

a.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z>z0)=0.5 is 0.0000.

Explanation of Solution

Calculation:

The value of z₀ such that P(Z>z0)= 0.5 is obtained below:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z>z0)=A(z0)A(z0)=A(0.50)

From Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas), the following is obtained:

Locate the area 0.5000 From Table 4: Appendix 3.
Move left until the first column and note the values as 0.0.
Move upward until the top row is reached and note the values as 0.00.

The value of z₀ such that P(Z>z0)= 0.5 is 0.000.

b.

To determine

Find the value of z₀ such that P(Z<z0)=0.8643

b.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z<z0)=0.86431−P(Z≥z0)=0.8643

Here, the following is obtained:

P(Z>z0)=A(z0)1−A(z0)=0.8643A(z0)=0.1357

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.8643 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.1
Move upward until the top row is reached and note the value as 0.0.

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

c.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.90.

c.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.90P(Z>z0)=A(z0)1−2A(z0)=0.90A(z0)=0.05

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.05 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.6
Move upward until the top row is reached and note the value as 0.04 and 0.05

The required value is as follows:

z0=1.64+1.652=1.645

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

d.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.99

d.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.99P(Z>z0)=A(z0)1−2A(z0)=0.99A(z0)=0.005

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.005 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 2.5
Move upward until the top row is reached and note the value as 0.07 and 0.08

The required value is as follows:

z0=2.57+2.582=2.575

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

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Answer 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 4.5, Problem 59E

a.

To determine

Find the value of z₀ such that P(Z>z0)=0.5

a.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z>z0)=0.5 is 0.0000.

Explanation of Solution

Calculation:

The value of z₀ such that P(Z>z0)= 0.5 is obtained below:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z>z0)=A(z0)A(z0)=A(0.50)

From Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas), the following is obtained:

Locate the area 0.5000 From Table 4: Appendix 3.
Move left until the first column and note the values as 0.0.
Move upward until the top row is reached and note the values as 0.00.

The value of z₀ such that P(Z>z0)= 0.5 is 0.000.

b.

To determine

Find the value of z₀ such that P(Z<z0)=0.8643

b.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z<z0)=0.86431−P(Z≥z0)=0.8643

Here, the following is obtained:

P(Z>z0)=A(z0)1−A(z0)=0.8643A(z0)=0.1357

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.8643 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.1
Move upward until the top row is reached and note the value as 0.0.

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

c.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.90.

c.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.90P(Z>z0)=A(z0)1−2A(z0)=0.90A(z0)=0.05

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.05 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.6
Move upward until the top row is reached and note the value as 0.04 and 0.05

The required value is as follows:

z0=1.64+1.652=1.645

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

d.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.99

d.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.99P(Z>z0)=A(z0)1−2A(z0)=0.99A(z0)=0.005

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.005 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 2.5
Move upward until the top row is reached and note the value as 0.07 and 0.08

The required value is as follows:

z0=2.57+2.582=2.575

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

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Answer 5

Chapter 4.5, Problem 59E

a.

To determine

Find the value of z₀ such that P(Z>z0)=0.5

a.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z>z0)=0.5 is 0.0000.

Explanation of Solution

Calculation:

The value of z₀ such that P(Z>z0)= 0.5 is obtained below:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z>z0)=A(z0)A(z0)=A(0.50)

From Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas), the following is obtained:

Locate the area 0.5000 From Table 4: Appendix 3.
Move left until the first column and note the values as 0.0.
Move upward until the top row is reached and note the values as 0.00.

The value of z₀ such that P(Z>z0)= 0.5 is 0.000.

b.

To determine

Find the value of z₀ such that P(Z<z0)=0.8643

b.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z<z0)=0.86431−P(Z≥z0)=0.8643

Here, the following is obtained:

P(Z>z0)=A(z0)1−A(z0)=0.8643A(z0)=0.1357

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.8643 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.1
Move upward until the top row is reached and note the value as 0.0.

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

c.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.90.

c.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.90P(Z>z0)=A(z0)1−2A(z0)=0.90A(z0)=0.05

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.05 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.6
Move upward until the top row is reached and note the value as 0.04 and 0.05

The required value is as follows:

z0=1.64+1.652=1.645

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

d.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.99

d.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.99P(Z>z0)=A(z0)1−2A(z0)=0.99A(z0)=0.005

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.005 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 2.5
Move upward until the top row is reached and note the value as 0.07 and 0.08

The required value is as follows:

z0=2.57+2.582=2.575

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

Answer 6

Chapter 4.5, Problem 59E

a.

To determine

Find the value of z₀ such that P(Z>z0)=0.5

a.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z>z0)=0.5 is 0.0000.

Explanation of Solution

Calculation:

The value of z₀ such that P(Z>z0)= 0.5 is obtained below:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z>z0)=A(z0)A(z0)=A(0.50)

From Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas), the following is obtained:

Locate the area 0.5000 From Table 4: Appendix 3.
Move left until the first column and note the values as 0.0.
Move upward until the top row is reached and note the values as 0.00.

The value of z₀ such that P(Z>z0)= 0.5 is 0.000.

Answer 7

Chapter 4.5, Problem 59E

a.

To determine

Find the value of z₀ such that P(Z>z0)=0.5

Answer 8

a.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z>z0)=0.5 is 0.0000.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

The value of z₀ such that P(Z>z0)= 0.5 is obtained below:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z>z0)=A(z0)A(z0)=A(0.50)

From Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas), the following is obtained:

Locate the area 0.5000 From Table 4: Appendix 3.
Move left until the first column and note the values as 0.0.
Move upward until the top row is reached and note the values as 0.00.

The value of z₀ such that P(Z>z0)= 0.5 is 0.000.

Answer 13

b.

To determine

Find the value of z₀ such that P(Z<z0)=0.8643

b.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z<z0)=0.86431−P(Z≥z0)=0.8643

Here, the following is obtained:

P(Z>z0)=A(z0)1−A(z0)=0.8643A(z0)=0.1357

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.8643 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.1
Move upward until the top row is reached and note the value as 0.0.

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

Answer 14

b.

To determine

Find the value of z₀ such that P(Z<z0)=0.8643

Answer 15

b.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z<z0)=0.86431−P(Z≥z0)=0.8643

Here, the following is obtained:

P(Z>z0)=A(z0)1−A(z0)=0.8643A(z0)=0.1357

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.8643 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.1
Move upward until the top row is reached and note the value as 0.0.

The value of z₀ such that P(Z<z0)=0.8643 is 1.10.

Answer 20

c.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.90.

c.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.90P(Z>z0)=A(z0)1−2A(z0)=0.90A(z0)=0.05

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.05 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.6
Move upward until the top row is reached and note the value as 0.04 and 0.05

The required value is as follows:

z0=1.64+1.652=1.645

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

Answer 21

c.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.90.

Answer 22

c.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.90P(Z>z0)=A(z0)1−2A(z0)=0.90A(z0)=0.05

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.05 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 1.6
Move upward until the top row is reached and note the value as 0.04 and 0.05

The required value is as follows:

z0=1.64+1.652=1.645

The value of z₀ such that P(−z0<Z<z0)=0.90 is 1.645.

Answer 27

d.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.99

d.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.99P(Z>z0)=A(z0)1−2A(z0)=0.99A(z0)=0.005

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.005 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 2.5
Move upward until the top row is reached and note the value as 0.07 and 0.08

The required value is as follows:

z0=2.57+2.582=2.575

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

Answer 28

d.

To determine

Find the value of z₀ such that P(−z0<Z<z0)=0.99

Answer 29

d.

Expert Solution

Answer to Problem 59E

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Calculation:

In general, the standard normal distribution has μ=0 and σ=1.

P(Z≤z0)+P(−z0<Z<z0)+P(Z≥z0)=1

Here,

P(−z0<Z<z0)+2P(Z≥z0)=1P(−z0<Z<z0)=1−2P(Z≥z0)

P(−z0<Z<z0)=0.99P(Z>z0)=A(z0)1−2A(z0)=0.99A(z0)=0.005

Use Table 4: Appendix 3, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the following:

Locate the area 0.005 in Table 4: Appendix 3.
Move left until the first column is reached and note the value as 2.5
Move upward until the top row is reached and note the value as 0.07 and 0.08

The required value is as follows:

z0=2.57+2.582=2.575

The value of z₀ such that P(−z0<Z<z0)=0.99 is 2.575.

Answer 34

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Find the value of z 0 such that P ( Z > z 0 ) = 0.5

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Find the value of z₀ such that P(Z>z0)=0.5

Answer to Problem 59E

Explanation of Solution

Find the value of z₀ such that P(Z<z0)=0.8643

Answer to Problem 59E

Explanation of Solution

Find the value of z₀ such that P(−z0<Z<z0)=0.90.

Answer to Problem 59E

Explanation of Solution

Find the value of z₀ such that P(−z0<Z<z0)=0.99

Answer to Problem 59E

Explanation of Solution

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Find the value of z 0 such that P ( Z &gt; z 0 ) = 0.5

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