Use Table 4, Appendix 3, to find the following probabilities for a standard normal random variable Z: a P (0 ≤ Z ≤ 1 . 2) b P (−.9 ≤ Z ≤ 0) c P (.3 ≤ Z ≤ 1 . 56) d P (−.2 ≤ Z ≤ .2) e P (−1 . 56 ≤ Z ≤ −.2) f Applet Exercise Use the applet Normal Probabilities to obtain P (0 ≤ Z ≤ 1.2). Why are the values given on the two horizontal axes identical?

Question

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Answer 1

Textbook Question

Chapter 4.5, Problem 58E

Use Table 4, Appendix 3, to find the following probabilities for a standard normal random variable Z:

a P(0 ≤ Z ≤ 1.2)

b P(−.9 ≤ Z ≤ 0)

c P(.3 ≤ Z ≤ 1.56)

d P(−.2 ≤ Z ≤ .2)

e P(−1.56 ≤ Z ≤ −.2)

f Applet Exercise Use the applet Normal Probabilities to obtain P(0 ≤ Z ≤ 1.2). Why are the values given on the two horizontal axes identical?

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.

a.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2).

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

It is given that Z is normally distributed with mean 0 and standard deviation 1.

The following can be observed for Z:

P(Z<0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z>0)=P(Z<0).

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.1151.

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1P(0≤Z≤1.2)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1151=0.3849

Thus, the value of P(0≤Z≤1.2) is 0.3849.

b.

Expert Solution

To determine

Find the value of P(−0.9≤Z≤0).

Answer to Problem 58E

The value of P(−0.9≤Z≤0) is 0.3159.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.9)=P(Z>0.9).

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.9 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.1841.

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1P(−0.9≤Z≤0)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1841=0.3159

Thus, the value of P(−0.9≤Z≤0) is 0.3159.

c.

Expert Solution

To determine

Find the value of P(0.3≤Z≤1.56).

Answer to Problem 58E

The value of P(0.3≤Z≤1.56) is 0.3227.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=11−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.3 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.3821.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

1−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1P(0.3≤Z≤1.56)=1−1+P(Z>0.3)−P(Z>1.56)=P(Z>0.3)−P(Z>1.56)

=A(0.3)−A(1.56)=0.3821−0.0594=0.3227

Thus, The value of P(−0.9≤Z≤0) is 0.3227.

d.

Expert Solution

To determine

Find the value of P(−0.2≤Z≤0.2).

Answer to Problem 58E

The value of P(−0.2≤Z≤0.2) is 0.1586.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.2)=P(Z>0.2).

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1P(−0.2≤Z≤0.2)=1−2×P(Z>0.2)=1−2×A(0.2)

=1−2(0.4207)=1−0.8414=0.1586

Thus, the value of P(−0.2≤Z≤0.2) is 0.1586.

e.

Expert Solution

To determine

Find the value of P(−1.56≤Z≤−0.2).

Answer to Problem 58E

The value of P(−1.56≤Z≤−0.2) is 0.3613.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−1.56)+P(−1.56≤Z≤−0.2)+P(Z>−0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−1.56)=P(Z>1.56). Again, P(Z>−0.2)=1−P(Z<−0.2)=1−P(Z>0.2).

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1P(−1.56≤Z≤−0.2)=P(Z>0.2)−P(Z>1.56)=A(0.2)−A(1.56)=0.4207−0.0594=0.3613

Thus, The value of P(−1.56≤Z≤−0.2) is 0.3613.

f.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2) by using Applet.

Explain the values given on the two horizontal axes identical or not.

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

The following can be observed for Z:

Step-by-step procedure to obtain the probability value using Applets as follows:

Enter Mean = 0 and StDev = 1 values in the provided boxes.
Enter Start = 0 and End = 1.2 value in the provided boxes.

Output obtained using Applets software is represented as follows:

Mathematical Statistics with Applications, Chapter 4.5, Problem 58E

From the above output, the value of P(0≤Z≤1.2) is 0.3849.

Thus, the value of P(0≤Z≤1.2) is 0.3849.

Explanation:

The values given on the two horizontal axes are identical. It is because, in the above output the top Z axes represent the standard normal variable and they y axes represents the variable of interest. In the given situation the standard normal variable is only used. Hence as a result the both axes provide the same Z value of P(0≤Z≤1.2) as 0.3849.

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Answer 2

Textbook Question

Chapter 4.5, Problem 58E

Use Table 4, Appendix 3, to find the following probabilities for a standard normal random variable Z:

a P(0 ≤ Z ≤ 1.2)

b P(−.9 ≤ Z ≤ 0)

c P(.3 ≤ Z ≤ 1.56)

d P(−.2 ≤ Z ≤ .2)

e P(−1.56 ≤ Z ≤ −.2)

f Applet Exercise Use the applet Normal Probabilities to obtain P(0 ≤ Z ≤ 1.2). Why are the values given on the two horizontal axes identical?

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.

a.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2).

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

It is given that Z is normally distributed with mean 0 and standard deviation 1.

The following can be observed for Z:

P(Z<0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z>0)=P(Z<0).

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.1151.

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1P(0≤Z≤1.2)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1151=0.3849

Thus, the value of P(0≤Z≤1.2) is 0.3849.

b.

Expert Solution

To determine

Find the value of P(−0.9≤Z≤0).

Answer to Problem 58E

The value of P(−0.9≤Z≤0) is 0.3159.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.9)=P(Z>0.9).

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.9 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.1841.

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1P(−0.9≤Z≤0)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1841=0.3159

Thus, the value of P(−0.9≤Z≤0) is 0.3159.

c.

Expert Solution

To determine

Find the value of P(0.3≤Z≤1.56).

Answer to Problem 58E

The value of P(0.3≤Z≤1.56) is 0.3227.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=11−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.3 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.3821.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

1−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1P(0.3≤Z≤1.56)=1−1+P(Z>0.3)−P(Z>1.56)=P(Z>0.3)−P(Z>1.56)

=A(0.3)−A(1.56)=0.3821−0.0594=0.3227

Thus, The value of P(−0.9≤Z≤0) is 0.3227.

d.

Expert Solution

To determine

Find the value of P(−0.2≤Z≤0.2).

Answer to Problem 58E

The value of P(−0.2≤Z≤0.2) is 0.1586.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.2)=P(Z>0.2).

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1P(−0.2≤Z≤0.2)=1−2×P(Z>0.2)=1−2×A(0.2)

=1−2(0.4207)=1−0.8414=0.1586

Thus, the value of P(−0.2≤Z≤0.2) is 0.1586.

e.

Expert Solution

To determine

Find the value of P(−1.56≤Z≤−0.2).

Answer to Problem 58E

The value of P(−1.56≤Z≤−0.2) is 0.3613.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−1.56)+P(−1.56≤Z≤−0.2)+P(Z>−0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−1.56)=P(Z>1.56). Again, P(Z>−0.2)=1−P(Z<−0.2)=1−P(Z>0.2).

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1P(−1.56≤Z≤−0.2)=P(Z>0.2)−P(Z>1.56)=A(0.2)−A(1.56)=0.4207−0.0594=0.3613

Thus, The value of P(−1.56≤Z≤−0.2) is 0.3613.

f.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2) by using Applet.

Explain the values given on the two horizontal axes identical or not.

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

The following can be observed for Z:

Step-by-step procedure to obtain the probability value using Applets as follows:

Enter Mean = 0 and StDev = 1 values in the provided boxes.
Enter Start = 0 and End = 1.2 value in the provided boxes.

Output obtained using Applets software is represented as follows:

Mathematical Statistics with Applications, Chapter 4.5, Problem 58E

From the above output, the value of P(0≤Z≤1.2) is 0.3849.

Thus, the value of P(0≤Z≤1.2) is 0.3849.

Explanation:

The values given on the two horizontal axes are identical. It is because, in the above output the top Z axes represent the standard normal variable and they y axes represents the variable of interest. In the given situation the standard normal variable is only used. Hence as a result the both axes provide the same Z value of P(0≤Z≤1.2) as 0.3849.

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Answer 3

Textbook Question

Chapter 4.5, Problem 58E

Use Table 4, Appendix 3, to find the following probabilities for a standard normal random variable Z:

a P(0 ≤ Z ≤ 1.2)

b P(−.9 ≤ Z ≤ 0)

c P(.3 ≤ Z ≤ 1.56)

d P(−.2 ≤ Z ≤ .2)

e P(−1.56 ≤ Z ≤ −.2)

f Applet Exercise Use the applet Normal Probabilities to obtain P(0 ≤ Z ≤ 1.2). Why are the values given on the two horizontal axes identical?

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.

a.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2).

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

It is given that Z is normally distributed with mean 0 and standard deviation 1.

The following can be observed for Z:

P(Z<0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z>0)=P(Z<0).

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.1151.

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1P(0≤Z≤1.2)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1151=0.3849

Thus, the value of P(0≤Z≤1.2) is 0.3849.

b.

Expert Solution

To determine

Find the value of P(−0.9≤Z≤0).

Answer to Problem 58E

The value of P(−0.9≤Z≤0) is 0.3159.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.9)=P(Z>0.9).

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.9 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.1841.

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1P(−0.9≤Z≤0)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1841=0.3159

Thus, the value of P(−0.9≤Z≤0) is 0.3159.

c.

Expert Solution

To determine

Find the value of P(0.3≤Z≤1.56).

Answer to Problem 58E

The value of P(0.3≤Z≤1.56) is 0.3227.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=11−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.3 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.3821.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

1−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1P(0.3≤Z≤1.56)=1−1+P(Z>0.3)−P(Z>1.56)=P(Z>0.3)−P(Z>1.56)

=A(0.3)−A(1.56)=0.3821−0.0594=0.3227

Thus, The value of P(−0.9≤Z≤0) is 0.3227.

d.

Expert Solution

To determine

Find the value of P(−0.2≤Z≤0.2).

Answer to Problem 58E

The value of P(−0.2≤Z≤0.2) is 0.1586.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.2)=P(Z>0.2).

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1P(−0.2≤Z≤0.2)=1−2×P(Z>0.2)=1−2×A(0.2)

=1−2(0.4207)=1−0.8414=0.1586

Thus, the value of P(−0.2≤Z≤0.2) is 0.1586.

e.

Expert Solution

To determine

Find the value of P(−1.56≤Z≤−0.2).

Answer to Problem 58E

The value of P(−1.56≤Z≤−0.2) is 0.3613.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−1.56)+P(−1.56≤Z≤−0.2)+P(Z>−0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−1.56)=P(Z>1.56). Again, P(Z>−0.2)=1−P(Z<−0.2)=1−P(Z>0.2).

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1P(−1.56≤Z≤−0.2)=P(Z>0.2)−P(Z>1.56)=A(0.2)−A(1.56)=0.4207−0.0594=0.3613

Thus, The value of P(−1.56≤Z≤−0.2) is 0.3613.

f.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2) by using Applet.

Explain the values given on the two horizontal axes identical or not.

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

The following can be observed for Z:

Step-by-step procedure to obtain the probability value using Applets as follows:

Enter Mean = 0 and StDev = 1 values in the provided boxes.
Enter Start = 0 and End = 1.2 value in the provided boxes.

Output obtained using Applets software is represented as follows:

Mathematical Statistics with Applications, Chapter 4.5, Problem 58E

From the above output, the value of P(0≤Z≤1.2) is 0.3849.

Thus, the value of P(0≤Z≤1.2) is 0.3849.

Explanation:

The values given on the two horizontal axes are identical. It is because, in the above output the top Z axes represent the standard normal variable and they y axes represents the variable of interest. In the given situation the standard normal variable is only used. Hence as a result the both axes provide the same Z value of P(0≤Z≤1.2) as 0.3849.

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Answer 4

Textbook Question

Chapter 4.5, Problem 58E

Use Table 4, Appendix 3, to find the following probabilities for a standard normal random variable Z:

a P(0 ≤ Z ≤ 1.2)

b P(−.9 ≤ Z ≤ 0)

c P(.3 ≤ Z ≤ 1.56)

d P(−.2 ≤ Z ≤ .2)

e P(−1.56 ≤ Z ≤ −.2)

f Applet Exercise Use the applet Normal Probabilities to obtain P(0 ≤ Z ≤ 1.2). Why are the values given on the two horizontal axes identical?

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.

a.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2).

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

It is given that Z is normally distributed with mean 0 and standard deviation 1.

The following can be observed for Z:

P(Z<0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z>0)=P(Z<0).

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.1151.

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1P(0≤Z≤1.2)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1151=0.3849

Thus, the value of P(0≤Z≤1.2) is 0.3849.

b.

Expert Solution

To determine

Find the value of P(−0.9≤Z≤0).

Answer to Problem 58E

The value of P(−0.9≤Z≤0) is 0.3159.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.9)=P(Z>0.9).

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.9 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.1841.

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1P(−0.9≤Z≤0)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1841=0.3159

Thus, the value of P(−0.9≤Z≤0) is 0.3159.

c.

Expert Solution

To determine

Find the value of P(0.3≤Z≤1.56).

Answer to Problem 58E

The value of P(0.3≤Z≤1.56) is 0.3227.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=11−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.3 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.3821.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

1−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1P(0.3≤Z≤1.56)=1−1+P(Z>0.3)−P(Z>1.56)=P(Z>0.3)−P(Z>1.56)

=A(0.3)−A(1.56)=0.3821−0.0594=0.3227

Thus, The value of P(−0.9≤Z≤0) is 0.3227.

d.

Expert Solution

To determine

Find the value of P(−0.2≤Z≤0.2).

Answer to Problem 58E

The value of P(−0.2≤Z≤0.2) is 0.1586.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.2)=P(Z>0.2).

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1P(−0.2≤Z≤0.2)=1−2×P(Z>0.2)=1−2×A(0.2)

=1−2(0.4207)=1−0.8414=0.1586

Thus, the value of P(−0.2≤Z≤0.2) is 0.1586.

e.

Expert Solution

To determine

Find the value of P(−1.56≤Z≤−0.2).

Answer to Problem 58E

The value of P(−1.56≤Z≤−0.2) is 0.3613.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−1.56)+P(−1.56≤Z≤−0.2)+P(Z>−0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−1.56)=P(Z>1.56). Again, P(Z>−0.2)=1−P(Z<−0.2)=1−P(Z>0.2).

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1P(−1.56≤Z≤−0.2)=P(Z>0.2)−P(Z>1.56)=A(0.2)−A(1.56)=0.4207−0.0594=0.3613

Thus, The value of P(−1.56≤Z≤−0.2) is 0.3613.

f.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2) by using Applet.

Explain the values given on the two horizontal axes identical or not.

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

The following can be observed for Z:

Step-by-step procedure to obtain the probability value using Applets as follows:

Enter Mean = 0 and StDev = 1 values in the provided boxes.
Enter Start = 0 and End = 1.2 value in the provided boxes.

Output obtained using Applets software is represented as follows:

Mathematical Statistics with Applications, Chapter 4.5, Problem 58E

From the above output, the value of P(0≤Z≤1.2) is 0.3849.

Thus, the value of P(0≤Z≤1.2) is 0.3849.

Explanation:

The values given on the two horizontal axes are identical. It is because, in the above output the top Z axes represent the standard normal variable and they y axes represents the variable of interest. In the given situation the standard normal variable is only used. Hence as a result the both axes provide the same Z value of P(0≤Z≤1.2) as 0.3849.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2).

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

It is given that Z is normally distributed with mean 0 and standard deviation 1.

The following can be observed for Z:

P(Z<0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z>0)=P(Z<0).

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.1151.

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1P(0≤Z≤1.2)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1151=0.3849

Thus, the value of P(0≤Z≤1.2) is 0.3849.

b.

Expert Solution

To determine

Find the value of P(−0.9≤Z≤0).

Answer to Problem 58E

The value of P(−0.9≤Z≤0) is 0.3159.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.9)=P(Z>0.9).

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.9 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.1841.

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1P(−0.9≤Z≤0)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1841=0.3159

Thus, the value of P(−0.9≤Z≤0) is 0.3159.

c.

Expert Solution

To determine

Find the value of P(0.3≤Z≤1.56).

Answer to Problem 58E

The value of P(0.3≤Z≤1.56) is 0.3227.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=11−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.3 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.3821.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

1−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1P(0.3≤Z≤1.56)=1−1+P(Z>0.3)−P(Z>1.56)=P(Z>0.3)−P(Z>1.56)

=A(0.3)−A(1.56)=0.3821−0.0594=0.3227

Thus, The value of P(−0.9≤Z≤0) is 0.3227.

d.

Expert Solution

To determine

Find the value of P(−0.2≤Z≤0.2).

Answer to Problem 58E

The value of P(−0.2≤Z≤0.2) is 0.1586.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.2)=P(Z>0.2).

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1P(−0.2≤Z≤0.2)=1−2×P(Z>0.2)=1−2×A(0.2)

=1−2(0.4207)=1−0.8414=0.1586

Thus, the value of P(−0.2≤Z≤0.2) is 0.1586.

e.

Expert Solution

To determine

Find the value of P(−1.56≤Z≤−0.2).

Answer to Problem 58E

The value of P(−1.56≤Z≤−0.2) is 0.3613.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−1.56)+P(−1.56≤Z≤−0.2)+P(Z>−0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−1.56)=P(Z>1.56). Again, P(Z>−0.2)=1−P(Z<−0.2)=1−P(Z>0.2).

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1P(−1.56≤Z≤−0.2)=P(Z>0.2)−P(Z>1.56)=A(0.2)−A(1.56)=0.4207−0.0594=0.3613

Thus, The value of P(−1.56≤Z≤−0.2) is 0.3613.

f.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2) by using Applet.

Explain the values given on the two horizontal axes identical or not.

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

The following can be observed for Z:

Step-by-step procedure to obtain the probability value using Applets as follows:

Enter Mean = 0 and StDev = 1 values in the provided boxes.
Enter Start = 0 and End = 1.2 value in the provided boxes.

Output obtained using Applets software is represented as follows:

Mathematical Statistics with Applications, Chapter 4.5, Problem 58E

From the above output, the value of P(0≤Z≤1.2) is 0.3849.

Thus, the value of P(0≤Z≤1.2) is 0.3849.

Explanation:

The values given on the two horizontal axes are identical. It is because, in the above output the top Z axes represent the standard normal variable and they y axes represents the variable of interest. In the given situation the standard normal variable is only used. Hence as a result the both axes provide the same Z value of P(0≤Z≤1.2) as 0.3849.

Answer 8

a.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2).

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

It is given that Z is normally distributed with mean 0 and standard deviation 1.

The following can be observed for Z:

P(Z<0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z>0)=P(Z<0).

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.1151.

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1P(0≤Z≤1.2)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1151=0.3849

Thus, the value of P(0≤Z≤1.2) is 0.3849.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Find the value of P(0≤Z≤1.2).

Answer 13

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Answer 14

Explanation of Solution

Calculation:

It is given that Z is normally distributed with mean 0 and standard deviation 1.

The following can be observed for Z:

P(Z<0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z>0)=P(Z<0).

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.1151.

P(Z>0)+P(0≤Z≤1.2)+P(Z>1.2)=1P(0≤Z≤1.2)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1151=0.3849

Thus, the value of P(0≤Z≤1.2) is 0.3849.

Answer 15

b.

Expert Solution

To determine

Find the value of P(−0.9≤Z≤0).

Answer to Problem 58E

The value of P(−0.9≤Z≤0) is 0.3159.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.9)=P(Z>0.9).

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.9 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.1841.

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1P(−0.9≤Z≤0)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1841=0.3159

Thus, the value of P(−0.9≤Z≤0) is 0.3159.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Find the value of P(−0.9≤Z≤0).

Answer 20

Answer to Problem 58E

The value of P(−0.9≤Z≤0) is 0.3159.

Answer 21

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.9)=P(Z>0.9).

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.0 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.5000.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.9 along the z column.
Locate 0 along the Second decimal place of z.
The intersection of row and column gives the probability value of 0.1841.

P(Z>0.9)+P(−0.9≤Z≤0)+P(Z>0)=1P(−0.9≤Z≤0)=1−P(Z>0.0)−P(Z>1.2)=1−A(0.0)−A(1.2)=1−0.5000−0.1841=0.3159

Thus, the value of P(−0.9≤Z≤0) is 0.3159.

Answer 22

c.

Expert Solution

To determine

Find the value of P(0.3≤Z≤1.56).

Answer to Problem 58E

The value of P(0.3≤Z≤1.56) is 0.3227.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=11−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.3 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.3821.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

1−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1P(0.3≤Z≤1.56)=1−1+P(Z>0.3)−P(Z>1.56)=P(Z>0.3)−P(Z>1.56)

=A(0.3)−A(1.56)=0.3821−0.0594=0.3227

Thus, The value of P(−0.9≤Z≤0) is 0.3227.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Find the value of P(0.3≤Z≤1.56).

Answer 27

Answer to Problem 58E

The value of P(0.3≤Z≤1.56) is 0.3227.

Answer 28

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=11−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.3 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.3821.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

1−P(Z>0.3)+P(0.3≤Z≤1.56)+P(Z>1.56)=1P(0.3≤Z≤1.56)=1−1+P(Z>0.3)−P(Z>1.56)=P(Z>0.3)−P(Z>1.56)

=A(0.3)−A(1.56)=0.3821−0.0594=0.3227

Thus, The value of P(−0.9≤Z≤0) is 0.3227.

Answer 29

d.

Expert Solution

To determine

Find the value of P(−0.2≤Z≤0.2).

Answer to Problem 58E

The value of P(−0.2≤Z≤0.2) is 0.1586.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.2)=P(Z>0.2).

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1P(−0.2≤Z≤0.2)=1−2×P(Z>0.2)=1−2×A(0.2)

=1−2(0.4207)=1−0.8414=0.1586

Thus, the value of P(−0.2≤Z≤0.2) is 0.1586.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

Find the value of P(−0.2≤Z≤0.2).

Answer 34

Answer to Problem 58E

The value of P(−0.2≤Z≤0.2) is 0.1586.

Answer 35

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−0.2)=P(Z>0.2).

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

P(Z>0.2)+P(−0.2≤Z≤0.2)+P(Z>0.2)=1P(−0.2≤Z≤0.2)=1−2×P(Z>0.2)=1−2×A(0.2)

=1−2(0.4207)=1−0.8414=0.1586

Thus, the value of P(−0.2≤Z≤0.2) is 0.1586.

Answer 36

e.

Expert Solution

To determine

Find the value of P(−1.56≤Z≤−0.2).

Answer to Problem 58E

The value of P(−1.56≤Z≤−0.2) is 0.3613.

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−1.56)+P(−1.56≤Z≤−0.2)+P(Z>−0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−1.56)=P(Z>1.56). Again, P(Z>−0.2)=1−P(Z<−0.2)=1−P(Z>0.2).

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1P(−1.56≤Z≤−0.2)=P(Z>0.2)−P(Z>1.56)=A(0.2)−A(1.56)=0.4207−0.0594=0.3613

Thus, The value of P(−1.56≤Z≤−0.2) is 0.3613.

Answer 37

e.

Expert Solution

Answer 38

e.

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

Find the value of P(−1.56≤Z≤−0.2).

Answer 41

Answer to Problem 58E

The value of P(−1.56≤Z≤−0.2) is 0.3613.

Answer 42

Explanation of Solution

Calculation:

The following can be observed for Z:

P(Z<−1.56)+P(−1.56≤Z≤−0.2)+P(Z>−0.2)=1

Since, the standard normal distribution is symmetric about 0, P(Z<−1.56)=P(Z>1.56). Again, P(Z>−0.2)=1−P(Z<−0.2)=1−P(Z>0.2).

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 0.2 along the z column
Locate 0 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.4207.

Use Table 4: Normal Curve Areas Standard normal probability in right-hand tail to obtain the probability as follows:

Locate 1.5 along the z column
Locate 6 along the Second decimal place of z
The intersection of row and column gives the probability value of 0.0594.

P(Z>1.56)+P(−1.56≤Z≤−0.2)+1−P(Z>0.2)=1P(−1.56≤Z≤−0.2)=P(Z>0.2)−P(Z>1.56)=A(0.2)−A(1.56)=0.4207−0.0594=0.3613

Thus, The value of P(−1.56≤Z≤−0.2) is 0.3613.

Answer 43

f.

Expert Solution

To determine

Find the value of P(0≤Z≤1.2) by using Applet.

Explain the values given on the two horizontal axes identical or not.

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Explanation of Solution

Calculation:

The following can be observed for Z:

Step-by-step procedure to obtain the probability value using Applets as follows:

Enter Mean = 0 and StDev = 1 values in the provided boxes.
Enter Start = 0 and End = 1.2 value in the provided boxes.

Output obtained using Applets software is represented as follows:

Mathematical Statistics with Applications, Chapter 4.5, Problem 58E

From the above output, the value of P(0≤Z≤1.2) is 0.3849.

Thus, the value of P(0≤Z≤1.2) is 0.3849.

Explanation:

The values given on the two horizontal axes are identical. It is because, in the above output the top Z axes represent the standard normal variable and they y axes represents the variable of interest. In the given situation the standard normal variable is only used. Hence as a result the both axes provide the same Z value of P(0≤Z≤1.2) as 0.3849.

Answer 44

f.

Expert Solution

Answer 45

f.

Expert Solution

Answer 46

Expert Solution

Answer 47

To determine

Find the value of P(0≤Z≤1.2) by using Applet.

Explain the values given on the two horizontal axes identical or not.

Answer 48

Answer to Problem 58E

The value of P(0≤Z≤1.2) is 0.3849.

Answer 49

Explanation of Solution

Calculation:

The following can be observed for Z:

Step-by-step procedure to obtain the probability value using Applets as follows:

Enter Mean = 0 and StDev = 1 values in the provided boxes.
Enter Start = 0 and End = 1.2 value in the provided boxes.

Output obtained using Applets software is represented as follows:

Mathematical Statistics with Applications, Chapter 4.5, Problem 58E

From the above output, the value of P(0≤Z≤1.2) is 0.3849.

Thus, the value of P(0≤Z≤1.2) is 0.3849.

Explanation:

The values given on the two horizontal axes are identical. It is because, in the above output the top Z axes represent the standard normal variable and they y axes represents the variable of interest. In the given situation the standard normal variable is only used. Hence as a result the both axes provide the same Z value of P(0≤Z≤1.2) as 0.3849.