Statistics for the Behavioral Sciences
Statistics for the Behavioral Sciences
2nd Edition
ISBN: 9781452286907
Author: Gregory J. Privitera
Publisher: SAGE Publications, Inc
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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.

(a)

Expert Solution
Check Mark

Answer to Problem 13CAP

The value of P(0z1.96) is 0.4750.

Explanation of Solution

Calculation:

The given z scores are 0 and 1.96.

Calculate the value of P(0z1.96).

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(0z1.96) is 0.4750.

(b)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 13CAP

The value of P(0<z<0) 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.

(c)

Expert Solution
Check Mark

Answer to Problem 13CAP

The value of P(1.50z1.50) is 0.8664.

Explanation of Solution

Calculation:

The given z scores are –1.50 and 1.50.

Calculate the value of P(1.50z1.50)=P(z1.50)P(z1.50).

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(z1.50) is,

P(z1.50)=P(z<0)+P(0<z<1.50)=0.5000+0.4332=0.9332

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(z1.50)=P(z1.50).

From the table value P(z1.50) is 0.0668.

The value of P(z1.50) is 0.0668.

P(1.50z1.50)=0.93320.0668=0.8664

Thus, the value of P(1.50z1.50) 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.

(d)

Expert Solution
Check Mark

Answer to Problem 13CAP

The value of P(0.30z0.10) is 0.0781.

Explanation of Solution

Calculation:

The given z scores are –0.30 and –0.10.

Calculate the value of P(0.30z0.10)=P(z0.10)P(z0.30).

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(z0.30)=P(z0.30).

From the table value P(z0.30) is 0.4602.

The value of P(z0.10) 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(z0.10)=P(z0.10).

From the table value P(z0.10) is 0.3821.

The value of P(z0.30) is 0.3821.

P(0.30z0.10)=0.46020.3821=0.0781

Thus, the value of P(0.30z0.10) 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.

(e)

Expert Solution
Check Mark

Answer to Problem 13CAP

The value of P(1.00z2.00) is 0.1359.

Explanation of Solution

Calculation:

The given z scores are 1.00 and 2.00.

Calculate the value of P(1.00z2.00)=P(z2.00)P(z1.00).

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(z2.00) is,

P(z2.00)=P(z<0)+P(0<z<2.00)=0.5000+0.4772=0.9772

Thus, the value of P(z2.00) 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(z1.00) is,

P(z1.00)=P(z<0)+P(0<z<1.00)=0.5000+0.3413=0.8413

Thus, the value of P(z1.00) is 0.8413.

P(1.00z2.00)=0.97720.3413=0.1359

Thus, the value of P(1.00z2.00) is 0.1359.

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