Explanation Approach: The bell-shaped normal curve is symmetric with respect to its mean (denoted as μ ). Thus, 50% of the data values lies to the left of mean and remaining 50% of the data lies to the right of the mean. In a standard bell-shaped normal curve, the weightage of data near the mean is high. Figure 1 Calculation: The z -score corresponding to mean is 0. The positive z-score lies to the right of the mean and the negative z-score lies to the left of the mean. The bell-shaped curve indicates that a greater number of data lie near the mean. Figure 2 illustrates the area under the normal curve between z = 1.3 and z = 2.0 . Figure 2 Figure 3 illustrates the area under the normal curve between z = 0 and z = 0

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ISBN: 9780134434681

Author: Tom Pirnot

Publisher: PEARSON

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Chapter 14.4, Problem 79E

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The reason why it is incorrect to subtract 1.3 from 2.0 and then use the z-score of the resultant to find the area between z=1.3 and z=2.0.

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Chapter 14.4, Problem 78E

Chapter 14.4, Problem 80E

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The reason why it is incorrect to subtract 1.3 from 2.0 and then use the z -score of the resultant to find the area between z = 1.3 and z = 2.0 .

Solution Summary: The author explains that it is incorrect to subtract 1.3 from 2.0 and then use the z -score of the resultant to find the area between

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To explain:

The reason why it is incorrect to subtract 1.3 from 2.0 and then use the z-score of the resultant to find the area between z=1.3 and z=2.0.

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Chapter 14 Solutions

The reason why it is incorrect to subtract 1.3 from 2.0 and then use the z -score of the resultant to find the area between z = 1.3 and z = 2.0 .

Solution Summary: The author explains that it is incorrect to subtract 1.3 from 2.0 and then use the z -score of the resultant to find the area between

Concept explainers

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To explain: The reason why it is incorrect to subtract 1.3 from 2.0 and then use the z-score of the resultant to find the area between z=1.3 and z=2.0.

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Chapter 14 Solutions

To explain:

The reason why it is incorrect to subtract 1.3 from 2.0 and then use the z-score of the resultant to find the area between z=1.3 and z=2.0.