Explanation Approach: Assume a normal distribution has a mean of μ and a standard deviation of σ . We use the equation z = x − μ σ to convert a value x in the nonstandard distribution to a z -score. Given Information: The mean is 3.6, standard deviation is 1.2 and z = − 1 .3 . Calculation: Substitute μ = 3.6 , σ = 1.2 and z = − 1.3 in z = x − μ σ

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ISBN: 9780134434681

Author: Tom Pirnot

Publisher: PEARSON

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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 14.4, Problem 60E

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To find:

The corresponding raw score if mean is 3.6, standard deviation is 1.2 and z=−1.3.

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Chapter 14.4, Problem 59E

Chapter 14.4, Problem 61E

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Chapter 14 Solutions

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The corresponding raw score if mean is 3.6, standard deviation is 1.2 and z = − 1 .3 .

Solution Summary: The author explains how the equation z=x-musigma converts a value x in the nonstandard distribution to the corresponding raw score.

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To find:

The corresponding raw score if mean is 3.6, standard deviation is 1.2 and z=−1.3.

Want to see the full answer?

Chapter 14 Solutions

The corresponding raw score if mean is 3.6, standard deviation is 1.2 and z = − 1 .3 .

Solution Summary: The author explains how the equation z=x-musigma converts a value x in the nonstandard distribution to the corresponding raw score.

Concept explainers

Videos

To find: The corresponding raw score if mean is 3.6, standard deviation is 1.2 and z=−1.3.

Want to see the full answer?

Chapter 14 Solutions

To find:

The corresponding raw score if mean is 3.6, standard deviation is 1.2 and z=−1.3.