(A) Explanation Given: The mean of the normal distribution is 100 and the standard deviation is 10. Theorem used: Z-scores Theorem : Suppose a normal distribution has mean μ and standard deviation σ . The area under the associated normal curve that is to the left of the value x is exactly the same as the area to the left of z = x − μ σ for the standard normal curve. Calculation: The mean is μ = 100 and the standard deviation σ = 10 . The applicant’s scores on a screening test is 92. Substitute x = 92 , μ = 100 and σ = 10 in z = x − μ σ , z = 92 − 100 10 = − 8 10 = − 0.80 Thus, the z -score is − 0.80 . The applicant’s scores on a screening test is 108 (B) To determine To find: The approximate percentage of applicants scores 115 or higher.

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ISBN: 9780134856667

Author: Stocker

Publisher: VST

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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 11, Problem 43RE

(A)

To determine

To find: The approximate percentage of applicants who scored between 92 and 108.

(B)

To determine

To find: The approximate percentage of applicants scores 115 or higher.

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Chapter 11, Problem 42RE

Chapter 11, Problem 44RE

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The approximate percentage of applicants who scored between 92 and 108.

Solution Summary: The approximate percentage of applicants who scored between 92 and 108 is 57.62%.