Explanation Given info: The breaking distances of a sedan follows normal distribution with mean 132 ft and standard deviation 4.53 ft. Calculation: The random variable x represents the breaking distances of a sedan. The breaking distance of a sedan that corresponds to 90 th percentile represents the probability value of 0.90. Use Table 4: Standard normal distribution to find z score. Procedure: Locate the area of 0.90 in the body of the Table 4. Move left until the first column and note the value as 1.2. Move upward until the top row is reached and note the value as 0.08. Thus, the z score is 1.28...

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ISBN: 9781323839577

Author: Larson

Publisher: Pearson Custom Publishing

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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 5, Problem 5.3.47RE

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To obtain: The breaking distance of a sedan that corresponds to 90^th percentile.

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Chapter 5, Problem 5.3.46RE

Chapter 5, Problem 5.3.48RE

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To obtain: The breaking distance of a sedan that corresponds to 90 th percentile.

Solution Summary: The author explains that the breaking distance of a sedan that corresponds to 90 th percentile is 137.80 ft.

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To obtain: The breaking distance of a sedan that corresponds to 90^th percentile.

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To obtain: The breaking distance of a sedan that corresponds to 90 th percentile.

Solution Summary: The author explains that the breaking distance of a sedan that corresponds to 90 th percentile is 137.80 ft.

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To obtain: The breaking distance of a sedan that corresponds to 90th percentile.

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Chapter 5 Solutions

To obtain: The breaking distance of a sedan that corresponds to 90^th percentile.