a. Explanation Given information: The average September rainfall is a random variable X with mean μ = 102 mm . The amount of rainfall is normally distributed having standard deviation σ = 48. Formula used: If X has a normal distribution with mean μ and standard deviation σ , then for all a ≤ b , P ( X ≤ b ) = F ( b − μ σ ) P ( a ≤ X ≤ b ) = F ( b − μ σ ) − F ( a − μ σ ) The standard cumulative distributive function is F ( z ) = 1 2 π ∫ − ∞ z e − x 2 2 d x Calculation: μ = 102 mm and σ = 48 b. To determine To express: P as an integral of an appropriate density function and also calculate its value numerically.

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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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 8.8, Problem 20E

To determine

To express: P(128≤X≤150) in terms of F(z) and also calculate its value.

To determine

To express: P as an integral of an appropriate density function and also calculate its value numerically.

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Chapter 8.8, Problem 19E

Chapter 8.8, Problem 21E

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To express : P ( 128 ≤ X ≤ 150 ) in terms of F ( z ) and also calculate its value.

Solution Summary: The author calculates the average September rainfall in terms of F(z) and the standard cumulative distributive function.