Explanation Given information: The provided probability is P ( X ≥ a ) = 0.03 when μ = 10 and σ = 5 . Formula used: The standard form of probability P ( X ≥ a ) when X has a normal distribution with mean μ and standard deviation σ is P ( Z ≥ a − μ σ ) where, Z is the standard normal variable. Calculation: Consider the probability, P ( X ≥ a ) = 0.03 when μ = 10 and σ = 5 . Then, the probability in the standard form is, P ( X ≥ a ) = P ( Z ≥ a − 10 5 ) As the value of P ( Z ≥ a − 10 5 ) is obtained by the subtracting the value of P ( 0 ≤ Z ≤ a − 10 5 ) from 0.5 . So, P ( X ≥ a ) = P ( Z ≥ a − 10 5 ) = 0

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Author: Stefan Waner, Steven Costenoble

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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 8.5, Problem 22E

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To calculate: The value of a such that P(X≥a)=0.03 when μ=10 and σ=5 where X is a normal random variable.

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Chapter 8.5, Problem 21E

Chapter 8.5, Problem 23E

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Chapter 8 Solutions

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The value of a such that P ( X ≥ a ) = 0.03 when μ = 10 and σ = 5 where X is a normal random variable.

Solution Summary: The author calculates that the value of P(Xge a)=0.03 when mu =10 and Sigma =5 is 19.4.

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To calculate: The value of a such that P(X≥a)=0.03 when μ=10 and σ=5 where X is a normal random variable.

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Chapter 8 Solutions