Explanation Given information: The provided probability is P ( 30 ≤ X ≤ 62 ) when μ = 50 and σ = 10 . Formula used: The standard form of probability P ( a ≤ X ≤ b ) when X has a normal distribution with mean μ and standard deviation σ is P ( a − μ σ ≤ Z ≤ b − μ σ ) where, Z is the standard normal variable. Calculation: Consider the probability, P ( 30 ≤ X ≤ 62 ) when μ = 50 and σ = 10 . Compare the probability P ( 30 ≤ X ≤ 62 ) with the general form P ( a ≤ X ≤ b ) , it will give a = 30 and b = 62 . Then, the probability in the standard form is, P ( 30 ≤ X ≤ 62 ) = P ( 30 − 50 10 ≤ Z ≤ 62 − 50 10 ) = P ( − 20 10 ≤ Z ≤ 12 10 ) = P ( − 2 ≤ Z ≤ 1.2 ) Simplify the probability as, P ( − 2 ≤ Z ≤ 1

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Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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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 11E

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To calculate: The value of P(30≤X≤62) when μ=50 and σ=10 where X represents the normal distribution.

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Chapter 8.5, Problem 10E

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The value of P ( 30 ≤ X ≤ 62 ) when μ = 50 and σ = 10 where X represents the normal distribution .

Solution Summary: The author explains how to calculate the value of P(30le Xl 62) when mu =50 and sigma =10.

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To calculate: The value of P(30≤X≤62) when μ=50 and σ=10 where X represents the normal distribution.

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