Explanation Given information: f ( x ) = 1 8 2 π e − ( 1 / 2 ) [ ( x − 50 ) / 8 ] 2 From the above function, we can write μ = 50 and σ = 8 Explanation: The random variable X be the life spans (in months) of a certain product μ = 50 and σ = 8 If X be the random variable, then the normal density function is f ( x ) = 1 σ 2 π e − ( 1 / 2 ) [ ( x − μ ) / σ ] 2 To find the probability that scores were between 30 and 50 months is written as Pr ( 30 ≤ X ≤ 50 ) . To find, Pr ( 30 ≤ X ≤ 50 ) , use the formula below, Pr ( a ≤ X ≤ b ) = ∫ a b f ( x ) d x Therefore, Pr ( 30 ≤ X ≤ 50 ) = ∫ 30 50 1 8 2 π e − ( 1 / 2 ) [ ( x − 50 ) / 8 ] 2 d x To find this integral, use substitution z = x − 50 8 Take the derivative of z with respect to x d z = ( 1 8 ) d x ⇒ 8 d z = d x Substitute the values of the lower and upper limits in z = x − 50 8 . If x = 30 , then z = 30 − 50 8 = − 2.5 and if x = 50 , then z = 50 − 50 8 = 0

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Chapter 12, Problem 21RE

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The probability that the life span of such a product is between 30 and 50 months.

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Chapter 12, Problem 20RE

Chapter 12, Problem 22RE

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Chapter 12 Solutions

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The probability that the life span of such a product is between 30 and 50 months.

Solution Summary: The author explains that the probability that a product's life span is between 30 and 50 months is 0.4938.

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