Explanation Formula used: “The number of standard deviations a measurement x is from a mean μ is given by z = x − μ σ where μ is the mean, σ is the standard deviation and x − μ is the distance between x and μ ”. Calculation: Given that the mean is − 10 , the standard deviation is 5 and the given interval is ( − ∞ , − 14 ] . The standard form of the probability density function of this normal distribution is f ( x ) = 1 σ 2 π e − ( x − μ ) 2 2 σ 2 where μ is the mean and σ is the standard deviation. Substitute μ = − 10 and σ = 5 in f ( x ) = 1 σ 2 π e − ( x − μ ) 2 2 σ 2 as, f ( x ) = 1 σ 2 π e − ( x − μ ) 2 2 σ 2 = 1 ( 5 ) 2 π e − ( x − ( − 10 ) ) 2 2 ( 5 ) 2 = 1 5 2 π e − ( x + 10 ) 2 50 Thus, the probability density function of this normal distribution becomes f ( x ) = 1 5 2 π e − ( x + 10 ) 2 50 . The area under the normal curve and f ( x ) = 1 5 2 π e − ( x + 10 ) 2 50 above the interval ( − ∞ , − 14 ] is as shown below in Figure 1

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

15th Edition

ISBN: 9781323047620

Author: Barnett, Ziegler, Byleen

Publisher: Pearson Education

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Chapter 11.4, Problem 31E

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To find: The area under the normal curve and above the interval (−∞,−14] on the horizontal axis.

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Chapter 11.4, Problem 30E

Chapter 11.4, Problem 32E

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Chapter 11 Solutions

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

Ch. 11.1 - Evaluate the following, if it converges: 3dx(x1)2.Ch. 11.1 - Prob. 2MP Ch. 11.1 - Prob. 3MP Ch. 11.1 - Prob. 4MP Ch. 11.1 - Prob. 5MP Ch. 11.1 - Prob. 6MP Ch. 11.1 - Prob. 1ED Ch. 11.1 - Prob. 2ED Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E

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The area under the normal curve and above the interval ( − ∞ , − 14 ] on the horizontal axis.

Solution Summary: The author calculates the area under the normal curve and above the interval (-infty,-14) on the horizontal axis.

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To find: The area under the normal curve and above the interval (−∞,−14] on the horizontal axis.

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Chapter 11 Solutions