Explanation Definition used: The probability density function that represents for a continuous random variable X is if 1. f ( x ) ≥ 0 , for all x ∈ ( − ∞ , ∞ ) . 2. ∫ − ∞ ∞ f ( x ) = 1 . 3. The probability that X lies in the interval [ c , d ] is given by P ( c ≤ X ≤ d ) = ∫ c d f ( x ) d x . Calculation: The given function is f ( x ) = 2 2 π e − 2 ( x − 1.5 ) 2 . Use online graphing calculator to draw the function f ( x ) is shown below in Figure 1. From Figure 1, it is observed that the maximum value of the function is approximately equal to 0.7979 and occurs at the point x = 1.5 . Evaluate the integral ∫ − N N f ( x ) d x when N = 1 , 2 and 3 . Calculate the value ∫ − 1 1 2 2 π e − 2 ( x − 1.5 ) 2 d x as follows. ∫ − 1 1 2 2 π e − 2 ( x − 1.5 ) 2 d x = 2 2 π ∫ − 1 1 e − 2 ( x − 1.5 ) 2 d x Use online calculator to find the integral value of ∫ − 1 1 e − 2 ( x − 1.5 ) 2 d x . From the above screenshot, it is observed that the value of the integral ∫ − 1 1 e − 2 ( x − 1.5 ) 2 d x is approximately equal to 0.198845 . Substitute ∫ − 1 1 e − 2 ( x − 1.5 ) 2 d x ≈ 0.198845 in 2 2 π ∫ − 1 1 e − 2 ( x − 1.5 ) 2 d x as, 2 2 π ∫ − 1 1 e − 2 ( x − 1.5 ) 2 d x ≈ 2 ( 2 ) 3.1415 ( 0.198845 ) ≈ 2 6.283 ( 0.198845 ) ≈ 0.198845 2.5066 ≈ 0.15866 Thus, the integral ∫ − 1 1 2 2 π e − 2 ( x − 1.5 ) 2 d x is approximately equal to 0.15866. Calculate the value ∫ − 2 2 2 2 π e − 2 ( x − 1.5 ) 2 d x as follows. ∫ − 2 2 2 2 π e − 2 ( x − 1.5 ) 2 d x = 2 2 π ∫ − 2 2 e − 2 ( x − 1.5 ) 2 d x Use online calculator to find the integral value of ∫ − 2 2 e − 2 ( x − 1.5 ) 2 d x . From the above screenshot, it is observed that the value of the integral ∫ − 2 2 e − 2 ( x − 1.5 ) 2 d x is approximately equal to 1.05447 . Substitute ∫ − 2 2 e − 2 ( x − 1.5 ) 2 d x ≈ 1.05447 in 2 2 π ∫ − 2 2 e − 2 ( x − 1.5 ) 2 d x as, 2 2 π ∫ − 2 2 e − 0.5 x 2 d x ≈ 2 ( 2 ) 3.1415 ( 1.05447 ) ≈ 2 6.283 ( 1.05447 ) ≈ 2.10894 2.5066 ≈ 0.8414 Thus, the integral ∫ − 2 2 2 2 π e − 2 ( x − 1.5 ) 2 d x is approximately equal to 0.8414. Calculate the value ∫ − 3 3 2 2 π e − 2 ( x − 1.5 ) 2 d x as follows. ∫ − 3 3 2 2 π e − 2 ( x − 1.5 ) 2 d x = 2 2 π ∫ − 3 3 e − 2 ( x − 1.5 ) 2 d x Use online calculator to find the integral value ∫ − 3 3 e − 2 ( x − 1.5 ) 2 d x

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

14th Edition

ISBN: 9780134668574

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen, Christopher J. Stocker

Publisher: PEARSON

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Chapter 11.2, Problem 48E

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To sketch: The graph of the function f(x); also compute the integral ∫−NNf(x)dx for N=1,2 and 3 and explain the proof for f being a probability density function.

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Chapter 11.2, Problem 47E

Chapter 11.2, Problem 49E

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Chapter 11 Solutions

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

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 graph of the function f ( x ) ; also compute the integral ∫ − N N f ( x ) d x for N = 1 , 2 and 3 and explain the proof for f being a probability density function.

Solution Summary: The author illustrates the graph of the function f(x) and the proof for it being a probability density function.

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To sketch: The graph of the function f(x); also compute the integral ∫−NNf(x)dx for N=1,2 and 3 and explain the proof for f being a probability density function.

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Chapter 11 Solutions