Explanation Given: The exponential distribution of a = 0.5 . (a) The value of the integral ∫ 0 35 0.5 e − 0.5 x d x is 1.00000. (b) The value of the integral ∫ 0 35 0.5 x e − 0.5 x d x is 1.99999. (c) The value of the integral ∫ 0 35 0.5 x 2 e − 0.5 x d x is 7.999998. Definition used: Exponential Distribution : If X is a random variable with probability density function f ( x ) = a e − a x for x in [ 0 , ∞ ) . Then, μ = 1 a and σ = 1 a . Proof: Find the total probability of an exponential distribution is 1. Compare the given integrals with the exponential distribution f ( x ) = a e − a x . That is, the function in (b) and (c) are not exponential distribution function. The ∫ 0 35 0.5 x e − 0.5 x d x and ∫ 0 35 0.5 x 2 e − 0.5 x d x are not exponential distribution because it does not satisfies the exponential distribution function f ( x ) = a e − a x . The function f ( x ) = 0.5 e − 0.5 x satisfies the definition of exponential distribution function f ( x ) = a e − a x

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Chapter 18.3, Problem 24E

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To verify: The total probability of an exponential distribution is 1 and both the mean and the standard deviation are equal to 1a for the exponential distribution.

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Chapter 18.3, Problem 23E

Chapter 18.3, Problem 25E

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Chapter 18 Solutions

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Ch. 18.1 - Prob. 1YT Ch. 18.1 - Prob. 2YT Ch. 18.1 - Using the probability density function of Example...Ch. 18.1 - Use part (a) of Example 5 to calculate the...Ch. 18.1 - Evaluate each of the following integrals. (Sec....Ch. 18.1 - Prob. 2WE Ch. 18.1 - Prob. 3WE Ch. 18.1 - Decide whether the functions defined as follows...Ch. 18.1 - Prob. 2E Ch. 18.1 - Prob. 3E

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To verify: The total probability of an exponential distribution is 1 and both the mean and the standard deviation are equal to 1 a for the exponential distribution.

Solution Summary: The author compares the given integrals with the exponential distribution f(x)=ae-ax

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To verify: The total probability of an exponential distribution is 1 and both the mean and the standard deviation are equal to 1a for the exponential distribution.

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Chapter 18 Solutions