a. Explanation Given info: The random variable x is a hypergeometric random variable with N = 10 , n = 5 , r = 7 . Calculation: Hypergeometric distribution: A random variable x is said to follow a Hypergeometric distribution, if the probability distribution is P ( x ) = ( r x ) ( N − r n − x ) ( N n ) Where, x = Maximum [ 0 , n − ( N − r ) ] , .... , Minimum ( r , n ) = Number of S ' s drawn in the n elements N = Total number of elements . r = Number of S ' s in the N elements . n = Number of elements drawn . As X follows hypergeometric distribution with N = 10 , n = 5 , r = 7 , thus X can take only values 2, 3, 4 or 5. Thus, for x = 2 , p ( 2 ) = ( 7 2 ) ( 10 − 7 5 − 2 ) ( 10 5 ) = 7 ! 2 ! 5 ! 3 ! 3 ! 0 ! 10 ! 5 ! 5 ! = 21 × 1 252 = 0 b. To determine To compute: The mean and variance of X . c. To determine To graph: p ( x ) and locate μ and μ ± 2 σ on the graph. d. To determine To find: The probability that X will fall in the interval μ ± 2 σ .

Statistics for Business and Economics (13th Edition)

13th Edition

ISBN: 9780134506593

Author: James T. McClave, P. George Benson, Terry Sincich

Publisher: PEARSON

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Chapter 4.4, Problem 4.68LM

To determine

To display: The probability distribution of x in tabular form.

To determine

To compute: The mean and variance of X.

To determine

To graph: p(x) and locate μ and μ±2σ on the graph.

To determine

To find: The probability that X will fall in the interval μ±2σ .

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Chapter 4.4, Problem 4.67LM

Chapter 4.4, Problem 4.69ACB

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To display: The probability distribution of x in tabular form.

Solution Summary: The author explains that the probability distribution of x is a hypergeometric random variable with N=10,n=5, r=7.

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To display: The probability distribution of x in tabular form.

To compute: The mean and variance of X.

To graph: p(x) and locate μ and μ±2σ on the graph.

To find: The probability that X will fall in the interval μ±2σ .

Want to see the full answer?

Chapter 4 Solutions