HammondLA5 - probabilities using the formula on page 208...

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1. P (0) =.001 P (1) =.014 P (2) =.073 P (3) =.021 P (4) =.328 P (5) =.279 P (6) =.099 2. P (4) has the highest probability at .328 3. What is the probability of getting more than 3 families to say that their children have an influence on their vacation destinations? Explain and show how you get your answer. P(x>3)= P(4)+P(5)+P(6)=.328+.279+.099=.706 4. What is the probability that no more than 3 families say that their children have an influence on their vacation destinations? Explain and show how you get your answer. P ( x ≤3 )= 1-.706=.294 5. Find the mean, variance, and standard deviation of this binomial distribution. Explain and show all steps to get your answers. n=6 p=.68 µ=np = 6(.68)=4.08 variance = npq= 4.08(.32)=1.3056 standard deviation= 1.1426 Extra Credit: You can receive 10% extra credit on this assignment if you show how to compute the binomial
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Unformatted text preview: probabilities using the formula on page 208 and as demostrated in the example 5 on page 211 3. N=6, P=.68, q=.32 and x=4,5,6 > = ! - ! !. . ( - )= . . =. Px 3 6 6 4 4 684 32 6 4 15∙ 2138∙ 1024 328 > = ! - ! !. . ( - )= . . =. Px 3 6 6 5 5 685 32 6 5 6∙ 1454∙ 32 279 > = ! - ! !. . ( - )= . =. Px 3 6 6 6 6 686 32 6 6 1∙ 0989∙1 099 > =. +. +. =. Px 3 328 279 099 706 N=6, P=.68, q=.32 and x=0, 1, 2, 3 ≤ = ! - ! !. . ( - )= . =. Px 3 6 6 0 0 680 32 6 0 1∙1∙ 0011 011 ≤ = ! - ! !. . ( - )= . . =. Px 3 6 6 1 1 681 32 6 1 6∙ 68∙ 0034 013 ≤ = ! - ! !. . ( - )= . . . =. Px 3 6 6 2 2 682 32 6 2 1 25∙ 4624∙ 0104 073 ≤ = ! - ! !. . ( - )= . . =. Px 3 6 6 3 3 683 32 6 3 20∙ 3144∙ 0328 021 ≤ =. +. +. +. = Px 3 001 013 073 021...
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This note was uploaded on 02/03/2011 for the course ACCT 1100 taught by Professor Matter during the Spring '08 term at Metropolitan Community College- Omaha.

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HammondLA5 - probabilities using the formula on page 208...

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