This is a written question, worth 14 points. DO NOT place the problem code on th

Question

This is a written question, worth 14 points. DO NOT place the problem code on the answer sheet. A proctor will fill this out after exam submission. Show all steps (work) on your answer sheet for full credit. Problem Code: 944 Suppose that in a large metropolitan area, 75% of all households have a flat screen television. Suppose you are interested in selecting a group of six households from this area. Let X be the number of households in a group of six households from this area that have a flat screen television. Show that this problem satisfies the requirements to be a binomial distribution. For what proportion of groups will exactly four of the six households have a flat screen television? For what proportion of groups will at most two of the households have a flat screen television? Part d: What is the expected number of households that have a flat screen television?

Explanation / Answer

Requirements of binomial distribution:

1. Must have n trials - there are 6 trials( 6 houselhold ) , each not effecting other
2.must have 2 outcomes - haas tv, doesn't have tv
3. Events must be independent - yes they are
4.Probability doesn't change from trial to trial - remains .75

a.There are 6 households, thus 6 trials

b. P = 6C4*(.75^4)(.25^2) = 15*(.75^4) *.(25^3) = .0742

c.P(X>=2)

=6C2*(.75^2)(.25^4)+6C3*(.75^3)(.25^3)+6C4*(.75^4)(.25^2)+6C5*(.75^5)(.25^1)+6C6*(.75^6)(.25^0)

= 0.9953

d. E(X) = np

= 6*.75

= 4.5

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