concomitance
New member
- Joined
- Sep 12, 2012
- Messages
- 2
I've already solved 3/5 of the problems. I'm not even asking for the answer; if someone could just show me which equations to use to solve these, I would be grateful. It is an online course. I can't figure out 2 and 3. For 2, do I just add the probabilities of 2, 1, and 0 occurring? And for 3, do I add all probabilities over 4? Or do I have to use the Z factor?
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In a recent study, twenty-two percent of the homes in the United States were found to own a Joe Biden action figure. In a sample of ten homes:
1. What is the probability that exactly four homes own a Biden action figure?
P(X=4) = (10 choose 4)
= (10! / (4!(10-4)!) ) * .22^4 * (1-.22)^6
= (10! / (4!*6!) ) * .22^4 * .78^6
= 210 * .22^4 * .78^6
= .110784
2. What is the probability that two or less homes own a Biden action figure?
3. What is the probability that more than four homes own a Biden action figure?
4. What is the expected value of the probability distribution?
μ = np
μ = 10(.22)
μ = 2.2
5. What is the standard deviation of the probability distribution?
σ = √(npq)
σ = √ (np(1-p))
σ = √ (10 * .22 * .78)
σ = √ 1.716
σ = 1.3099
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In a recent study, twenty-two percent of the homes in the United States were found to own a Joe Biden action figure. In a sample of ten homes:
1. What is the probability that exactly four homes own a Biden action figure?
P(X=4) = (10 choose 4)
= (10! / (4!(10-4)!) ) * .22^4 * (1-.22)^6
= (10! / (4!*6!) ) * .22^4 * .78^6
= 210 * .22^4 * .78^6
= .110784
2. What is the probability that two or less homes own a Biden action figure?
3. What is the probability that more than four homes own a Biden action figure?
4. What is the expected value of the probability distribution?
μ = np
μ = 10(.22)
μ = 2.2
5. What is the standard deviation of the probability distribution?
σ = √(npq)
σ = √ (np(1-p))
σ = √ (10 * .22 * .78)
σ = √ 1.716
σ = 1.3099
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