I'm stuck on this probability problem

concomitance

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Sep 12, 2012
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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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Is this the answer?

I think I figured it out...

Can someone let me know if I'm correct?

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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?





= probability of 2 + probability of 1 + probability of 0


P(X=2) = (10 choose 2)
= (10! / (2!(10-2)!) ) * .22^2 * (1-.22)^8
= (10! / (2!*8!) ) * .22^2 * .78^8
= 45 * .22^2 * .78^8
= .2984


P(X=1) = (10 choose 1)
= (10! / (1!(10-1)!) ) * .22^1 * (1-.22)^9
= (10! / (1!*9!) ) * .22^1 * .78^9
= 10 * .22^2 * .78^8
= .2351


P(X=0) = (10 choose 0)
= (10! / (0!(10-0)!) ) * .22^0 * (1-.22)^10
= (10! / (0!*10!) ) * .22^0 * .78^10
= 1 * .22^0 * .78^10
= .0833


.0833 + .2351 + .2984


= .6168 chance that two or less have a Biden action figure.






3. What is the probability that more than four homes own a Biden action figure?


More than four homes = Prob of 5 + 6 + 7 + 8 + 9 + 10


P(X=5) = (10 choose 5)
= (10! / (5!(10-5)!) ) * .22^5 * (1-.22)^5
= (10! / (5!*5!) ) * .22^5 * .78^5
= 252 * .22^5 * .78^5
= .0374


P(X=6) = (10 choose 6)
= (10! / (6!(10-6)!) ) * .22^6 * (1-.22)^4
= (10! / (6!*4!) ) * .22^6 * .78^4
= 210 * .22^6 * .78^4
= .0088


P(X=7) = (10 choose 7)
= (10! / (7!(10-7)!) ) * .22^7 * (1-.22)^3
= (10! / (7!*3!) ) * .22^7 * .78^3
= 120 * .22^7 * .78^3
= .0014


P(X=8) = (10 choose 8)
= (10! / (8!(10-8)!) ) * .22^8 * (1-.22)^2
= (10! / (8!*2!) ) * .22^8 * .78^2
= 45 * .22^8 * .78^2
= 1.5023 E -4


P(X=9) = (10 choose 9)
= (10! / (9!(10-1)!) ) * .22^9 * (1-.22)^1
= (10! / (9!*1!) ) * .22^9 * .78^1
= 10 * .22^9 * .78^1
= 9.4166 E -6


P(X=10) = (10 choose 10)
= (10! / (10!(10-10)!) ) * .22^10 * (1-.22)^0
= (10! / (10!*0!) ) * .22^10 * .78^0
= 1 * .22^10 * .78^0
= 2.6559 E -7


Probs of 5 + 6 + 7 + 8 + 9 + 10
= .0477








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
 
Hey concomitance,

Sorry your professor hasn't been too helpful.

I didn't actually do out all the calculations to check the individual results, but the methods do look correct, assuming we are indeed dealing with a proper binomial distribution (I assume that's the topic you've been covering). You seem to have figured things out quite nicely in the time between your two posts!

Just as an aside, another way to approach the question about P(more than 4 homes) is to think of it as 1 - P(3 or less homes). This works because the total probability must be 1, so if we subtract the probability of 3 or less homes (which is 0, 1, 2, or 3 homes) then we'll be left with 4 or more homes. You'll get the same answer but with slightly fewer computations (and this would especially come in handy if we were dealing with, say, 100 homes instead of 10. Then we would much prefer to deal with 1 - P(3 or less) than P(4 through 100) ). If I haven't explained this clearly enough, definitely let me know. Probability can be quite confusing at times!

Hope that helps...
 
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