statistics help?

Question

I don't understand how to do this problem:

43% of home computers are online. Suppose 25 people with home computers were randomly sampled. Find the probability that more than 21 do not have online service.

Thanks for the help

Answer 1

So, you're trying to find the probability that probability that more than 21 do not have online service.

Let x = number of people with online service
n= number of trials (25 in this case)
p = probabily of ppl with online service (43% = 0.43)
q = 1-p

If more than 21 people do not have online service, it is the same thing to say that 21 or less people do have online service. So, we want x to be less or equal to 21 according to the question.

P(x ≤ 21) = ∑ (from x=0 to 21) of all the probabilities that x = 0, 1, 2, ... , 21


So, to find a single probability:

P(x = 0) = nCx (p^x ) (q^(n-x)) = [25!/(0!25!)] * .43^0 *.57^25

P(x = 1) = nCx (p^x ) (q^(n-x)) = [25!/(1!24!)] * .43^1 *.57^24

and so on... add all these up including P(x = 21) to get your answer.


Excel has a wonderful formula (BINOMDIST) that will calculate this for you if you can figure out how to work it...

and I get P(x ≤ 21) = 0.999995936...


This can be done a different way but you should get the same answer...

You could calculate

1 - P(X = 25) - P(X = 24) - P(X = 23) - P(X = 22)

and you should get the same answer.

Answer 2

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