SOLUTION: Many U.S. households still do not have Internet access. Suppose 25 out of 80 households in a small southern town do not have Internet access. A company that provides high-speed Int

Question 1171891: Many U.S. households still do not have Internet access. Suppose 25 out of 80 households in a small southern town do not have Internet access. A company that provides high-speed Internet has recently entered the market. As part of the marketing campaign, the company decides to randomly select ten households and offer them red laptops along with a brochure that describes their services. The aim is to build goodwill and, with a free laptop, tempt nonusers into getting Internet access.
a. What is the probability that six laptop recipients do not have Internet access?
(2 Marks)
b. What is the probability that at least five recipients do not have Internet access?
(2 Marks)
c. What is the probability that two or fewer laptop recipients do not have Internet access?
(2 Marks)
d. What is the expected number of laptop recipients who do not have Internet access?
(1 Marks)
e. Calculate the expected value, the variance, and the standard deviation for the recipients who do not have Internet access.

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Note: I am assuming the 80 households is the entire population and the sampling of one item changes the distribution significantly for the next item. We have a fixed number of houses sampled (10), but not a constant probability for each one, with the way the problem is worded. This is different from saying 25 of every 80 households. This is a hypergeometric distribution in other words.
80C10 are the number of ways to choose 10 from 80. That is the denominator
The numerator for a is 25C6*55C4
The probability is 0.0367
-
at least 5 do not have access. Look at 7 do not have:0.0077 using the same approach. And for 8 0.0001
9 and 10 are <0.0001, 5 do not have access is 0.1123. The sum (using 6 above as well) is 0.1568, the answer.
-
probability of 2 not having is 25C2*55C8/80C10 or 0.2218
of 1 it is 25C1*55C9/80C10=0.0965
of 0 it is 55C10/80C10=0.0178
That sum is 0.3361
------------
To do the expected value, one needs the probability * the random variable.
9 and 10 are both 0
3 and 4 are 25C3*55C7/80C10=0.2835
and p(4)=0.2228
so
0*p(0)=0 as well as 9*p(9) and 10 p(10)
p(1)=0.0965 and *1=0.0965
p(2)=0.2218 and *2=0.4436
p(3)=0.2835 and *3=0.8505
p(4)=0.2228 and *4=0.8912
p(5)=0.1123 and *5=0.5615
p(6)=0.0367 and *6=0.0220
p(7)=0.0077 and *7=0.0539
p(8)=0.0001 and *8=0.0008
That sum is 2.92 and that is the expected value.
The variance and the sd (sqrt (V) ) can be done from that.
Note: by using a binomial with a larger pool, the expected value is 3.125, but that doesn't apply given the above assumptions.
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