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trclem40 · Apr 30, 2008, 07:48 PM

One medical textbook states that only 6% of th human population has blood type O-Negative. Assume that a simple random sample of 45 people are selected.

A. Find the probability that 5 of the selected people have blood type O-Negative. Would this be considered unusual. Why or why not?

B. Find the probability that at least 4 of the selected people have blood type O- Negative. Would this be considered unusual . Why or why not?

ebaines · May 1, 2008, 02:17 PM

For A: The probability that n people in a population of 45 have this blood type is $C_N ^{45}\cdot (.06)^n(1-.06)^{45-n}$ . For n = 5: $\frac {45*44*43*42*41} { 5*4*3*2*1} (0.06)^5 (.94)^{40}$ .

For B: P(at least 4) = 1 - P(0) - P(1) - P(2) - P(3).

Hope this helps.

iamthetman · May 2, 2008, 08:14 AM

To determine whether it's unusual, look at how many people have it vs. how many people you would expect to have it.

For A: 5 people actually have that blood type
You'd expect 45 x 6% of people to have it, meaning you'd expect 2.7 people to have it out of your group of 45 people.

Hope this helps.

ebaines · May 2, 2008, 08:52 AM

The best way to determine whether 5 is an unusual number to have for this blood type out of a sample size of 45 people is to determine how many standard deviations 5 is from the mean. As IATM points out, the mean is 2.7. The formula for standard deviation of a binomial distribution like this is:

$<br /> Variance = \sigma ^2 = np(1-p)<br />$

where for this problem n = 45 and p = 0.06. So here the variance is 45(/06)(.94)= 2.54, and the standard deviation is $\sigma = \sqrt 2.54 = 1.6$ . So having 5 people with this blood type is (5-2.7)/1.6 = 1.44 standard deviations from the mean. So now you need to decide whether a Z-score of 1.44 is "unusual" or not. Hope this helps.