Posted by Anonymous on .
Data shows that 88% of the people in a certain population are right-handed. A group of 7 people from this population are selected at random.
(a) What would be the expected value for the number of right-handed people in the group?
(b) What is the probability that exactly 5 of the people in the group are right-handed?
(c) What is the probability that at least 5 of the people in the group are right-handed?
(a) Take sample size times .88 for expected value.
(b) Use a binomial probability table. Values are: n = 7, x = 5, and p = .88
(c) Use the table again. Find x = 5,6,7. Use same values for n and p. Add together for total probability.
I hope this will help get you started.