In June 2003, approximately 45% of a workforce were women. In a random sample of 20 workers:
a) What is the probability that six are women?
b) What is the probability that there were 12 or more men?
Please help
Thanks in advance.
As Mr. Fantastic said, this follows a binomial distribution. For problem a we have a probability of success (p) = 0.45 where p is the probability of a person being a woman. The number of trials is 20 which we denote as (n). Finally, we want (k), where k is the number of successes. In problem a, a success is a person being a woman. So we want P(X = 6) where X denotes the binomial distribution.
Go here to my binomial lesson and plug in 20, 0.45, 6 and press the appropriate button:
Binomial Distribution
Follow the math on the green chalkboard section. I get an answer of 0.0746 or 7.46% that out of 20 people, EXACTLY 6 are women.
Problem b is similar. N = 20. But now, we want men, not women. So 1 - 0.45 = 0.65. If the probability of being a woman is 0.45 on each person chosen, then it's 1 - 0.45 = 0.65 to be a man. Now, the problem asks for P(X >= 12) This is the same as asking for P(X > 11). 12 or more men is the same as greater than 11 men.
So our k for this problem becomes 11, since we want 12 or more. Enter 11 for k and press the P(X > k) button. I get 0.7624 or 76.24. There is a lot of math on the chalkboard. It calculates P(X=0) all the way through P(X = 11). Since it is a greater than probability, it takes 1 - P(X<=11) to get our answer.
Read through the math and let me know if you have questions.