Permutations, probability and standard normals

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Question from Katrina, a student:

I have a few problems that i seem to be stuck on or can not start. Can you please help me ?

1) There are 20 people on an event planning committee. How many differnt ways can a chairperson and assistant to the chairperson be selected?

2) An unprepared student makes random guesses for 10 true or false questions on a quiz. Find the probability that the student passes the quiz by guessing 7 of the questions correctly.

P(7 problems correct)= 7-P(3 problems wrong)= 7-(1/2)^10
Thats as far as i got because i kept getting a different answer.

3) The heights of 18 year old men are normally distributed with a mean of 68 inches and a standard deviation of 3 inches. If a random sample of 25 18-year old men is selected what is the probability that the mean height is between 68.5 and 72 inches?

mean:
standard deviation:3 inches
n: 25

Thats as far as i got with this one my brain decided to go blank

Hi Katrina,

1) A chairperson can be selected in 20 different ways, each of the 20 people could be selected. Once a chairperson has been selected there are 19 ways to choose an assistant. Thus there are 20 * 19 ways to select a chairperson and an assistant.

2) This is a binomial probability question with n = 10, p = 1/2 and X the number of questions the student gets correct. You are asked to find Pr(X > 7)

3) You are given a normal random variable X with mean 68 inches and standard deviation 3 inches. From a random sample of size n = 25 you are asked to find Pr(68.5 < X-bar < 72). X-bar is also a normal random variable with mean 68 inches but its standard deviation is 3/root(25). Convert the probability statement to a statement concerning the standard normal distribution and look up the probabilities in the standard normal table.

I hope this helps,
Penny

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