Problem 1 in the book a guide to the development and use of...

Problem 1: In the book "A Guide to the Development and Use of the Myers-Briggs Type Indicators" by Myers and McCaully, it was reported that approximately 45% of all university professors are extroverted. Suppose you have classes with six different professors. (a) What is the probability that all six are extroverts? (b) What is the probability that none of your professors is an extrovert? (c) What is the probability that at least two of your professors are extroverts? (d) In a group of six professors selected at random, what is the expected number of extroverts? What is the standard deviation of the distribution? (e) Suppose you were assigned to write an article for the student newspaper and you were given a quota (by the editor) of interviewing at least three extroverted professors. How many professors selected at random would you need to interview to be at least 90% sure of filling the quota?

Transcript

00:01 All right, so we were told in this study or whatever that 45 % or proportion of 0 .45 of university professors are extroverted.

00:10 We have six different professors that we're looking at.

00:14 We want, first of all, what's the probability that all six are extroverts? so this one should be easy.

00:20 We don't even need to use the binomial probability.

00:22 We just take 0 .45 the probability that one is an extrovert, and we do it to the sixth power.

00:30 So this is, let's see, 0 .45 to the 6th power, gives us a probability of 0 .0083 that all 6 are extroverts.

00:43 Okay? for b, what's the probability that none of them are extroverts? for that, we would just take kind of the opposite.

00:51 We would take 0 .55, which is the probability of one not being an extrovert because it's 1 minus 0 .4 or 5.

00:59 And we take that to the sixth power, that would be the probability that none of them are extroverts.

01:06 So 0 .55 to the 6 power gives us a 0 .077 chance of that happening.

01:13 For c, what's the probability that at least two of your professors are extroverts? so that's the x is greater than are equal to 2.

01:22 All right, so one way of finding this is to take 100 % minus the probability that none of them are extroverts.

01:33 Excuse me minus the probability that only one is an extrovert and that will give me the probability that two or more are so i already know the probability of none of them being an extrovert is 0 .077 for this probability that one is an extrovert i need to use the binomial theorem so that's in this case six choose one right and then the probability of one being an extrovert is 0 .45, and then times the probability that the rest of them, the other five, are introverts.

02:13 So taking that gives me, all right, so a 0 .1359 chance that only one of them is.

02:23 So now to get that total probability that i'm looking for, i take 1 minus 0 .07, or 0277, minus 0 .1359, which gives me a 0 .8364 chance that at least two of my professors are extroverts.

02:41 D asks in the group of six professors at random, what is the expected number of extroverts? so that would just be n times p.

02:50 Right? out of six, i just take 6 times 0 .45.

02:55 That should give me the expected number.

02:59 So that is 2 .7.

03:01 Okay...

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