In the book A Guide to the Development and Use of the Myers-BriggsType Indicator

Question

In the book A Guide to the Development and Use of the Myers-BriggsType Indicators by Myers and McCaully, it was reported thatapproximately 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 anextrovert?
c.) what is the probability that at least two of yourprofessors are extroverts?
d.) in a group of six professors selected at random, what isthe expected number of extroverts? what is the standard deviationof the distribution?
a.) what is the probability that all six are extroverts?
b.) what is the probability that none of your professors is anextrovert?
c.) what is the probability that at least two of yourprofessors are extroverts?
d.) in a group of six professors selected at random, what isthe expected number of extroverts? what is the standard deviationof the distribution?

Explanation / Answer

PMF - P(X=x)nCxpx(1-p)n-x a)P(X=6) =6C60.456(0.55)6-6 = 0.008304 b)P(X=0) = 6C00.450(0.55)6-0 = 0.02768064 c)P(X=>2) = 1 - [P(X=0) + P(X=1)] = 1 -[6C0 0.450(0.55)6-0 +6C1 0.451(0.55)6-1] = 1- [0.02768064 + 0.135886781] = 0.163567421 You could have also done: P(X=>2) = P(X=2) + P(X=3) +P(X=4) + P(X=5) + P(X=6) d)The expectation of a Binomial distribution - E(X) = np =6(0.45) = 2.7 The standard deviation of a Binomial Distribution - SD(X) =Var(X) = (np(1-p) =(6)(0.45)(0.55) =1.485 = 1.2186 a)P(X=6) =6C60.456(0.55)6-6 = 0.008304 b)P(X=0) = 6C00.450(0.55)6-0 = 0.02768064 c)P(X=>2) = 1 - [P(X=0) + P(X=1)] = 1 -[6C0 0.450(0.55)6-0 +6C1 0.451(0.55)6-1] = 1- [0.02768064 + 0.135886781] = 0.163567421 You could have also done: P(X=>2) = P(X=2) + P(X=3) +P(X=4) + P(X=5) + P(X=6) d)The expectation of a Binomial distribution - E(X) = np =6(0.45) = 2.7 The standard deviation of a Binomial Distribution - SD(X) =Var(X) = (np(1-p) =(6)(0.45)(0.55) =1.485 = 1.2186

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