a) OK, this is a binomial distribution, with Pr(extrovert) = 0.65 and 15 cases (n=15). So the probability of 10 or more extroverts in a sample of 15 can be found using 1 minus the cdf of the binomial distribution at k=9 - since that is 1 minus the total probability of 9 or fewer extroverts in this group. So n=15, k=9 and p=0.65 (k=outcomes, n=trials, p=prob. each trial is the outcome).

For example using binom.dist function in Excel you get: 1-binom.dist(9,15,0.65,1) = 1-0.436 = 0.564.

The second part is identical, but now we set k=4 ("below 5") instead of k=9: 1-binom.dist(4,15,0.65,1) = 1-0.003 = 0.997.

Finally, the probability that all are extroverts is given by the pdf at k=15: binom.dist(15,15,0.65,0) = 0.002 (to 3 decimal places).

b) Now we have Pr(introvert) = 70%, so p=0.7. And we just have 5 programmers, so n=5. The probability of none being introverts is the pdf at k=0: binom.dist(0,5,0.7,0) = 0.002.

The probability of 3 or more is 1 minus the probability of 2 or fewer (i.e. 1 minus the cdf of the binomial distribution at 2): 1-binom.dist(2,5,0.7,1) = 1-0.163 = 0.837.

The probability of all being introverted is simply the pdf at k=5: binom.dist(5,5,0.7,0) = 0.168.

Hope this helped! If you can't use the Excel functions I've given, try googling an online calculator for the binomial distribution - if you plug in the correct values for k, n and p each time you should get identical answers. Be careful and remember sometimes you are using the cdf (total sum of probabilities for a range) and other times you might use the pdf (probability for one exact outcome).

Feb 27th, 2015

Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.