Let N be the number that are greater than 12 out of the 100 sampled items. Then N is a binomial with pi = .1587 and n = 100. Using H&O page 797 we can approximate this by the pi = .15 column which give an answer of .0402 + .0270 + .0171 + .0103 + ... = .09.
Or we can use the fact that N is approximately a normal with a mean of 15.87 ( = n x pi) and a standard deviation of 3.65 = sqrt(n x pi x (1-pi)). So, P(N >= 20) = P(Z >= 1.15) = .5 - .3749 = .1251. Unfortunately, we also could do the calculation as P(N >= 20) = P(N > 19) = P(Z > .85) = .5 - .3023 = .2977.
Complex: Yes. Similar story.
Last modified: Wed Dec 22 14:57:39 EST 1999
-------------------------------------------------------------------------------- 5A) Most got the right. A few said yes, but also emphasised that it was skewed in which case they lost 1 or 3 points depending on their argument. 5B) above 5C) Most argued better sd. If they also argued complex "more normal looking" I took off a point since the test states the distribution is lognormal and so using a normal appearance is an incorrect criterion. 6A) Everyone had the mean correct. I accepted .189 or .194 for the sd. If they got the sd wrong, I subtracted two (+2 for the mean and +1 partial credit for the sd). 6B) I graded this pretty easy. I gave full credit if they had temp increased significantly and/or out of control.