Problem 3.2

a) For a standard normal, Z>-1.13, 87.08% is under the curve.

b) For a standard normal, Z<.18, 57.14% is under the curve.

c) For a standard normal, Z>8, nearly 0% is under the curve.

d) For a standard normal, |Z| < 0.5, 38.3% is under the curve.

Problem 3.4

a) Men, Ages 30 - 34~N(4313 seconds,583 seconds)
Women, Ages 25 - 29 ~N(5261 seconds,807 seconds)
b) Leo’s Z-score is 1.0892, which means 86.21% in his bracket had faster times, according to the distribution.
Mary’s Z-score was .3123, which means 62.17% in his bracket had faster times, according to the distribution.
c) Mary did better in her bracket because her Z-score was lower, and lower times are more desirable.
d) Leo was faster than 13.79% in his bracket.
e) Mary was faster than 37.83% in her bracket.
f) If the distribution of finishing times were not normal, we could not tell what kind of comparison to make.
We might know that it’s above the mean, but wouldn’t be able to tell how many came in between without knowing more.

Problem 3.18

a) Based on our theorized distribution, the 68-95-99.7% Rule would put our 3 ranges at: 56.94-66.1
52.36-70.68
47.78-75.26
We have 68%,96% and 100% of our empirical distribution between our first 3 s.d. intervals. That is close to a standard normal and less close toward the tails, where it’s more succeptible to small changes.
b) Our distribution is almost symmetric, but it appears from both graphs to be skewed slightly to the right. On the Q-Q plot, the slope is less than 1 to the left and greater than 1 to the right. Compared to the normal curve, out empirical distribution has a slightly enlarged right tail. This distribution is close to normal, but a little skewed.

Problem 3.22

a) The probability that the 10th is first defective part, with each being 2% likely to fail, is a geometric distribution: .989*.02, which equals 0.016675
b) For a batch of 100, the probability they are all working is .98100, which equals 0.1326196.
c) On average, 50 would be expected before the first defect (for a geometric dist. - 1/p). The standard deviation is sqrt(1-p/p), which is 7.
d) With a 5% defective rate, the mean is 1/.05, which equals 20.
The standard deviation, under those conditions, is sqrt(1-p/p), which is 4.36.
e) The mean and the standard deviation both go down after raising the probability of an event. If something is more likely, it is probably going to be reached earlier.

Problem 3.38

a) The probability of two out of three for a binomial with p=.51 = 3C2 .512 .49 = .382

b) The possible sets are:
Boy-Girl-Boy—>.1274 probability
Girl-Boy-Boy—>.1274 probability
Boy-Boy-Girl—>.1274 probability

These add up to .382

(c) 3C1=3, whereas 8C3=56. There would have to be 56 disjoint events counted up to do it piecemeal. The Binomial distribution has a choose (or n factorial/(n factorial*n-k factorial)) function to account for the possible outcomes with the same number of “successes” for trials.

Problem 3.42

This question is based on the negative binomial distribution
a)Pr(3rd success on 10th try) = 9c2 * .153 * .857 = .03895
b) Her probability of being successful is .15 c) Parts a and b are asking different questions. Part a contains the probability of 10 different tries. Part b is just a single Bernoulli with a probability for only one event.

Code used to shade the normal plot was modified from: http://t-redactyl.io/blog/2016/03/creating-plots-in-r-using-ggplot2-part-9-function-plots.html