3.2 Area under the curve, Part II. What percent of a standard normal distribution N(μ = 0, SD = 1) is found in each region?


  1. Z > 1.13
1-.1292
## [1] 0.8708

or

1 - pnorm(-1.13, mean = 0, sd = 1)
## [1] 0.8707619


  1. Z <0.18
.5714
## [1] 0.5714


(c) Z >8

1
## [1] 1


(d) |Z|<0.5

.6915-.3085
## [1] 0.383


3.4 Triathlon times, Part I. In triathlons, it is common for racers to be placed into age and gender groups. Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages 30 - 34 group while Mary competed in the Women, Ages 25 - 29 group. Leo completed the race in 1:22:28 (4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster, but they are curious about how they did within their respective groups. Can you help them? Here is some information on the performance of their groups: • The finishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds with a standard deviation of 583 seconds. • The finishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds. • The distributions of finishing times for both groups are approximately Normal. Remember: a better performance corresponds to a faster finish.


  1. Write down the short-hand for these two normal distributions.


Men 30-34 – N(Mean=4313, St.dev = 583) Women 25-39 – N(Mean=5261, St.dev = 807)


  1. What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?
(4948-4313)/583
## [1] 1.089194
(5513-5261)/807
## [1] 0.3122677

Leo’s finishing time was 1.09 st. dev above the mean where as Mary’s finishing time was 0.31 st. dev above the mean.


  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning.

Mary ranked better by finishing faster than a bigger portion of her group.


  1. What percent of the triathletes did Leo finish faster than in his group?
Leo_perc <-pnorm(1.09,lower.tail=FALSE)
Leo_perc
## [1] 0.1378566
# 13.8%


  1. What percent of the triathletes did Mary finish faster than in her group?
Mary_perc <- pnorm(0.31,lower.tail=FALSE)
Mary_perc
## [1] 0.3782805
# 37.8%


  1. If the distributions of finishing times are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning.

Z-score would still be calculated without a normal distribution but other than (b), it would not be possible to calculate percentages since we cannot use normal probability tables.


3.18 (a) The mean height is 61.52 inches with a standard deviation of 4.58 inches. Use this information to determine if the heights approximately follow the 68-95-99.7% Rule.

wh <- c(54, 55, 56, 56, 57, 58, 58, 59, 60, 60, 60, 61, 61, 62, 62, 63, 63, 63, 64, 65, 65, 67, 67, 69, 73)
pnorm(mean(wh) + sd(wh), mean = 61.52, sd = 4.58, lower.tail = T) - 
  pnorm(mean(wh) + sd(wh), mean = 61.52, sd = 4.58, lower.tail = F)
## [1] 0.6830768
pnorm(mean(wh) + 2*sd(wh), mean = 61.52, sd = 4.58, lower.tail = T) - 
  pnorm(mean(wh) + 2*sd(wh), mean = 61.52, sd = 4.58, lower.tail = F)
## [1] 0.9546724
pnorm(mean(wh) + 3*sd(wh), mean = 61.52, sd = 4.58, lower.tail = T) - 
  pnorm(mean(wh) + 3*sd(wh), mean = 61.52, sd = 4.58, lower.tail = F)
## [1] 0.9973214

They do approximately follow the 68-95.99.7 rule.


  1. Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below.

The distribution seems to be symmetric. The points on the normal probability plot does not deviate from the line much. It looks like the plots with simulated data sets below. The distribution appears to be normal.

library(DATA606)
## 
## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics 
## This package is designed to support this course. The text book used 
## is OpenIntro Statistics, 3rd Edition. You can read this by typing 
## vignette('os3') or visit www.OpenIntro.org. 
##  
## The getLabs() function will return a list of the labs available. 
##  
## The demo(package='DATA606') will list the demos that are available.
## 
## Attaching package: 'DATA606'
## The following object is masked from 'package:utils':
## 
##     demo
qqnormsim(wh)


3.22 Defective rate. A machine that produces a special type of transistor (a component of computers) has a 2% defective rate. The production is considered a random process where each transistor is independent of the others.


  1. What is the probability that the 10th transistor produced is the first with a defect?
(1-.02)^9*.2
## [1] 0.1667496


  1. What is the probability that the machine produces no defective transistors in a batch of 100?
.98^100
## [1] 0.1326196


  1. On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?
1/0.02
## [1] 50


p = .02
sqrt((1-p)/p^2)
## [1] 49.49747


  1. Another machine that also produces transistors has a 5% defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the first with a defect? What is the standard deviation?
p2 = .05
1/p2
## [1] 20
sqrt((1-p2)/p2^2)
## [1] 19.49359


  1. Based on your answers to parts (c) and (d), how does increasing the probability of an event affect the mean and standard deviation of the wait time until success?

Increasing the probability of the event decreased the mean and standard devidation of the wait time until success.


3.38 Male children. While it is often assumed that the probabilities of having a boy or a girl are the same, the actual probability of having a boy is slightly higher at 0.51. Suppose a couple plans to have 3 kids.


  1. Use the binomial model to calculate the probability that two of them will be boys.
ncol(combn(3,2))*.51^2*.49
## [1] 0.382347


  1. Write out all possible orderings of 3 children, 2 of whom are boys. Use these scenarios to calculate the same probability from part (a) but using the addition rule for disjoint outcomes. Confirm that your answers from parts (a) and (b) match.

M M F M F M F M M

.51^2*.49 + .51*.49*.51 + .49*.51*.51
## [1] 0.382347
# They do match


  1. If we wanted to calculate the probability that a couple who plans to have 8 kids will have 3 boys, briefly describe why the approach from part (b) would be more tedious than the approach from part (a).

Writing all the possible orders for 3 boys out of 8 kids would be quite long. If we adopted the approach (a), we would simply calculate the combination (8,3), then calculate the probabilities for boys and girls depending on their numbers, and finally multiply them all.


3.42 Serving in volleyball. A not-so-skilled volleyball player has a 15% chance of making the serve, which involves hitting the ball so it passes over the net on a trajectory such that it will land in the opposing team’s court. Suppose that her serves are independent of each other.


  1. What is the probability that on the 10th try she will make her 3rd successful serve?
ncol(combn(9,2)) * .15^3 * .85^7
## [1] 0.03895012


  1. Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful?
.15
## [1] 0.15


  1. Even though parts (a) and (b) discuss the same scenario, the probabilities you calculated should be different. Can you explain the reason for this discrepancy?

For the part (b), the previous serves won’t affect the last serve since serves are independent. Also for part (b), conditions for the previous serves were already given. For each shot, the probability of serving succesfully will still be .15.