Distributions of Normal Variables

3.2

Area under the curve, Part II. What percent of a standard normal distribution N(μ = 0, σ = 1) is found in each region? Be sure to draw a graph.

  1. Z >−1.13

Using a Normal Probability Distribution Table for all percentages, percent of Z > -1.13 is

z <- 1-.1292
z
## [1] 0.8708
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
normalPlot(mean = 0, sd = 1, bounds = c(-1.13, Inf), tails=FALSE)

  1. Z <0.18

0.5714 or 57.14%

normalPlot(mean = 0, sd = 1, bounds = c(.18, Inf), tails=FALSE)

  1. Z >8

Probability is greater than or equal to .0002 or .02%

normalPlot(mean = 0, sd = 1, bounds = c(8, Inf), tails = FALSE)

  1. |Z|<0.5
z_point5 <- 0.6915
z_negpoint5 <- 0.3085 

righttail <- 1-z_point5
absz <- righttail+z_negpoint5
absz
## [1] 0.617
normalPlot(mean = 0, sd = 1, bounds = c(-.5,.5), tails = TRUE)

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.

Leo: μ = 4313 σ = 583

Mary μ = 5261 σ = 807

  1. What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?
zLeo = (4948 - 4313)/583
zLeo
## [1] 1.089194
zMary = (5513 - 5261)/807
zMary
## [1] 0.3122677
  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning.

Leo ranked better because he ranked 1.089 standard deviations above the mean, while Mary only ranked .312 standard devations above the mean.

  1. What percent of the triathletes did Leo finish faster than in his group?

Using the normal probability distribution chart, Leo finished faster than 0.8621 or 86.21% of all people in his group.

  1. What percent of the triathletes did Mary finish faster than in her group?

Using the normal probability distribution chart, Mary finished faster than 0.6217 or 62.17% of all people in her group.

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

Z scores can be used even if the distribution is not normal, so b and c are find. However, the other questions are dependent on the distribution table. Also, we are unaware of how many people are in each group. If it is larger than 30 then we can safely assume times are normal and our answers should not need to change.

3.18

Heights of female college students. (use qqnormsim from lab 3)

Below are heights of 25 female college students. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 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

  1. 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.
fheights <- 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)

quantile(fheights, c(.67, .95, .997))
##    67%    95%  99.7% 
## 63.000 68.600 72.712
avgfheights <- mean(fheights)
sdfheights <- sd(fheights)

z67 = (63-avgfheights)/sdfheights
z67
## [1] 0.3228856
z95 = (68.6-avgfheights)/sdfheights
z95
## [1] 1.544615
z95 = (72.712-avgfheights)/sdfheights
z95
## [1] 2.441713

This is approximately 62%, 94%, and 99.3%. Therefore in general, I believe the heights 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.

Data appears roughly normal, potentially more-so than the simulation

qqnorm(fheights)
qqline(fheights)

qqnormsim(fheights)

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?
pro_10tran <- .98^(9)*.02
pro_10tran
## [1] 0.01667496
  1. What is the probability that the machine produces no defective transistors in a batch of 100?
pro_100tran <- .98^(100)
pro_100tran
## [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?

First defect at 50 transitors on average with a standard deviation of approximately 49.5 transitors.

#expected value
first_defect <- 1/.02
first_defect
## [1] 50
#standard deviation
tran_sd = sqrt(.98/(.02)^2)
tran_sd
## [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?

First defect at 20 transitors on average with a standard deviation of approximately 19.5 transitors.

#expected value
first_defect <- 1/.05
first_defect
## [1] 20
#standard deviation
tran_sd = sqrt(.95/(.05)^2)
tran_sd
## [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 decreases the mean and standard deviation.

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.
n <- 3
k <- 2
p <- .51

two_boys <- factorial(n)/(factorial(k)*(factorial(n-k)))*(p^k)*((1-p)^(n-k))
two_boys
## [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.

bbg bgb gbb

bbg <- sum((.51)*(.51)*(1-.51)) 
bgb <- sum((.51)*(1-.51)*(.51)) 
gbb <- sum((1-.51)*(.51)*(.51)) 

two_boys_add <- bbg+bgb+gbb
two_boys_add
## [1] 0.382347
  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).

Part A is much easier because it is formulaic and we can change variables n, k, or p much quicker.

n <- 8
k <- 3
p <- .51

three_boys <- factorial(n)/(factorial(k)*(factorial(n-k)))*(p^k)*((1-p)^(n-k))
three_boys
## [1] 0.2098355

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?
serve <- .15
n <- 10
k <- 3

success3 <- (factorial(n-1)/(factorial(k-1)*factorial(n-k)))*p^k*((1-p)^(n-k))
success3
## [1] 0.03238811
  1. Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful?

15% because they are independent

  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?

The first time we used the negative binomial distribution formula, which describes the probability of observing the kth success on the nth trial.