Elina Azrilyan

Homework Chapter 3
10/30/2018

Chapter 3 - Distributions of Random Variables Practice: 3.1 (see normalPlot), 3.3, 3.17 (use qqnormsim from lab 3), 3.21, 3.37, 3.41 Graded: 3.2 (see normalPlot), 3.4, 3.18 (use qqnormsim from lab 3), 3.22, 3.38, 3.42

3.2. What percent of a standard normal distribution N(μ = 0, sd = 1) is found in each region? Be sure to draw a graph.

  1. Z >-1.13
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.
normalPlot(mean = 0, sd = 1, bounds = c(-1.13, 4))

  1. Z <0.18
normalPlot(mean = 0, sd = 1, bounds = c(-4, 0.18))

  1. Z >8
normalPlot(mean = 0, sd = 1, bounds = c(8, 8))

  1. |Z|<0.5
normalPlot(mean = 0, sd = 1, bounds = c(-4, -.5))

normalPlot(mean = 0, sd = 1, bounds = c(0.5, 4))

0.309*2
## [1] 0.618

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. (a) Write down the short-hand for these two normal distributions.

Men: Mean = 4313, sd = 583 Women: Mean = 5261, sd = 801

  1. What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?

Leo’s Z = (x−μ)/σ =

(4948 - 4313)/583
## [1] 1.089194

Mary’s Z = (x−μ)/σ =

(5513 - 5261)/801
## [1] 0.3146067
  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning.

Leo did better in his respective group since his Z-score is higher.

  1. What percent of the triathletes did Leo finish faster than in his group?
normalPlot(mean = 4313, sd = 583, bounds = c(0, 4948))

86.2%

  1. What percent of the triathletes did Mary finish faster than in her group?
normalPlot(mean = 5261, sd = 801, bounds = c(0, 5513))

62.3%

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

Part (b) answer would not change as Z-scores can be calculated for distributions that are not normal. However, we could not answer parts (d)-(e) since we could not use the normal probability table to calculate probabilities and percentiles.

3.18 Heights of female college students. Below are heights of 25 female college students.

femcsht<-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)
  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.
#Range withit 1 standard deviation: 61.52 - 4.58 and 61.52 + 4.58
61.52 - 4.58
## [1] 56.94
61.52 + 4.58
## [1] 66.1
#Percentage of observation within this range:
17/25
## [1] 0.68
#Range withit 2 standard deviations: 
61.52 - 2*4.58
## [1] 52.36
61.52 + 2*4.58
## [1] 70.68
#Percentage of observation within this range:
24/25
## [1] 0.96
#Range withit 3 standard deviations:
61.52 - 3*4.58
## [1] 47.78
61.52 + 3*4.58
## [1] 75.26
#Percentage of observation within this range:
25/25
## [1] 1

Yes, the height follows the 68-95-99.7% Rule.

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

Yes, the graphs suggest that data follows normal distribution.

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. (a) What is the probability that the 10th transistor produced is the first with a defect?

(0.98^9)*0.02
## [1] 0.01667496
  1. What is the probability that the machine produces no defective transistors in a batch of 100?
0.98^100
## [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?
#Mean
1/0.02
## [1] 50
#Standard deviation
sqrt((1-0.02)/(0.02^2))
## [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?
#Mean
1/0.05
## [1] 20
#Standard deviation
sqrt((1-0.05)/(0.05^2))
## [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?

When p is higher, the event is going to occur more often, meaning the expected number of trials before a success and the standard deviation of the waiting time is lower.

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. (a) Use the binomial model to calculate the probability that two of them will be boys.

dbinom(2,3,0.51)
## [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 GBB BGB

((0.51^2)*0.49)+((0.51^2)*0.49)+((0.51^2)*0.49)
## [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).

That approach would be extremely tedious since the number of possible orderings would be higher.

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. (a) What is the probability that on the 10th try she will make her 3rd successful serve?

choose(10, 3)*(0.15^3)*(0.85^7)
## [1] 0.1298337
  1. Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful?

The probability that her 10th serve will be successful is 0.15, since her serves are independent of each other.

  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 probabilities are different since in question (a) we are calculating the probability of having 3 successes out of 10, and in question (b) we are calculating the probability of 1 success and since the serves are independent it is irrelevant what happened in earlier serves.