DATA606 HW5

1. Area under the curve, Part I. (4.1, p. 142) What percent of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region? Be sure to draw a graph.

1. \(Z < -1.35\) 0.0885*100= 8.85%

=8.85%

2. \(Z > 1.48\) 0.9306 1-0.9306= 0.0694*100= 6.94%

=6.94%

3. \(-0.4 < Z < 1.5\) 0.3446 0.9332 0.9332-0.3346=0.5886 *100= 58.86

=58.86%

4. \(|Z| > 2\) 0.9772 1-0.9772=0.02282=0.0456100= 4.56%

=4.56%

2. Triathlon times, Part I (4.4, p. 142) 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 group has a mean of 4313 seconds with a standard deviation of 583 seconds.
• The finishing times of the 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): u=4313 sd= 583 Women(25-29): u=5261 sd= 807

2. What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you? Leo: Z-score (4948-4313)/583=1.09 Leo is within the 0.8621 percentile. Mary: Z-score (5513-5261)/807=0.31 Mary is within the 0.6217 percentile.

3. Did Leo or Mary rank better in their respective groups? Explain your reasoning. Mary rank better in their respective group because shes within the 62 percentile, opposed to Leo being part of the 86 percentile.
4. What percent of the triathletes did Leo finish faster than in his group? 1-0.8621=0.1379 He is faster than 13.8% of the triathletes.
5. What percent of the triathletes did Mary finish faster than in her group? 1-0.6217=0.3783 Mary finish faster than 38% of the triatheletes in her group.

6. If the distributions of finishing times are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning. Yes, this calculations will only work on normal distributions.

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

\[ \stackrel{1}{54}, \stackrel{2}{55}, \stackrel{3}{56}, \stackrel{4}{56}, \stackrel{5}{57}, \stackrel{6}{58}, \stackrel{7}{58}, \stackrel{8}{59}, \stackrel{9}{60}, \stackrel{10}{60}, \stackrel{11}{60}, \stackrel{12}{61}, \stackrel{13}{61}, \stackrel{14}{62}, \stackrel{15}{62}, \stackrel{16}{63}, \stackrel{17}{63}, \stackrel{18}{63}, \stackrel{19}{64}, \stackrel{20}{65}, \stackrel{21}{65}, \stackrel{22}{67}, \stackrel{23}{67}, \stackrel{24}{69}, \stackrel{25}{73} \]

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.

No, it does not follow the 68-95-99.7% Rule because it does not look like a bell-curve graph. The values are not under the red line which depicts the normal distribution.

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

The red line on the barplot shows the normal distributions. The values of the mean heights fall out of the red line. Also in the qqnorm graph, a lot of the points do not touch the line.

library(openintro)

## Please visit openintro.org for free statistics materials

## 
## Attaching package: 'openintro'

## The following objects are masked from 'package:datasets':
## 
##     cars, trees

heights <- 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)
par(mar=c(3.7,2.2,1,1), las=1, mgp=c(2.5,0.7,0), mfrow = c(1,1), cex.lab = 1.5, cex.axis = 1.5)
histPlot(heights, col = COL[1], xlab = "Heights", ylab = "", probability = TRUE, axes = FALSE, ylim = c(0,0.085))
axis(1)
x <- heights
xfit <- seq(min(x)-5, max(x)+5, length = 400)
yfit <- dnorm(xfit, mean = mean(x), sd = sd(x))
lines(xfit, yfit, col = COL[4], lwd = 2)

par(mar=c(3.7,3.7,1,1), las=1, mgp=c(2.5,0.7,0), mfrow = c(1,1), cex.lab = 1.5, cex.axis = 1.5)
qqnorm(heights, col = COL[1], pch = 19, main = "", axes = FALSE)
axis(1)
axis(2)
qqline(heights, col = COL[1])

qqnorm(heights)

4. Defective rate. (4.14, p. 148) 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?

#0.98 rate have no defects, 9 transistor with no defects with the 10th one is results the following:
((0.98)^9)*(0.02)

## [1] 0.01667496

2. What is the probability that the machine produces no defective transistors in a batch of 100?

#Using the same equation from above we input 100
((0.98)^100)

## [1] 0.1326196

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

Standard deviation is 49.49747.

#the mean
1/0.02

## [1] 50

#standard deviation
sqrt((1-0.02)/(0.02^2))

## [1] 49.49747

4. 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? u=1/p p=0.05 20 transistors will be produced with this machine before the first with a defect. The standard deviation is 19.49359.

#the mean
1/0.05

## [1] 20

#standard deviation
sqrt((1-0.05)/(0.05^2))

## [1] 19.49359

5. 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 an event decreases value of mean and standard deviation until success.

5. 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.

Using the binomial model, the probability that two of them will be boys is 0.382347.

0.51*0.51*0.49

## [1] 0.127449

0.49*0.51*0.51

## [1] 0.127449

0.51*0.49*0.51

## [1] 0.127449

0.127449*3

## [1] 0.382347

2. 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. Using the formula, (n k)pk(1−p)(n−k) n=3 k=2 p=0.51 The answers from part a and b matches.

choose(3,2)*(.51^2)*(.49^1)

## [1] 0.382347

3. 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 b will be more tedious because you can just use the same formula over and over again. But for part a, you have to create more scenarios with the numbers get bigger.

choose(8,3)*(.51^3)*(.49^5)

## [1] 0.2098355

6. Serving in volleyball. (4.30, p. 162) 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?

choose(10,3)*(0.15^3)*(0.85^7)

## [1] 0.1298337

2. Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful?

Her 10th serve is still going to be 15%.

3. 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?

Part a includes the total for 10 serves however part b includes for 9 serves and the last one being independent.