Homework 3

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

*(a) - Z > !1.13

The area is 1 − 0.1292381=0.8707619 or 87.08%.
normalPlot(mean = 0, sd = 1, bounds=c(-1.13,4), tails = FALSE)

*(b) - Z < 0.18

Z<0.18: The area is 0.5714237 or 57.14%.
normalPlot(mean = 0, sd = 1, bounds=c(-4,0.18), tails = FALSE)

*(c)- Z > 8

The probability of a value in normal distribution being 8 standard deviations away from mean is well under 0.01%, so the area is almost nearly 0.

*(d)- |Z| < 0.5

The area is "0.6914625−0.3085375"=0.3829249 or 38.3%.
normalPlot(mean = 0, sd = 1, bounds=c(-0.5,0.5), tails = FALSE)

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.

The short-hands are: For men’s group, N(μ=4313,σ=583) For women’s group, N(μ=5261,σ=807)


These Z-scores tell us relatively how Leo and Mary did in their respective groups. Their Z-scores are both higher than their respective group means, meaning they are slower than at most people in their groups. Leo is 1.09 standard deviation higher than his group mean, while Mary is 0.31 higher than hers.

*(b)- What are the Z-scores for Leo’s and Mary’s finishing times? What do these Z-scores tell you?

\({ Z }_{ leo }=\frac { x-\mu }{ \sigma } =\frac { 4948-4313 }{ 583 } \approx 1.0892\) \({ Z }_{ mary }=\frac { x-\mu }{ \sigma } =\frac { 5513-5261 }{ 807 } \approx 03123\)

These Z-scores tell us relatively how Leo and Mary did in their respective groups. Their Z-scores are both higher than their respective group means, meaning they are slower than at most people in their groups. Leo is 1.09 standard deviation higher than his group mean, while Mary is 0.31 higher than hers.

*(c)- Did Leo or Mary rank better in their respective groups? Explain your reasoning.

Mary did better, because the lower tail area represents the percentage of people finish better (faster, lesser time). Mary’s lower tail is smaller than Leo’s. Therefore, less people finish better than Mary in her group than Leo’s in his group.

*(d)- What percent of the triathletes did Leo finish faster than in his group?

(percent <- pnorm((4948-4313)/583, lower.tail = FALSE))
## [1] 0.1380342
1−0.8619672=0.1380328 or 13.8%.

*(e)- What percent of the triathletes did Mary finish faster than in her group?

(percent <- pnorm((5513-5261)/807, lower.tail = FALSE))
## [1] 0.3774186
1−0.6225937=0.3774063 or 37.74%.

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

If distributions are not nearly normal, then part (b) will remain the same since Z-scores can still be calculated. However, parts (d) and (e) rely on the normal model for calculations, so the results would change.

3.18 Heights of female college students.

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)
summary(heights)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   54.00   58.00   61.00   61.52   64.00   73.00

*(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.

heightsm <- mean(heights)
heightssd <- sd(heights)
For 68%
limitup <- heightsm + heightssd
limitdm <- heightsm - heightssd
interval <- heights[heights >= limitdm & heights <= limitup]
length(interval)/length(heights)
## [1] 0.68

For 95%

limitup <- heightsm + 2*heightssd
limitdm <- heightsm - 2*heightssd
interval <- heights[heights >= limitdm & heights <= limitup]
length(interval)/length(heights)
## [1] 0.96

For 99.7%

limitup <- heightsm + 3*heightssd
limitdm <- heightsm - 3*heightssd
interval <- heights[heights >= limitdm & heights <= limitup]
length(interval)/length(heights)
## [1] 1

*(b)- Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below.

qqnormsim(heights)

qqnorm(heights)
qqline(heights)

Comparing the true Q-Q plot with the simulated normal Q-Q plots, it is apparent that the true data is close to normal

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?

\(P(10th\quad Transistor\quad is\quad the\quad first\quad with\quad defect)=(1-p)^{ n-1 }\\ p=(1-0.02)^{ 9 }*0.02=0.016675\\\)

*(b)- What is the probability that the machine produces no defective transistors in a batch of 100?

\(P(no\quad defects\quad in\quad a\quad batch\quad of\quad 100)=(1-p)^{ 100 }\quad =0.98^{ 100 }=0.016675\\ \\\)

*(c)- On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?

\(\mu \quad =\frac { 1 }{ p } =\frac { 1 }{ 0.02 } =\quad 50\quad and\quad \sigma \quad =\quad \sqrt { \frac { 1-p }{ { p }^{ 2 } } } =\sqrt { \frac { 1-0.02 }{ { 0.02 }^{ 2 } } } =\sqrt { 2450 } =\quad 49.4974747\\ \\\)

*(d)- 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?

\(if\quad p=0.05,\quad then\quad \mu =\frac { 1 }{ 0.05 } =20\quad and\quad \sigma =\sqrt { \frac { 1-p }{ { p }^{ 2 } } } =\quad \sqrt { \frac { 0.95 }{ 0.0025 } } =\quad 19.4935887\\\)

*(e)- Based on your answers to parts (c) and (d), how does increasing the probability of an event a↵ect the mean and standard deviation of the wait time until success?

When probability of an event is higher, the event is more common, so the expected number of trials before it occurs and the standard deviation are 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.

\(if\quad p=0.51,\quad n=3,\quad and\quad k=2,\quad then\quad P(two\quad boys\quad out\quad of\quad three\quad kids)=\frac { n! }{ k!{ (n-k)! } } { p }^{ k }{ (1-p) }^{ n-k }\\ \frac { 3! }{ 2! } *{ 0.51 }^{ 2 }*0.49=\quad 0.382347\)

*(b)- 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.

Possible Combinations are:

boy,boy, girl
boy,girl,boy
girl, boy, boy

\(P(two\quad boys\quad out\quad of\quad three\quad kids)=\quad (P(boy)*P(boy)*P(girl))+(P(boy)*P(girl)*P(boy))+(P(girl)*P(boy)*P(boy))=\quad 3*0.51*0.51*.049=\quad 0.382347\\\)

*(c)- 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).

If using approach b to calculate the probability that a couple with 8 kids has 3 boys, it is first necessary to determine the number of combinations of having 3 boys among 8 kids. The list will be significantly longer than with 3 kids (in fact, there are 56 combinations). With approach a, all that is necessary is just to plug in the numbers into the formula.

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?

\(\\ P(3rd\quad success\quad on\quad the\quad 10th\quad try)=\left( \frac { n-1 }{ k-1 } \right) { p }^{ k }{ (1-p) }^{ n-k }=\frac { 9! }{ 2!*7! } *{ 0.15 }^{ 3 }*{ 0.85 }^{ 7 }=0.0389501\)

*(b)- Suppose she has made two successful serves in nine attempts. What is the probability that her 10th serve will be successful?

Serves are independent events and previous outcomes have no effect on future events. The probability of the success on the 10th serve is 0.15.

*(c)- Even though parts (a) and (b) discuss the same scenario, the probabilities you calculated should be di↵erent. Can you explain the reason for this discrepancy?

Part a is looking for the probability of a specific combination of successes withing 10 serves. Although each serve is independent, we are considering all 10 serves in determining the probability of the desired pattern. Contrary to this part b is only concerned with one serve. Previous outcomes are irrelevant because events are independent.