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.

Citation for drawing area under curve: https://www.youtube.com/watch?v=HtXeTsx9ka0

  1. \(Z < -1.35\)
mean <- 0
std <- 1
z <- -1.35

# Solving for x:
x <- z*std + mean
x
## [1] -1.35
# Finding probability
1 - pnorm(x, mean = 0, sd = 1)
## [1] 0.911492
# Graphing and shading area under normal curve

x = seq(from=-3, to=z, by=0.1)
y = dnorm(x, mean = 0, sd = 1)
plot(x,y, type = "l", col = "blue", main = "Normal Curve")
polygon(c(-3, x, z), c(0,y,0), col = 'red')

(b) \(Z > 1.48\)

mean <- 0
std <- 1
z <- 1.48

# Solving for x:
x <- z*std + mean
x
## [1] 1.48
# Finding probability
pnorm(x, mean = 0, sd = 1)
## [1] 0.9305634
# Graphing and shading area under normal curve

x = seq(from=-3, to=z, by=0.1)
y = dnorm(x, mean = 0, sd = 1)
plot(x,y, type = "l", col = "blue", main = "Normal Curve")
polygon(c(-3, x, 3), c(0,y,0), col = 'red')

(c) \(-0.4 < Z < 1.5\)

mean <- 0
std <- 1
z1 <- 1.48
z2 <- 1.5

# Solving for x:
x <- z*std + mean
x
## [1] 1.48
# Finding probability
pnorm(x, mean = 0, sd = 1)
## [1] 0.9305634
# Graphing and shading area under normal curve

x = seq(from=z1, to=z2, by=.1)
y = dnorm(x, mean = 0, sd = 1)
plot(x,y, type = "l", col = "blue", main = "Normal Curve")
polygon(c(-3, x, 3), c(0,y,0), col = 'red')

  1. \(|Z| > 2\)
mean <- 0
std <- 1
z <- 2


# Solving for x:
x <- z*std + mean
x
## [1] 2
# Finding probability
pnorm(x, mean = 0, sd = 1)
## [1] 0.9772499
# Graphing and shading area under normal curve

x = seq(from=z1, to=z2, by=.1)
y = dnorm(x, mean = 0, sd = 1)
plot(x,y, type = "l", col = "blue", main = "Normal Curve")
polygon(c(-3, x, 3), c(0,y,0), col = 'red')

clearpage

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:

Remember: a better performance corresponds to a faster finish.

  1. Write down the short-hand for these two normal distributions. N(μ=4313, sd=583) → Men, Ages 30 - 34

N(μ=5261, sd=807) → Women, Ages 25 - 29

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

For Leo, the Z score is 1.089; the number of standard deviations is above the mean.As for Mary her Z score is 0.312, which is the number of standard deviations below the mean.

  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning. > Mary got a lower Zscore, 0.312, than Leo: 1.089. This indicates that Mary has a better rank in her group.

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

Percent_Leo <- 1-pnorm(1.089)
Percent_Leo
## [1] 0.1380769
  1. What percent of the triathletes did Mary finish faster than in her group?
Percent_Mary <- 1-pnorm(0.312)
Percent_Mary
## [1] 0.3775203
  1. If the distributions of finishing times are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning. >There would be no difference in parts a-c, thus the ranks will not change and the Zscore will still reflect its position relative to the mean. However, parts d & e will change, as “Pnorm” is only used for the normal distribution.

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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, becuause the probability of falling within one standard deviation of the mean is about 0.997–thus, the distribution of the heights do not follow the 68-95-99.7 rule, as proved below.
height <- 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(height)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   54.00   58.00   61.00   61.52   64.00   73.00
pnorm(61.52+1*4.58,mean=61.52,sd=4.58)
## [1] 0.8413447
pnorm(61.52+2*4.58,mean=61.52,sd=4.58)
## [1] 0.9772499
pnorm(61.52+3*4.58,mean=61.52,sd=4.58)
## [1] 0.9986501
  1. Do these data appear to follow a normal distribution? Explain your reasoning using the graphs provided below. > Yes, the data appear to follow a normal distribution. These data plots show that the data follows the line of fit on the QQplot with some deviations/semi-outliers on both the high and low ends.

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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?
pgeom(10-1,0.02)
## [1] 0.1829272
  1. What is the probability that the machine produces no defective transistors in a batch of 100?
pgeom(100,0.02)
## [1] 0.8700328
  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/.02)
mean
## [1] 50
stand_dev <- sqrt((1-.02)/.02^2)
stand_dev
## [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_d <- (1/.05)
mean_d
## [1] 20
stand_dev_d <- sqrt((1-.05)/0.05^2)
stand_dev_d
## [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? > By increasing the probability of an event, we decrease the success wait time, and decrease the spread within the distribution.

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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.
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. > Orderings: {Boy, Boy, Girl}; {Girl, Boy, Boy}; {Boy, Girl, Boy};
(0.51^2)*0.49*3
## [1] 0.382347
t1 <- dbinom(2,3,0.51)
t1
## [1] 0.382347
t2 <- (0.51^2)*0.49*3
t2
## [1] 0.382347
t1 == t2
## [1] FALSE

For some reason, comparing these values returns FALSE, indicating they are not the same number. Yet, by looking at both values for t1 and t2, we can clearly see these are the same numbers.

  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).

The approach in part B would be much more tedious because we would have to write out 64 total possibilities, and count how many possible outcomes have 3 boys. Whereas, in part A, the function computes the probability automatically and with ease.

clearpage

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(9,2)*0.15^3*0.85^7
## [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? > Because the serves are independent of each other, the probability of her 10th serve being successful is still 15%.

  2. 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? > A is asking for the probability of the THIRD successful serve happening on the 10th serve, whereas B is asking whether or not the 10th serve would be successful. But because of independence, B has a probability of 0.15 since ALL serves are independent of each other.