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## Welcome to CUNY DATA606 Statistics and Probability for Data Analytics 
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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. \(Z < -1.35\)
round(((pnorm(-1.35))*100),2)
## [1] 8.85
normalPlot(mean = 0, sd = 1, bounds = c(-1.35, Inf), tails = TRUE)

8.85% of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region.

  1. \(Z > 1.48\)
round((1-pnorm(1.48))*100,2)
## [1] 6.94
normalPlot(mean = 0, sd = 1, bounds = c(1.48,Inf), tails = FALSE)

6.94% of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region.

  1. \(-0.4 < Z < 1.5\)
z<- c(-.4,1.5)
pnorm(z)
## [1] 0.3445783 0.9331928
round((0.9331928-0.3445783)*100,2)
## [1] 58.86
normalPlot(mean = 0, sd = 1, bounds = c(-0.4, 1.5), tails = FALSE)

58.86% of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region.

  1. \(|Z| > 2\)
z<- c(-2,2)
pnorm(z)
## [1] 0.02275013 0.97724987
round(((0.97724987-0.02275013)*100),2)
## [1] 95.45
normalPlot(mean = 0, sd = 1, bounds = c(-2,2), tails = FALSE)

95.45% of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region.

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.

Men, Ages 30 - 34 N(μ = 4313, σ = 583)
Women, Ages 25 - 29 N(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?
#Z-scores for Leo's finishing time
#(Leo = 4948, mean = 4313, Standard deviation = 583)
Leo_x <- 4948
Leo_mu <- 4313
Leo_sd <- 583
Leo_z <- (Leo_x - Leo_mu)/(Leo_sd)
Leo_z
## [1] 1.089194

Leo’s finishing time is 1.089194 standard deviations above the mean. We know it must be above the mean since Z is positive.

#Z-scores for Mary's finishing times
#(Mary = 5513, mean = 5261, Standard deviation = 807)
Mary_x <- 5513
Mary_mu <- 5261
Mary_sd <- 807
Mary_z <- (Mary_x - Mary_mu)/(Mary_sd)
Mary_z
## [1] 0.3122677

Mary’s finishing time is 0.3122677 standard deviations above the mean. We know it must be above the mean since Z is positive.

  1. Did Leo or Mary rank better in their respective groups? Explain your reasoning.
    Based on their Z scores, Mary was more close to her respective age category mean by only .03122677 standard deviation. On the other hand Leo is at 1.089194 standard deviation above from mean. In racing senario the less you are away from mean is better raning. Therefor Mary rank better in her respective group.

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

round( (1-pnorm(Leo_z))*100,2)
## [1] 13.8

Answer: 13.8%

  1. What percent of the triathletes did Mary finish faster than in her group?
round((1-pnorm(Mary_z))*100,2)
## [1] 37.74

Answer: 37.74%

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

Z-scores for non normal distributions are relevant for analysis in a multiple group. We will not be able to compare two people from different groups based on Z-scores of non normal distributions. We cannot use the normal probability table to calculate the probabililties and percentiles without a normal model for parts (d) through (e).

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.

# Use the DATA606::qqnormsim function
qqnormsim(heights)

check weather 68% lie within 1 standard deviation

# height mean
heights_mean <- mean(heights)
# height standard deviation
heights_sd <- sd(heights)
1-2*pnorm(heights_mean+heights_sd, heights_mean, heights_sd, lower.tail = FALSE)
## [1] 0.6826895

check weather 95% lie within 2 standard deviation

1-2*pnorm(heights_mean+2*heights_sd, heights_mean, heights_sd, lower.tail = FALSE)
## [1] 0.9544997

check weather 99.7% lie within 3 standard deviation

1-2*pnorm(heights_mean+3*heights_sd, heights_mean, heights_sd, lower.tail = FALSE)
## [1] 0.9973002

Yes, from above we can say 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.
summary(heights)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   54.00   58.00   61.00   61.52   64.00   73.00

This histogram shows that data is not perfectly symmetric or in bel-shape and more of right-skewed. Mean is little bit more than the median. But it is unimodal. Normal probability plot also shows that points follows a straight line. We can conclude that this is nearly normal distribution.

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?
# using actual formula (1-p)^(n-1) * p
defect <- 0.02
success <- 1-defect
n <- 10
Pof_defect <- (success)^(n-1) * defect
Pof_defect
## [1] 0.01667496
# using dgeom function
dgeom((n-1), defect)
## [1] 0.01667496

Answer: 0.01667496

  1. What is the probability that the machine produces no defective transistors in a batch of 100?
n <- 100
(1-defect)**n
## [1] 0.1326196

Answer: 0.1326196 (c) On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?

#first defect
1/defect
## [1] 50
#standard deviation
sqrt((1-defect)/(defect^2))
## [1] 49.49747

Answer: 50 transistors would you expect to be produced before the first with a defect. standard deviation is 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?
defect2 <- 0.05
#first defect
1/defect2
## [1] 20
#standard deviation
sqrt((1-defect2)/(defect2^2))
## [1] 19.49359

Answer: 20 transistors would you expect to be produced before the first with a defect. standard deviation is 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 of an event decreases value of mean and standard deviation until success.

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 dbinom function
dbinom(2, 3, prob = 0.51)
## [1] 0.382347
#using mathamatical function of binomial distribution.
n <- 3
k <- 2
p <- 0.51

binomial_d <- factorial(n)/(factorial(k)*factorial(n-k))

prob_boy <- binomial_d*(p^k)*((1-p)^(n-k))

prob_boy
## [1] 0.382347

Answer: 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.
# boy boy girls
senario1 <- .51 * .51 * .49
# boys girl boy
senario2 <- .51 * .49 * .51
# girl boy boy
senario3 <- .49 * .51 * .51
senario1 + senario2 + senario3
## [1] 0.382347

Answer from part (a) matches with answer from part (b)

  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
  2. would be more tedious than the approach from part (a).
choose(8,3)
## [1] 56

Above we are calculating just saying 3 success in 8 independant trials.if you do it by hand dealing with factorial part is very tedious. Using addition rule of disjoint outcomes for this senario is possible but not easy as we need to find out all possible senarios that give us 3 boys out of 8 trial.

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?
#using dbinom function
dbinom(3, 10, prob = 0.15)
## [1] 0.1298337
#using mathamatical function of binomial distribution.
n <- 10
k <- 3
p <- 0.15

binomial_d <- factorial(n)/(factorial(k)*factorial(n-k))

prob_serve <- binomial_d*(p^k)*((1-p)^(n-k))

prob_serve
## [1] 0.1298337

Answer: 0.1298337

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

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

  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?

Answer: Part B is referencing a single shot and the probability is independant. Part A is combining successful shots (multiple shots) so the probability becomes joined.