Question 1

What is the variance of the distribution of the average an IID (independent and identically distributed) draw of \(n\) observations from a population with mean \(\mu\) and variance \(\sigma^2\).

  • \(\sigma / n\)
  • \(\sigma^2\)
  • \(\frac{\sigma^2}{n}\)
  • \(\sigma / \sqrt{n}\)

Answer

\[Var(\bar X) = \frac{\sigma^2}{n}\] Where:

  • \(\sigma^2\): population variance
  • \(n\): number of observation
  • \(Var(\bar X)\): Variance of the sample mean

Question 2

Suppose that diastolic blood pressures (DBPs) for men aged 35-44 are normally distributed with a mean of 80 (mm Hg) and a standard deviation of 10. About what is the probability that a random 35-44 year old has a DBP less than 70?

  • 22%
  • 16%
  • 8%
  • 32%

Answer

normally distributed

# option 1
pnorm(q = 70, mean = 80, sd = 10, lower.tail = TRUE)
## [1] 0.1586553
# option 2
pnorm(q = (70-80)/10, lower.tail = TRUE)
## [1] 0.1586553

Question 3

Brain volume for adult women is normally distributed with a mean of about 1,100 cc for women with a standard deviation of 75 cc. What brain volume represents the 95th percentile?

  • approximately 977
  • approximately 1175
  • approximately 1247
  • approximately 1223

Answer

normally distributed

qnorm(p = 0.95, mean = 1100, sd = 75)
## [1] 1223.364

Question 4

Refer to the previous question. Brain volume for adult women is about 1,100 cc for women with a standard deviation of 75 cc. Consider the sample mean of 100 random adult women from this population. What is the 95th percentile of the distribution of that sample mean?

  • approximately 1112 cc
  • approximately 1115 cc
  • approximately 1110 cc
  • approximately 1088 cc

Answer

normally distributed

qnorm(p = 0.95, mean = 1100, sd = 75/sqrt(100))
## [1] 1112.336

Question 5

You flip a fair coin 5 times, about what’s the probability of getting 4 or 5 heads?

  • 12%
  • 19%
  • 6%
  • 3%

Answer

Binomial Distribution

4 out of 5 coin tosses should be heads:

\[p1 = {5 \choose 4} \cdot (1-p)^{5-4} \cdot (p)^4\]

5 out of 5 coin tosses should be heads:

\[p2 = {5 \choose 5} \cdot (1-p)^{5-5} \cdot (p)^5\]

Thus:

\[R = p1 + p2\]

p <- 0.5
n <- 5

P1 <- (factorial(5) / (factorial(5-4) * factorial(4))) * (1 - p)^(n-4) * (p)^(4)
P2 <- (factorial(5) / (factorial(5-5) * factorial(5))) * (1 - p)^(n-5) * (p)^(5)

R <- P1 + P2

R
## [1] 0.1875

Question 6

The respiratory disturbance index (RDI), a measure of sleep disturbance, for a specific population has a mean of 15 (sleep events per hour) and a standard deviation of 10. They are not normally distributed. Give your best estimate of the probability that a sample mean RDI of 100 people is between 14 and 16 events per hour?

  • 47.5%
  • 34%
  • 95%
  • 68%

Answer

not normally distributed

p1 <- pnorm((14-15)/(10/sqrt(100)), lower.tail = TRUE)

p2 <- pnorm((16-15)/(10/sqrt(100)), lower.tail = TRUE)

p2 - p1
## [1] 0.6826895

Question 7

Consider a standard uniform density. The mean for this density is .5 and the variance is \(1/12\). You sample 1,000 observations from this distribution and take the sample mean, what value would you expect it to be near?

  • 0.25
  • 0.10
  • 0.75
  • 0.5

Answer

Law of large numbers in action (LLN)

\[Var(X) = \sigma^2 = \frac{1}{12} = \frac{(b-a)^2}{12}\] \[mean = 0.5 =\frac{a+b}{2} \] \[b + a = 1\] \[b - a = 1\]

\(b = 1\) and \(a = 0\)

library(ggplot2)

set.seed(2022)

n <- c(10,50,75,100,300,500,1000,2000,5000,10000, 20000)
df <- data.frame("mean" = rep(NA, length(n)),"n" = rep(NA, length(n)), "var" = rep(NA, length(n)))

for (i in 1:length(n)) {
     df$mean[i] <- mean(runif(n = n[i], min = 0, max = 1))
     df$var[i] <- sd(runif(n = n[i], min = 0, max = 1))^2
     df$n[i] <- n[i]
}

ggplot(data = df, aes(x = n, y = mean)) +
    geom_line() +
    scale_y_continuous(limits = c(0.375, 0.625), breaks = seq(0.25, 0.75, by = 0.125))

Question 8

The number of people showing up at a bus stop is assumed to be Poisson with a mean of 5 people per hour. You watch the bus stop for 3 hours. About what’s the probability of viewing 10 or fewer people?

  • 0.12
  • 0.03
  • 0.08
  • 0.06

Answer

Poisson Distribution

ppois(q = 10,lambda = 5 * 3)
## [1] 0.1184644