Problem 1: Commuting by Tube.

Transport for London reports that 68% of commuters use the Underground at least once per week.
You take a random sample of \(n = 120\) London residents.

Part A

We are dealing with sample proportions, since each person either uses or does not use the Underground weekly. The 68% is a clue (with proportion p=0.68). So the CLT for proportions is the way to go. We get that our sampling distribution should follow the normal model N(0.68, sqrt(0.68*0.32/120)), which rounds to N(0.68, 0.043). The calculations are below:

p <- 0.68          # population proportion
n <- 120           # sample size

mean_p <- p
sd_p <- sqrt(p*(1-p)/n)

mean_p
## [1] 0.68
sd_p
## [1] 0.04258325

Part B

The probability that there are fewer than 75 out of 120 sampled residents using the underground can be found by calculating the z-score 75/120 = 0.625. We then calculate the probability of a sample having a z-score as small, or smaller, than 0.625. It turns out that the z-score is -1.291588, which we round to -1.29. The probability that a random sample has a z-score less than -1.29 is equal to 0.09852533. The relevant calculations are below.

sample_p <- 75/120
sample_z <- (sample_p - mean_p)/sd_p

sample_p
## [1] 0.625
sample_z
## [1] -1.291588
pnorm(-1.29, mean=0, sd=1)
## [1] 0.09852533

Therefore: the probability that fewer than 75 of the sampled residents use the Underground weekly is equal to 0.09852533, or roughly 9.9%.

Problem 2: British Museum Visit Times.

Tourist time in the British Museum (minutes) has population mean \(\mu = 90\) and standard deviation \(\sigma = 30\).
We take a random sample of \(n = 50\).

Part A

# Given values
mu <- 90        # population mean
sigma <- 30     # population standard deviation
n <- 50         # sample size

# Central Limit Theorem
mu_xbar <- mu                      # mean of sampling distribution
sigma_xbar <- sigma / sqrt(n)      # standard error (standard deviation of sample mean)

# Display results
cat("Mean of sampling distribution (mu_xbar):", mu_xbar, "\n")
## Mean of sampling distribution (mu_xbar): 90
cat("Standard error (sigma_xbar):", sigma_xbar, "\n")
## Standard error (sigma_xbar): 4.242641
# If you want to define the sampling distribution as approximately normal:
# X̄ ~ N(90, 4.24)
cat("Sampling distribution: Xbar ~ N(", mu_xbar, ", ", round(sigma_xbar, 2), ")\n", sep = "")
## Sampling distribution: Xbar ~ N(90, 4.24)

Part B

mu <- 90
sigma <- 30
n <- 50
sigma_xbar <- sigma / sqrt(n)

# Probability sample mean > 100
p_gt_100 <- 1 - pnorm(100, mean = mu, sd = sigma_xbar)
p_gt_100
## [1] 0.009211063

Problem 3: The London Eye.

The number of selfies taken per capsule ride is approximately normal with mean \(\mu = 12\) and standard deviation \(\sigma = 5\).
We take a random sample of \(n = 16\) capsule rides and consider the average number of selfies taken in this sample.

Part A

Part B

Problem 4: Recycling Habits.

In the UK, about \(p = 0.74\) of households report recycling regularly.
Suppose you randomly sample \(n = 120\) households.

Part A

Part B

Problem 5: Bumblebee Flight Distances.

A scientist finds that bumblebee flight distances between flowers, in meters, follow a normal distribution with mean \(\mu = 3.2\) and standard deviation \(\sigma = 2.1\).
Suppose \(n = 64\).

Part A

Part B

Problem 6: Social Media Use.

Among university students, the population mean time on social media per day is \(\mu = 2.8\) hours with standard deviation \(\sigma = 1.1\) hours.
Let \(n = 50\).

Part A

Part B