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 (minutes)
sigma <- 30     # population standard deviation (minutes)
n <- 50         # sample size

# Sampling distribution parameters
mean_x <- mu
sd_x <- sigma / sqrt(n)

mean_x
## [1] 90
# [1] 90
sd_x
## [1] 4.242641
# [1] 4.242641

# Part B: Probability that the sample mean > 100 minutes
sample_mean <- 100

Part B

# Compute z-score
sample_z <- (sample_mean - mean_x) / sd_x
sample_z
## [1] 2.357023
# [1] 2.355

# Compute probability (area to the right of z = 2.355)
probability <- 1 - pnorm(sample_z)
probability
## [1] 0.009211063
# [1] 0.009322

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

# Given values
mu <- 12         # population mean (average selfies per ride)
sigma <- 5       # population standard deviation
n <- 16          # sample size

# Sampling distribution parameters
mean_x <- mu
sd_x <- sigma / sqrt(n)

mean_x
## [1] 12
# [1] 12
sd_x
## [1] 1.25
# [1] 1.25

Part B

sample_mean <- 15

# Compute z-score
sample_z <- (sample_mean - mean_x) / sd_x
sample_z
## [1] 2.4
# [1] 2.4

# Compute probability (area to the right of z = 2.4)
probability <- 1 - pnorm(sample_z)
probability
## [1] 0.008197536
# [1] 0.008198

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

# Given values
p <- 0.74      # population proportion (households that recycle regularly)
n <- 120       # sample size

# Sampling distribution parameters
mean_p <- p
sd_p <- sqrt(p * (1 - p) / n)

mean_p
## [1] 0.74
# [1] 0.74
sd_p
## [1] 0.04004164
# [1] 0.03999583

Part B

sample_p <- 90 / 120   # sample proportion = 0.75

# Compute z-score
sample_z <- (sample_p - mean_p) / sd_p
sample_z
## [1] 0.24974
# [1] 0.2501041

# Compute probability (area to the right of z = 0.25)
probability <- 1 - pnorm(sample_z)
probability
## [1] 0.4013942
# [1] 0.4012937

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

# Given values
mu <- 3.2       # population mean (meters)
sigma <- 2.1    # population standard deviation (meters)
n <- 64         # sample size

# Sampling distribution parameters
mean_x <- mu
sd_x <- sigma / sqrt(n)

mean_x
## [1] 3.2
# [1] 3.2
sd_x
## [1] 0.2625
# [1] 0.2625

Part B

sample_mean <- 3.5

# Compute z-score
sample_z <- (sample_mean - mean_x) / sd_x
sample_z
## [1] 1.142857
# [1] 1.142857

# Compute probability (area to the right of z = 1.142857)
probability <- 1 - pnorm(sample_z)
probability
## [1] 0.126549
# [1] 0.12654895

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

# Given values
mu <- 2.8       # population mean (hours per day)
sigma <- 1.1    # population standard deviation (hours)
n <- 50         # sample size

# Sampling distribution parameters
mean_x <- mu
sd_x <- sigma / sqrt(n)

mean_x
## [1] 2.8
# [1] 2.8
sd_x
## [1] 0.1555635
# [1] 0.1555635

Part B

sample_mean <- 3

# Compute z-score
sample_z <- (sample_mean - mean_x) / sd_x
sample_z
## [1] 1.285649
# [1] 1.285

# Compute probability (area to the right of z = 1.285)
probability <- 1 - pnorm(sample_z)
probability
## [1] 0.09928285
# [1] 0.09927499