Distributions review Problem set A One question at a time

The random variable X is the score when a spinner is spun. The table shows its probability distribution.

1. (a) Find the value of k.

   • The probabilities add to 1: 0.16 + k + 0.32 + 3k = 1
   • 0.48 + 4k = 1, so k = 0.13

2. (b) Find E(X).

   • \(E(X) = 1 \times 0.16 + 2 \times 0.13 + 3 \times 0.32 + 4 \times 0.39 = 2.94\)

3. (c) Find P(X ≥ 2).

   • P(X ≥ 2) = 0.13 + 0.32 + 0.39 = 0.84

A student guesses the answers to 40 true-or-false questions. Each guess is correct with probability 0.5. Let Y be the number correct, so Y follows a binomial distribution. Write down E(Y).

• \(Y \sim B(40, 0.5)\)
• \(E(Y) = np = 40 \times 0.5 = 20\)

A seed germinates with probability 0.35. Twelve seeds are planted. Let Y be the number that germinate, so \(Y \sim B(12, 0.35)\).

1. (a) Find the probability of exactly 5.

   • \(Y \sim B(12, 0.35)\)
   • From the calculator: binomial pdf with n = 12, p = 0.35, x = 5: 0.204

2. (b) Find the probability of at least 5.

   • \(P(Y \geq 5) = 1 - P(Y \leq 4)\)
   • From the calculator: binomial cdf with n = 12, p = 0.35, upper = 4: 0.5833
   • 1 − 0.5833 = 0.417

The mass of a pear is normally distributed with mean 160 g and standard deviation 12 g. Find the probability that a pear has mass between 150 g and 175 g.

• \(X \sim N(160, 12^2)\)
• From the calculator: normal cdf with lower = 150, upper = 175, μ = 160, σ = 12
• P = 0.692

The time a bus takes to complete a route is normally distributed with mean 42 minutes and standard deviation 5 minutes. A day has 80 journeys.

1. (a) Find the probability that a journey takes between 38 and 50 minutes.

   • From the calculator: normal cdf with lower = 38, upper = 50, μ = 42, σ = 5
   • P = 0.733

2. (b) Find the expected number of journeys in a day that take less than 35 minutes.

   • From the calculator: normal cdf with lower = −10^99, upper = 35, μ = 42, σ = 5: 0.08076
   • \(80 \times 0.08076 = 6.46\)
   • About 6

The height of a species of tree is normally distributed with mean 18.0 m and standard deviation 2.5 m. The tallest 12% of trees are marked for study. Find the least height of a marked tree.

• P(X > k) = 0.12, so P(X < k) = 0.88
• From the calculator: inverse normal with area = 0.88, μ = 18, σ = 2.5
• k = 20.9 m

The journey time to school is normally distributed with mean 25 minutes and standard deviation 4 minutes. A journey is late if it takes more than 32 minutes. Over 8 days, let Y be the number of late journeys.

1. (a) Find the probability that one journey is late.

   • From the calculator: normal cdf with lower = 32, upper = 10^99, μ = 25, σ = 4
   • p = 0.0401

2. (b) Find P(Y = 2).

   • \(Y \sim B(8, 0.04006)\)
   • From the calculator: binomial pdf with n = 8, p = 0.04006, x = 2: 0.0352

A basketball player scores a free throw with probability 0.7. She takes 15 free throws, independently. Let Y be the number she scores.

1. (a) Write down the expected number of free throws scored.

   • \(np = 15 \times 0.7 = 10.5\)

2. (b) Find the probability that exactly 12 are scored.

   • From the calculator: binomial pdf with n = 15, p = 0.7, x = 12: 0.170

3. (c) Find the probability that at least 12 are scored.

   • From the calculator: binomial cdf with n = 15, p = 0.7, upper = 11: 0.7031
   • 1 − 0.7031 = 0.297