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IB Mathematics: Applications and Interpretation HL · 13.2: Probability Distributions · 13.2.15

Probability distributions: exam review Exam review One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Section A: short questions

  1. The discrete random variable X has the probability distribution shown in the table, where k is a constant. [5]

    x12345P(X = x)0.05k2k0.250.1

    1. (a) Find the value of k. [2]

      • M1 for the probabilities adding to 1: 0.4 + 3k = 1
      • A1 k = 0.2

    2. (b) Find \(P(X \geq 4)\). [1]

      • A1 0.35

    3. (c) Find \(P(X \leq 3 \mid X \geq 2)\). [2]

      • M1 for P(2 ≤ X ≤ 3) ÷ P(X ≥ 2) = 0.6 ÷ 0.95
      • A1 12/19 (0.632)

  2. The discrete random variable X has probability distribution \(P(X = x) = k(5 - x)\) for \(x = 1, 2, 3, 4\), where k is a constant. [4]

    1. (a) Find the value of k. [2]

      • M1 for the probabilities adding to 1: \(4k + 3k + 2k + k = 1\)
      • A1 k = 0.1

    2. (b) Find \(P(X \geq 2 \mid X \leq 3)\). [2]

      • M1 for \(P(2 \leq X \leq 3) \div P(X \leq 3) = 0.5 \div 0.9\)
      • A1 5/9 (0.556)

  3. A fair six-sided die is rolled in a game at a fair. A score of 6 wins a prize of \$12. A score of 4 or 5 wins a prize of \$3. Any other score wins nothing. Let X be the prize in dollars. [5]

    1. (a) Find the expected prize. [2]

      • M1 for Σ x × P(X = x): 12 × 1/6 + 3 × 1/3 + 0 × 1/2
      • A1 3

    2. (b) The game costs \$4 to play. Find the expected gain of a player and state whether the game is fair. [2]

      • M1 for 3 − 4
      • A1 −1, which is not 0, so the game is not fair

    3. (c) Find the cost to play that would make the game fair. [1]

      • A1 3 (the cost equals the expected prize)

  4. A bag holds 4 red sweets and 5 white sweets. Two sweets are taken at random, without replacement. Let X be the number of red sweets taken. [6]

    1. (a) Find the probability distribution of X, as a table. [3]

      • M1 for P(X = 0) = 5/9 × 1/2
      • A1 for two correct probabilities
      • A1 5/18, 5/9, 1/6

    2. (b) Find E(X). [2]

      • M1 for 0 × 5/18 + 1 × 5/9 + 2 × 1/6
      • A1 8/9 (0.889)

    3. (c) Find \(P(X \geq 1)\). [1]

      • A1 13/18 (0.722) (1 − 5/18)

  5. Each bulb made by a factory is faulty with probability 0.08, independently of the others. A box holds 20 bulbs. Let X be the number of faulty bulbs in a box. [6]

    1. (a) Write down the distribution of X in the form \(X \sim B(n, p)\). [1]

      • A1 X ~ B(20, 0.08)

    2. (b) Find the probability that exactly 2 bulbs are faulty. [2]

      • M1 for binomial pdf with n = 20, p = 0.08, x = 2
      • A1 0.271

    3. (c) Find the probability that at most 3 bulbs are faulty. [1]

      • A1 0.929 (binomial cdf with n = 20, p = 0.08, lower = 0, upper = 3)

    4. (d) Find the probability that at least 3 bulbs are faulty. [2]

      • M1 for 1 − P(X ≤ 2), using binomial cdf with n = 20, p = 0.08, lower = 0, upper = 2
      • A1 0.212

  6. Some random variables are modelled by binomial distributions. [7]

    1. (a) Find E(X) and Var(X) for \(X \sim B(40, 0.35)\). [2]

      • A1 E(X) = 40 × 0.35 = 14
      • A1 Var(X) = 40 × 0.35 × 0.65 = 9.1

    2. (b) \(Y \sim B(n, p)\) has mean 18 and variance 12.6. Find the value of p and the value of n. [3]

      • M1 for np = 18 and np(1 − p) = 12.6
      • M1 for 1 − p = 12.6 ÷ 18, so p = 0.3
      • A1 n = 60

    3. (c) Find \(P(Y > 18)\). [2]

      • M1 for 1 − P(Y ≤ 18), using binomial cdf with n = 60, p = 0.3, lower = 0, upper = 18
      • A1 0.437

  7. A multiple-choice test has 10 questions. Each question has 4 answers, only one of which is correct. A student passes by answering at least 5 questions correctly. Mia guesses the answer to every question. [6]

    1. (a) State two assumptions that are needed to model the number of correct answers as a binomial distribution. [2]

      • R1 each answer is guessed independently of the others
      • R1 the probability of a correct guess is the same (0.25) for every question

    2. (b) Find the probability that a student who guesses every answer passes. [2]

      • M1 for X ~ B(10, 0.25) and 1 − P(X ≤ 4), using binomial cdf with n = 10, p = 0.25, lower = 0, upper = 4
      • A1 0.0781

    3. (c) Twelve students each guess every answer. Assuming each student passes independently, find the probability that exactly 2 students pass. [2]

      • M1 for Y ~ B(12, 0.07813), where p is the answer to part (b)
      • A1 0.179 (binomial pdf with n = 12, p = 0.07813, x = 2)

  8. A factory packs light bulbs in boxes. Each bulb is faulty with probability 0.05, independently of the others. [8]

    1. (a) A box contains 12 bulbs. Find the probability that at least one bulb in the box is faulty. [2]

      • M1 for 1 − P(X = 0) with \(X \sim B(12, 0.05)\), or \(1 - 0.95^{12}\)
      • A1 0.46

    2. (b) A larger box contains n bulbs. Find the smallest value of n for which the probability that at least one bulb is faulty is greater than 0.95. [3]

      • M1 for \(1 - 0.95^n > 0.95\), that is \(0.95^n < 0.05\) (or a table of \(P(X \geq 1)\) for \(X \sim B(n, 0.05)\))
      • A1 for n = 58 gives 0.9490, which is too small, and n = 59 gives 0.9515
      • A1 n = 59

    3. (c) An airline sells n tickets for a flight with 60 seats. Each passenger who buys a ticket turns up with probability 0.92, independently of the others. Find the largest value of n for which the probability that more passengers turn up than there are seats is less than 0.05. [3]

      • M1 for \(X \sim B(n, 0.92)\) and \(P(X > 60) < 0.05\), that is \(1 - P(X \leq 60) < 0.05\)
      • A1 for n = 62 gives 0.0363, which is less than 0.05, and n = 63 gives 0.1111, which is not
      • A1 n = 62

  9. The time that a player takes to finish a puzzle is normally distributed with mean 240 seconds and variance 225 seconds². [4]

    1. (a) Use the 68–95–99.7 rule to estimate the percentage of player times between 210 and 255 seconds. [2]

      • M1 for 210 is 2 standard deviations below the mean and 255 is 1 above: 47.5% + 34%
      • A1 81.5%

    2. (b) Use your calculator to find the probability that a player's time is longer than 270 seconds. [2]

      • M1 for normal cdf with lower = 270, upper = 10^99, μ = 240, σ = 15
      • A1 0.0228

  10. The masses of eggs laid on a farm are normally distributed with mean 58 g and standard deviation 3.2 g. [9]

    1. (a) Find the symmetric interval about the mean that contains the masses of the middle 90% of the eggs. [3]

      • M1 for 5% of the eggs lie below the interval and 5% above it
      • M1 for inverse normal with area 0.05 and with area 0.95, μ = 58, σ = 3.2
      • A1 52.7 g to 63.3 g

    2. (b) Find the lower quartile and the upper quartile of the masses. [2]

      • M1 for inverse normal with area 0.25 and with area 0.75, μ = 58, σ = 3.2
      • A1 55.8 g and 60.2 g

    3. (c) The masses of eggs from a second farm are normally distributed with mean \(\mu\) g and standard deviation \(\sigma\) g. On this farm 10% of the eggs have mass less than 54 g and 20% have mass greater than 62 g. Find the value of \(\mu\) and the value of \(\sigma\). [4]

      • M1 for the standardised values \(\frac{54 - \mu}{\sigma} = -1.2816\) and \(\frac{62 - \mu}{\sigma} = 0.8416\)
      • M1 for two equations in \(\mu\) and \(\sigma\), for example \(54 = \mu - 1.2816\sigma\) and \(62 = \mu + 0.8416\sigma\)
      • A1 \(\sigma = 3.77\)
      • A1 \(\mu = 58.8\)

Section B: long questions

  1. The mass of a loaf of bread made by a bakery is normally distributed with mean 500 g and variance 144 g². The diagram shows the distribution, with the loaves lighter than 484 g shaded. [14]

    460480500520540Mass (g)

    1. (a) Find the probability that a loaf weighs less than 484 g. This is the shaded area in the diagram. [2]

      • M1 for σ = 12 (the square root of 144) and normal cdf with lower = −10^99, upper = 484, μ = 500, σ = 12
      • A1 0.0912

    2. (b) Find the probability that a loaf is between 480 and 510 g. [2]

      • M1 for normal cdf with lower = 480, upper = 510, μ = 500, σ = 12
      • A1 0.750

    3. (c) The bakery rejects the lightest 4% of loaves. Find the least mass of a loaf that is not rejected. [2]

      • M1 for inverse normal with area = 0.04, μ = 500, σ = 12
      • A1 479

    4. (d) The bakery makes 800 loaves in a day. Find the expected number that weigh more than 520 g. [2]

      • M1 for 800 × P(X > 520), using normal cdf with lower = 520, upper = 10^99, μ = 500, σ = 12 = 0.04779
      • A1 38.2

    5. (e) Ten loaves are chosen at random. Find the probability that exactly 2 loaves weigh less than 484 g. [3]

      • M1 for the number Y ~ B(10, 0.09121), with p from part (a)
      • M1 for binomial pdf with n = 10, p = 0.09121, x = 2
      • A1 0.174

    6. (f) State one assumption that you made in part (e). [1]

      • R1 the items are independent of each other
      • R1 the probability that an item is in the shaded region is the same for each item

    7. (g) The bakery also makes small rolls. Their mass has mean 40 g and standard deviation 22 g. Explain why a normal distribution is not a good model for the mass of a roll. [2]

      • M1 for comparing the mean with 2 standard deviations: 40 < 44
      • R1 a normal curve with these values would give a large probability of negative values, which cannot happen

  2. At a fair, a player spins a spinner and receives X dollars, where X is the score. The probability distribution of X is shown in the table. [13]

    x1234P(X = x)0.15ab0.35

    1. (a) Write down an equation connecting a and b. [1]

      • A1 0.15 + a + b + 0.35 = 1, so a + b = 0.5

    2. (b) Given that E(X) = 2.75, find the value of a and the value of b. [3]

      • M1 for 1 × 0.15 + 2a + 3b + 4 × 0.35 = 2.75
      • M1 for solving the two equations together
      • A1 a = 0.3, b = 0.2

    3. (c) Find \(P(X \geq 3 \mid X \geq 2)\). [2]

      • M1 for P(X ≥ 3) ÷ P(X ≥ 2) = 0.55 ÷ 0.85
      • A1 11/17 (0.647)

    4. (d) A spin costs \$3. Find the expected profit of a player and state whether the game is fair. [2]

      • M1 for 2.75 − 3
      • A1 −0.25, which is not 0, so the game is not fair

    5. (e) The spinner is spun 8 times. Find the probability that exactly 3 of the 8 scores are 4. [3]

      • M1 for Y ~ B(8, 0.35)
      • M1 for binomial pdf with n = 8, p = 0.35, x = 3
      • A1 0.279

    6. (f) Find the probability that at least 2 of the 8 scores are 4. [2]

      • M1 for 1 − P(Y ≤ 1), using binomial cdf with n = 8, p = 0.35, lower = 0, upper = 1
      • A1 0.831

  3. The time that a courier takes to make a delivery, in minutes, is modelled by \(X \sim N(42, 8^2)\). [13]

    20253035404550556065Time (minutes)

    1. (a) The box plot shows the times of a random sample of 200 deliveries. Give two features of the box plot that suggest a normal distribution is a suitable model. [2]

      • R1 the median is in the middle of the box (the box is symmetric)
      • R1 the whiskers are of equal length, and there are no outliers

    2. (b) Find the lower quartile and the upper quartile of X. [2]

      • M1 for inverse normal with area = 0.25, μ = 42, σ = 8, or area = 0.75
      • A1 Q1 = 36.6, Q3 = 47.4

    3. (c) Hence find the interquartile range. [1]

      • A1 10.8 (from the unrounded quartiles)

    4. (d) The slowest 6% of deliveries are called very late. Find the time that separates these from the others. [2]

      • M1 for inverse normal with area = 0.94, μ = 42, σ = 8
      • A1 54.4

    5. (e) Find the value of k such that \(P(42 < X < k) = 0.4\). [3]

      • M1 for P(X < k) = 0.5 + 0.4 = 0.9
      • M1 for inverse normal with area = 0.9, μ = 42, σ = 8
      • A1 52.3

    6. (f) The courier is paid \$6 for a delivery that takes less than 35 minutes, \$4 for a delivery that takes between 35 and 50 minutes and \$2 for a delivery that takes more than 50 minutes. Find the expected pay for one delivery. [3]

      • M1 for normal cdf with lower = −10^99, upper = 35, μ = 42, σ = 8 = 0.1908, normal cdf with lower = 35, upper = 50, μ = 42, σ = 8 = 0.6506, normal cdf with lower = 50, upper = 10^99, μ = 42, σ = 8 = 0.1587
      • M1 for 6 × 0.1908 + 4 × 0.6506 + 2 × 0.1587
      • A1 4.06

  4. The number of phone calls that arrive at a school office is modelled by a Poisson distribution X with a constant average rate of 9 calls per hour. [10]

    1. (a) State two conditions that must hold for X to be modelled by a Poisson distribution, in the context of phone calls. [2]

      • A1 events are independent of each other (in context)
      • A1 events occur at a constant average rate (or occur singly)

    2. (b) Find the probability that exactly 7 calls arrive in one hour. [2]

      • M1 for Poisson pdf with \(\lambda = 9\), x = 7
      • A1 0.117

    3. (c) Find the probability that more than 12 calls arrive in one hour. [2]

      • M1 for 1 − P(X ≤ 12), using Poisson cdf with \(\lambda = 9\), lower = 0, upper = 12
      • A1 0.124

    4. (d) Find the probability that at least 2 calls arrive in a period of 10 minutes. [3]

      • M1 for rescaling the mean: 9 × 10/60 = 1.5, so \(\lambda = 1.5\)
      • M1 for 1 − P(Y ≤ 1), using Poisson cdf with \(\lambda = 1.5\), lower = 0, upper = 1
      • A1 0.442

    5. (e) Find the standard deviation of X. [1]

      • A1 3 (\(\sqrt{9}\), because Var(X) = 9)

  5. Cars enter a car park by two separate entrances. The number of cars per hour that enter by the north gate is modelled by \(X \sim \text{Po}(8)\). The number of cars that enter by the south gate averages 3 per 15 minutes, and is modelled by a Poisson distribution Y. X and Y are assumed to be independent. [10]

    1. (a) Show that the mean number of cars per hour at the south gate is 12. [1]

      • A1 3 per 15 minutes, so 3 × 4 = 12 per hour

    2. (b) Write down the distribution of Z, the total number of cars per hour at the north gate and the south gate together. [2]

      • M1 for adding the means: 8 + 12
      • A1 Z ~ Po(20)

    3. (c) Find \(P(Z \leq 15)\). [2]

      • M1 for Poisson cdf with \(\lambda = 20\), lower = 0, upper = 15
      • A1 0.157

    4. (d) Find E(Z) and the standard deviation of Z. [2]

      • A1 E(Z) = 20
      • A1 standard deviation = \(\sqrt{20}\) = 4.47

    5. (e) Explain why the number of cars at the north gate and the number at the south gate may not be independent. [1]

      • R1 a sensible link in the context (the same cause affects both counts)

    6. (f) Hana says that T, the time in minutes between one of the cars and the next, follows a Poisson distribution. Explain why she is wrong. [2]

      • R1 a Poisson variable counts the number of events in a given period
      • R1 T measures a time between events (so it is not a count)

  6. The number of requests that arrive at a school IT help desk is modelled by a Poisson distribution with a constant average rate of 3.4 requests per hour. [10]

    1. (a) Find the most likely number of requests in one hour. Justify your answer. [3]

      • M1 for the probabilities of neighbouring values: \(P(X = 2) = 0.1929\), \(P(X = 3) = 0.2186\), \(P(X = 4) = 0.1858\)
      • A1 \(P(X = 3)\) is the largest of these
      • R1 for concluding that the most likely number is 3

    2. (b) In the school holidays the number of requests per hour has mean \(\lambda\). The probability that at least one request arrives in an hour is 0.6. Find the value of \(\lambda\). [3]

      • M1 for \(P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-\lambda} = 0.6\) (or solving \(P(X \geq 1) = 0.6\) with the solver)
      • M1 for solving, for example \(e^{-\lambda} = 0.4\)
      • A1 0.916

    3. (c) On a Saturday the desk is open for 5 hours. Find the smallest number of printer cartridges the desk must hold so that the probability of running out is at most 0.05, if each request uses one cartridge. [4]

      • M1 for rescaling the mean: 3.4 × 5 = 17, so \(\lambda = 17\)
      • M1 for \(P(Y \leq s) \geq 0.95\) (equivalently \(P(Y > s) \leq 0.05\)) using Poisson cdf with \(\lambda = 17\)
      • A1 for s = 23 gives 0.9367, which is too small, and s = 24 gives 0.9594
      • A1 24

jmaths.xyz · AI HL · 13.2.15 Unit review