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

Probability distributions: exam review Exam review: second paper Worked solutions

Section A: short questions

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

    x01234P(X = x)0.20.1k2k0.25

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

      • M1 for the probabilities adding to 1: 0.55 + 3k = 1
      • A1 k = 0.15

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

      • A1 0.55

    3. (c) Find \(P(X \leq 2 \mid X \geq 1)\). [2]

      • M1 for P(1 ≤ X ≤ 2) ÷ P(X ≥ 1) = 0.25 ÷ 0.8
      • A1 5/16 (0.312)

  2. The discrete random variable Y has probability distribution \(P(Y = y) = ky\) for \(y = 1, 2, 3, 4\), where k is a constant. [4]

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

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

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

      • M1 for \(P(Y \geq 3) \div P(Y \geq 2) = 0.7 \div 0.9\)
      • A1 7/9 (0.778)

  3. A fair spinner has 10 equal sectors numbered 1 to 10. It is spun in a game at a school stall. A score of 10 wins a prize of \$20. A score of 7, 8 or 9 wins a prize of \$5. 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): 20 × 1/10 + 5 × 3/10 + 0 × 3/5
      • A1 3.5

    2. (b) The game costs \$5 to play. Find the expected gain of a player and state whether the game is fair. [2]

      • M1 for 3.5 − 5
      • A1 −1.5, 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 3.5 (the cost equals the expected prize)

  4. A box holds 3 faulty bulbs and 7 working bulbs. Two bulbs are taken at random, without replacement. Let X be the number of faulty bulbs taken. [6]

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

      • M1 for P(X = 0) = 7/10 × 2/3
      • A1 for two correct probabilities
      • A1 7/15, 7/15, 1/15

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

      • M1 for 0 × 7/15 + 1 × 7/15 + 2 × 1/15
      • A1 3/5 (0.6)

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

      • A1 8/15 (0.533) (1 − 7/15)

  5. Each seed in a tray fails to germinate with probability 0.12, independently of the others. A tray holds 15 seeds. Let X be the number of seeds in a tray that fail to germinate. [6]

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

      • A1 X ~ B(15, 0.12)

    2. (b) Find the probability that exactly 3 seeds fail to germinate. [2]

      • M1 for binomial pdf with n = 15, p = 0.12, x = 3
      • A1 0.170

    3. (c) Find the probability that at most 4 seeds fail to germinate. [1]

      • A1 0.974 (binomial cdf with n = 15, p = 0.12, lower = 0, upper = 4)

    4. (d) Find the probability that at least 4 seeds fail to germinate. [2]

      • M1 for 1 − P(X ≤ 3), using binomial cdf with n = 15, p = 0.12, lower = 0, upper = 3
      • A1 0.0959

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

    1. (a) Find E(X) and Var(X) for \(X \sim B(25, 0.6)\). [2]

      • A1 E(X) = 25 × 0.6 = 15
      • A1 Var(X) = 25 × 0.6 × 0.4 = 6

    2. (b) \(Y \sim B(n, p)\) has mean 20 and variance 15. Find the value of p and the value of n. [3]

      • M1 for np = 20 and np(1 − p) = 15
      • M1 for 1 − p = 15 ÷ 20, so p = 0.25
      • A1 n = 80

    3. (c) Find \(P(Y > 20)\). [2]

      • M1 for 1 − P(Y ≤ 20), using binomial cdf with n = 80, p = 0.25, lower = 0, upper = 20
      • A1 0.440

  7. A quiz has 12 questions. Each question has 5 answers, only one of which is correct. A student passes by answering at least 5 questions correctly. Omar 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 (0.2) for every question

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

      • M1 for X ~ B(12, 0.2) and 1 − P(X ≤ 4), using binomial cdf with n = 12, p = 0.2, lower = 0, upper = 4
      • A1 0.0726

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

      • M1 for Y ~ B(15, 0.07256), where p is the answer to part (b)
      • A1 0.208 (binomial pdf with n = 15, p = 0.07256, x = 2)

  8. A company packs memory cards in boxes. Each card is faulty with probability 0.03, independently of the others. [8]

    1. (a) A box contains 15 cards. Find the probability that at least one card in the box is faulty. [2]

      • M1 for 1 − P(X = 0) with \(X \sim B(15, 0.03)\), or \(1 - 0.97^{15}\)
      • A1 0.367

    2. (b) A larger box contains n cards. Find the smallest value of n for which the probability that at least one card is faulty is greater than 0.9. [3]

      • M1 for \(1 - 0.97^n > 0.9\), that is \(0.97^n < 0.1\) (or a table of \(P(X \geq 1)\) for \(X \sim B(n, 0.03)\))
      • A1 for n = 75 gives 0.8982, which is too small, and n = 76 gives 0.9012
      • A1 n = 76

    3. (c) A theatre sells n tickets for a show with 80 seats. Each person who buys a ticket turns up with probability 0.9, independently of the others. Find the largest value of n for which the probability that more people turn up than there are seats is less than 0.05. [3]

      • M1 for \(X \sim B(n, 0.9)\) and \(P(X > 80) < 0.05\), that is \(1 - P(X \leq 80) < 0.05\)
      • A1 for n = 84 gives 0.0264, which is less than 0.05, and n = 85 gives 0.0643, which is not
      • A1 n = 84

  9. The life of a type of battery is normally distributed with mean 300 hours and variance 400 hours². [4]

    1. (a) Use the 68–95–99.7 rule to estimate the percentage of battery lives between 280 and 360 hours. [2]

      • M1 for 280 is 1 standard deviation below the mean and 360 is 3 above: 34% + 49.85%
      • A1 83.85%

    2. (b) Use your calculator to find the probability that a battery's life is longer than 330 hours. [2]

      • M1 for normal cdf with lower = 330, upper = 10^99, μ = 300, σ = 20
      • A1 0.0668

  10. The heights of seedlings in a nursery are normally distributed with mean 120 mm and standard deviation 8.5 mm. [9]

    1. (a) Find the symmetric interval about the mean that contains the heights of the middle 80% of the seedlings. [3]

      • M1 for 10% of the seedlings lie below the interval and 10% above it
      • M1 for inverse normal with area 0.1 and with area 0.9, μ = 120, σ = 8.5
      • A1 109 mm to 131 mm

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

      • M1 for inverse normal with area 0.25 and with area 0.75, μ = 120, σ = 8.5
      • A1 114 mm and 126 mm

    3. (c) The heights of seedlings in a second nursery are normally distributed with mean \(\mu\) mm and standard deviation \(\sigma\) mm. In this nursery 5% of the seedlings have height less than 100 mm and 15% have height greater than 130 mm. Find the value of \(\mu\) and the value of \(\sigma\). [4]

      • M1 for the standardised values \(\frac{100 - \mu}{\sigma} = -1.6449\) and \(\frac{130 - \mu}{\sigma} = 1.0364\)
      • M1 for two equations in \(\mu\) and \(\sigma\), for example \(100 = \mu - 1.6449\sigma\) and \(130 = \mu + 1.0364\sigma\)
      • A1 \(\sigma = 11.2\)
      • A1 \(\mu = 118\)

Section B: long questions

  1. The time that a bus takes to travel between two towns is normally distributed with mean 35 minutes and variance 25 minutes². The diagram shows the distribution, with the journeys shorter than 30 minutes shaded. [14]

    152025303540455055Time (minutes)

    1. (a) Find the probability that a journey takes less than 30 minutes. This is the shaded area in the diagram. [2]

      • M1 for σ = 5 (the square root of 25) and normal cdf with lower = −10^99, upper = 30, μ = 35, σ = 5
      • A1 0.159

    2. (b) Find the probability that a journey is between 32 and 41 minutes. [2]

      • M1 for normal cdf with lower = 32, upper = 41, μ = 35, σ = 5
      • A1 0.611

    3. (c) The fastest 8% of journeys win a prize for the driver. Find the greatest time of a journey that wins a prize. [2]

      • M1 for inverse normal with area = 0.08, μ = 35, σ = 5
      • A1 28.0

    4. (d) The bus makes 600 journeys in a month. Find the expected number that take more than 45 minutes. [2]

      • M1 for 600 × P(X > 45), using normal cdf with lower = 45, upper = 10^99, μ = 35, σ = 5 = 0.02275
      • A1 13.7

    5. (e) Twelve journeys are chosen at random. Find the probability that exactly 3 journeys take less than 30 minutes. [3]

      • M1 for the number Y ~ B(12, 0.1587), with p from part (a)
      • M1 for binomial pdf with n = 12, p = 0.1587, x = 3
      • A1 0.186

    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) A short route has journey times with mean 6 minutes and standard deviation 4 minutes. Explain why a normal distribution is not a good model for the journey time on this route. [2]

      • M1 for comparing the mean with 2 standard deviations: 6 < 8
      • R1 a normal curve with these values would give a large probability of negative values, which cannot happen

  2. At a charity event, a player turns a wheel and receives X dollars, where X is the number the wheel shows. The probability distribution of X is shown in the table. [13]

    x1234P(X = x)0.2ab0.25

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

      • A1 0.2 + a + b + 0.25 = 1, so a + b = 0.55

    2. (b) Given that E(X) = 2.6, find the value of a and the value of b. [3]

      • M1 for 1 × 0.2 + 2a + 3b + 4 × 0.25 = 2.6
      • M1 for solving the two equations together
      • A1 a = 0.25, b = 0.3

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

      • M1 for P(X ≥ 3) ÷ P(X ≥ 2) = 0.55 ÷ 0.8
      • A1 11/16 (0.688)

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

      • M1 for 2.6 − 3
      • A1 −0.4, which is not 0, so the game is not fair

    5. (e) The wheel is turned 10 times. Find the probability that exactly 4 of the 10 scores are 4. [3]

      • M1 for Y ~ B(10, 0.25)
      • M1 for binomial pdf with n = 10, p = 0.25, x = 4
      • A1 0.146

    6. (f) Find the probability that at least 2 of the 10 scores are 4. [2]

      • M1 for 1 − P(Y ≤ 1), using binomial cdf with n = 10, p = 0.25, lower = 0, upper = 1
      • A1 0.756

  3. The time that a student takes to finish an online puzzle, in minutes, is modelled by \(X \sim N(38, 6^2)\). [13]

    15202530354045505560Time (minutes)

    1. (a) The box plot shows the times of a random sample of 150 students. 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 = 38, σ = 6, or area = 0.75
      • A1 Q1 = 34.0, Q3 = 42.0

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

      • A1 8.09 (from the unrounded quartiles; accept 8 from the rounded quartiles)

    4. (d) The slowest 8% of students are given an extra hint. Find the time that separates these from the others. [2]

      • M1 for inverse normal with area = 0.92, μ = 38, σ = 6
      • A1 46.4

    5. (e) Find the value of k such that \(P(38 < X < k) = 0.35\). [3]

      • M1 for P(X < k) = 0.5 + 0.35 = 0.85
      • M1 for inverse normal with area = 0.85, μ = 38, σ = 6
      • A1 44.2

    6. (f) A student scores 10 points for a time under 30 minutes, 6 points for a time between 30 and 45 minutes and 2 points for a time over 45 minutes. Find the expected score for one student. [3]

      • M1 for normal cdf with lower = −10^99, upper = 30, μ = 38, σ = 6 = 0.09121, normal cdf with lower = 30, upper = 45, μ = 38, σ = 6 = 0.7871, normal cdf with lower = 45, upper = 10^99, μ = 38, σ = 6 = 0.1217
      • M1 for 10 × 0.09121 + 6 × 0.7871 + 2 × 0.1217
      • A1 5.88

  4. The number of customers who arrive at a bicycle repair stall is modelled by a Poisson distribution X with a constant average rate of 7.5 customers per hour. [10]

    1. (a) State two conditions that must hold for X to be modelled by a Poisson distribution, in the context of customers arriving at the stall. [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 9 customers arrive in one hour. [2]

      • M1 for Poisson pdf with \(\lambda = 7.5\), x = 9
      • A1 0.114

    3. (c) Find the probability that more than 10 customers arrive in one hour. [2]

      • M1 for 1 − P(X ≤ 10), using Poisson cdf with \(\lambda = 7.5\), lower = 0, upper = 10
      • A1 0.138

    4. (d) Find the probability that at least 3 customers arrive in a period of 20 minutes. [3]

      • M1 for rescaling the mean: 7.5 × 20/60 = 2.5, so \(\lambda = 2.5\)
      • M1 for 1 − P(Y ≤ 2), using Poisson cdf with \(\lambda = 2.5\), lower = 0, upper = 2
      • A1 0.456

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

      • A1 2.74 (\(\sqrt{7.5}\), because Var(X) = 7.5)

  5. Two ticket machines at a station are used independently. The number of tickets per hour sold by machine A is modelled by \(X \sim \text{Po}(6)\). Machine B sells tickets at an average of 5 per 20 minutes, and the number it sells is modelled by a Poisson distribution Y. X and Y are assumed to be independent. [10]

    1. (a) Show that the mean number of tickets per hour at machine B is 15. [1]

      • A1 5 per 20 minutes, so 5 × 3 = 15 per hour

    2. (b) Write down the distribution of Z, the total number of tickets per hour at machine A and machine B together. [2]

      • M1 for adding the means: 6 + 15
      • A1 Z ~ Po(21)

    3. (c) Find \(P(Z \leq 17)\). [2]

      • M1 for Poisson cdf with \(\lambda = 21\), lower = 0, upper = 17
      • A1 0.227

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

      • A1 E(Z) = 21
      • A1 standard deviation = \(\sqrt{21}\) = 4.58

    5. (e) Explain why the number of tickets at machine A and the number at machine B 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 tickets 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 orders that arrive at a bakery counter is modelled by a Poisson distribution with a constant average rate of 2.8 orders per hour. [10]

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

      • M1 for the probabilities of neighbouring values: \(P(X = 1) = 0.1703\), \(P(X = 2) = 0.2384\), \(P(X = 3) = 0.2225\)
      • A1 \(P(X = 2)\) is the largest of these
      • R1 for concluding that the most likely number is 2

    2. (b) On a quiet day the number of orders per hour has mean \(\lambda\). The probability that at least one order arrives in an hour is 0.7. Find the value of \(\lambda\). [3]

      • M1 for \(P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-\lambda} = 0.7\) (or solving \(P(X \geq 1) = 0.7\) with the solver)
      • M1 for solving, for example \(e^{-\lambda} = 0.3\)
      • A1 1.2

    3. (c) On a Friday the counter is open for 4 hours. Find the smallest number of loaves the bakery must bake so that the probability of running out is at most 0.05, if each order uses one loaf. [4]

      • M1 for rescaling the mean: 2.8 × 4 = 11.2, so \(\lambda = 11.2\)
      • M1 for \(P(Y \leq s) \geq 0.95\) (equivalently \(P(Y > s) \leq 0.05\)) using Poisson cdf with \(\lambda = 11.2\)
      • A1 for s = 16 gives 0.9364, which is too small, and s = 17 gives 0.9628
      • A1 17

jmaths.xyz · AI HL · 13.2.15 Unit review