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IB Mathematics: Applications and Interpretation HL · AHL 8: Statistics 2 · AHL 8.9

Mixed practice: normal combinations, sample means and the central limit theorem Problem set B Worked solutions

Skills

  1. X and Y are independent, \(X\sim N(18,\,7)\) and \(Y\sim N(45,\,13)\). \(X_1,X_2\) are independent observations of X and \(Y_1,Y_2,Y_3\) are independent observations of Y. State the distribution of each.

    1. (a) \(W=3X+4Y-6\)

      • Mean: \(E(W)\) = 3 × 18 + 4 × 45 − 6 = 228
      • Variance: \(Var(W)\) = 3² × 7 + 4² × 13 = 271

    2. (b) \(W=X_1-X_2\)

      • Mean: 18 − 18 = 0
      • Variance: 1² × 7 + (−1)² × 7 = 14

    3. (c) \(W=Y_1+Y_2+Y_3\)

      • Mean: 3 × 45 = 135
      • Variance: 3 × 13 = 39 (three observations, not 3²)

  2. The mass of an item is a normal variable. Samples of independent items are taken and \(\bar X\) is the sample mean.

    1. (a) The mass has distribution \(N(120,\,30)\). State the distribution of \(\bar X\) for a sample of 5 items.

      • \(Var(\bar{X})\) = 30 ÷ 5 = 6

    2. (b) A second item has mass with mean 8.5 and standard deviation 1.2. Find the mean and the variance of the average mass of a batch of 15 such items.

      • Variance of one item = 1.2² = 1.44
      • Variance of the average = 1.44 ÷ 15 = 0.096

  3. A random variable X has mean 70 and standard deviation 16. The distribution of X is not known. Give the approximate distribution of each quantity, or say that it cannot be found.

    1. (a) \(\bar X\) for a sample of 36 observations

      • n = 36 > 30, so the central limit theorem applies
      • Variance = 16² ÷ 36 = 256 ÷ 36

    2. (b) \(\bar X\) for a sample of 10 observations

      • n = 10 is not more than 30 and X is not known to be normal

    3. (c) \(\sum_{i=1}^{50} X_i\)

      • n = 50 > 30, so the central limit theorem applies
      • Mean = 50 × 70 = 3500; variance = 50 × 16² = 12800

  4. The volume of juice in a carton is normally distributed with mean 1.5 litres and standard deviation 0.12 litres. Find the probability that a randomly chosen carton holds less than 1.4 litres.

    • \(X\sim N(1.5,\,0.12^2)\)
    • From the calculator: \(P(X<1.4)\) = 0.202 (normal cdf with μ = 1.5, σ = √0.0144 = 0.12)

In context

  1. An airline weighs a traveller and their carry-on bag. The mass of a traveller, in kg, is normally distributed with mean 72.5 and variance 9.6. The mass of the bag, in kg, is normally distributed with mean 6.2 and variance 1.4. The two masses are independent.

    1. (a) State the distribution of the total mass T and find any parameters.

      • Mean = 72.5 + 6.2 = 78.7
      • Variance = 9.6 + 1.4 = 11

    2. (b) Find the probability that the total mass is more than 85 kg.

      • From the calculator: \(P(T>85)\) = 0.0287 (normal cdf with μ = 78.7, σ = √11 = 3.32…)

  2. Two taxi firms serve the same route. The journey time with firm A, A minutes, is normally distributed with mean 24.3 and variance 0.64. The journey time with firm B, B minutes, is normally distributed with mean 23.8 and variance 0.36. The times are independent.

    1. (a) Find the mean and the standard deviation of D = A − B.

      • Mean = 24.3 − 23.8 = 0.5
      • Variance = 0.64 + 0.36 = 1
      • Standard deviation = √1 = 1

    2. (b) Find the probability that A is less than B.

      • A < B means D < 0
      • From the calculator: \(P(D<0)\) = 0.309 (normal cdf with μ = 0.5, σ = √1 = 1)

    3. (c) Find the probability that A is less than B by more than 0.5.

      • A is less than B by more than 0.5 means D < −0.5
      • From the calculator: \(P(D<-0.5)\) = 0.159 (normal cdf with μ = 0.5, σ = √1 = 1)

  3. The length of a plank, X metres, is normally distributed with mean 4.2 and variance 0.16. The length of a pole, Y metres, is normally distributed with mean 2.0 and variance 0.09. Planks and poles are chosen independently.

    1. (a) Find the probability that a randomly chosen plank is more than twice as long as a randomly chosen pole.

      • X > 2Y means X − 2Y > 0
      • Mean = 4.2 − 2 × 2 = 0.2
      • Variance = 1² × 0.16 + (−2)² × 0.09 = 0.52
      • From the calculator: \(P(X-2Y>0)\) = 0.609 (normal cdf with μ = 0.2, σ = √0.52 = 0.721…)

    2. (b) Find the probability that the total length of 3 randomly chosen planks and 2 randomly chosen poles is more than 17 metres.

      • Total = 3 observations of X and 2 of Y
      • Mean = 3 × 4.2 + 2 × 2 = 16.6
      • Variance = 3 × 0.16 + 2 × 0.09 = 0.66
      • From the calculator: \(P(T>17)\) = 0.311 (normal cdf with μ = 16.6, σ = √0.66 = 0.812…)

  4. The time a student spends on homework each evening is normally distributed with mean 55 minutes and standard deviation 8 minutes. A random sample of 6 evenings is taken. Find the probability that the mean time is more than 58 minutes.

    • \(\bar X\sim N(55,\,8^2/6)\)
    • From the calculator: \(P(\bar{X}>58)\) = 0.179 (normal cdf with μ = 55, σ = 3.27…)

  5. The number of emails a teacher receives in a day has mean 32 and standard deviation 9, with an unknown distribution. Find the probability that the mean number of emails over a random sample of 50 days is between 30 and 34.

    • n = 50 > 30, so the central limit theorem applies
    • \(\bar X\approx N(32,\,9^2/50)\)
    • From the calculator: \(P(30<\bar{X}<34)\) = 0.884 (normal cdf, μ = 32, σ = 1.27…)

  6. The mass of a bag of oranges has mean 1.8 kg and standard deviation 0.3 kg. A pallet carries 48 randomly chosen bags. Use the central limit theorem to find the probability that the total mass of the bags is less than 88 kg.

    • n = 48 > 30, so the central limit theorem applies
    • \(S\approx N(48\times 1.8,\,48\times 0.3^2)\) = \(N(86.4,\,4.32)\)
    • From the calculator: \(P(S<88)\) = 0.779 (normal cdf with μ = 86.4, σ = √4.32 = 2.08…)

Stretch

  1. The random variable X has mean 120 and standard deviation 30. A sample of 36 independent observations is taken. Find \(P(|\bar X-115|>9)\).

    • n = 36 > 30, so \(\bar X\approx N(120,\,30^2/36)\)
    • \(|\bar X-115|>9\) means \(\bar X>124\) or \(\bar X<106\)
    • From the calculator: \(P(\bar{X}>124)\) = 0.212 (normal cdf with μ = 120, σ = √25 = 5)
    • From the calculator: \(P(\bar{X}<106)\) = 0.00256 (normal cdf with μ = 120, σ = √25 = 5)
    • Add the tails: 0.2119 + 0.002555 = 0.214

  2. The mass of a bag of flour is normally distributed with mean 1000 g and variance 25 g². Bags are checked in batches of 5. A batch is rejected if the average mass is less than 996 g or more than 1005 g.

    1. (a) Find the mean and the variance of the average of a random sample of this size.

      • Mean = 1000
      • Variance = 25 ÷ 5 = 5

    2. (b) Hence find the probability that a batch is rejected.

      • From the calculator: \(P(\bar{X}<996)\) = 0.0368 (normal cdf with μ = 1000, σ = √5 = 2.24…)
      • From the calculator: \(P(\bar{X}>1005)\) = 0.0127 (normal cdf with μ = 1000, σ = √5 = 2.24…)
      • Rejected: add the two tails, 0.03682 + 0.01267 = 0.0495

  3. X and Y are independent, \(X\sim N(6,\,5)\) and \(Y\sim N(28,\,4)\). \(X_1,X_2\) are independent observations of X and \(Y_1,Y_2\) are independent observations of Y. Find \(P(Y_1-Y_2>X_1+X_2)\).

    • Write the inequality as \(Y_1-Y_2-X_1-X_2>0\) and call the left side W.
    • Mean = 28 − 28 − 6 − 6 = −12
    • Variance = 4 + 4 + 5 + 5 = 18
    • From the calculator: \(P(W>0)\) = 0.00234 (normal cdf with μ = −12, σ = √18 = 4.24…)

jmaths.xyz · AI HL · AHL 8.9 Mixed practice: normal combinations, sample means and the central limit theorem