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AI HL · AHL 8

Statistics 2

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AHL 8.7

Linear combinations of independent normal variables

AHL 8.7

Linear combinations of independent normal variables

Target: I can say that a linear combination of independent normal variables is normal, and write its distribution as N(mean, variance). (+3 more)

By the end of this section

Key points

The rule

  • If X and Y are independent and normally distributed, then W = aX + bY + c is also normally distributed.
  • \(W\sim N\left(a\mu_X+b\mu_Y+c,\; a^2\sigma_X^2+b^2\sigma_Y^2\right)\).
  • In N(μ, σ²) the second number is the variance, not the standard deviation. If you are given a standard deviation, square it first.
  • The rule extends to three or more variables: one more term in the mean and one more in the variance.

Where each part comes from

  • The mean uses the coefficients as they are, with their signs, and the constant c.
  • The variance uses the squared coefficients. A negative coefficient such as \(-3\) gives \((-3)^2=9\), so subtracting a variable still adds to the spread.
  • The constant c moves the mean and does not change the variance.
  • The rule needs X and Y to be independent. Check the question says so.

Where it is used

  • Totals: the time for a job made of several tasks, or the mass of a packed box and its contents.
  • Differences: the gap between two runners' times, or the amount left after taking one quantity from another.

On the calculator

Probability for a combination W

  • Work out the mean and the variance of W by hand first.
  • Use the normal cumulative distribution with the mean of W and the standard deviation of W.
  • The calculator asks for the standard deviation: take the square root of Var(W). Typing the variance is the commonest mistake.
  • For an upper tail \(P(W>k)\), put a very large number as the upper bound. Sketch the curve and shade the area to check it is sensible.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Mean and variance of W

X and Y are independent, \(X\sim N(20,\,9)\) and \(Y\sim N(7,\,4)\). Let \(W=X+3Y-5\). Find the variance of W.

Working
  1. Mean

    E(W) = 20 + 3 × 7 − 5 = 36

    Use the coefficients and the constant as they are.

  2. Variance

    Var(W) = 1² × 9 + 3² × 4 = 45

    Square each coefficient. The constant −5 adds nothing to the variance.

  3. State the distribution

    \(W\sim N(36,\,45)\)

    W is normal because X and Y are independent and normal. The second number is the variance.

Video walkthrough coming
A probability for W

X and Y are independent, \(X\sim N(30,\,8)\) and \(Y\sim N(12,\,5)\). Let \(W=2X-3Y+1\). Find \(P(W>28)\).

Working
  1. Mean

    E(W) = 2 × 30 − 3 × 12 + 1 = 25

    The −3 stays negative in the mean.

  2. Variance

    Var(W) = 2² × 8 + (−3)² × 5 = 77

    (−3)² = 9, so the negative coefficient adds to the variance.

  3. State

    \(W\sim N(25,\,77)\)

    Normal, because X and Y are independent and normal.

  4. From the calculator

    From the calculator: \(P(W>28)\) = 0.366 (normal cdf with μ = 25, σ = √77 = 8.77…)

    Enter the standard deviation √77, not 77.

Video walkthrough coming
Total time of a job

A decorator prepares a room and then paints it. The preparation time, X minutes, has distribution \(N(40,\,25)\). The painting time, Y minutes, has distribution \(N(95,\,64)\). The two times are independent. Let T = X + Y be the total time. State the distribution of T and find the probability that the job takes more than 150 minutes.

Working
  1. Mean

    E(T) = 40 + 95 = 135

    Add the two means.

  2. Variance

    Var(T) = 1² × 25 + 1² × 64 = 89

    Variances add for a sum of independent variables.

  3. State

    \(T\sim N(135,\,89)\)

    Total time is normal.

  4. From the calculator

    From the calculator: \(P(T>150)\) = 0.0559 (normal cdf with μ = 135, σ = √89 = 9.43…)

    The job is more likely to take less than 150 minutes than more.

Video walkthrough coming
A difference

Sam and Tia run 200 m. Sam's time, X seconds, has distribution \(N(26.4,\,0.64)\). Tia's time, Y seconds, has distribution \(N(25.9,\,0.49)\). The times are independent. Let D = X − Y. Find the variance of D, and then find the probability that Sam takes more than 1 second longer than Tia.

Working
  1. Mean

    E(D) = 26.4 − 25.9 = 0.5

    Sam is 0.5 s slower on average.

  2. Variance

    Var(D) = 1² × 0.64 + (−1)² × 0.49 = 1.13

    Subtracting does not subtract the variances. The spread of a difference is bigger than either spread alone.

  3. State

    \(D\sim N(0.5,\,1.13)\)

    D is normal.

  4. From the calculator

    From the calculator: \(P(D>1)\) = 0.319 (normal cdf with μ = 0.5, σ = √1.13 = 1.06…)

    Sam is more than 1 s slower than Tia in about a third of races.

Video walkthrough coming