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AI HL · AHL 8

Statistics 2

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AHL 8.9

Mixed practice: normal combinations, sample means and the central limit theorem

AHL 8.9

Mixed practice: normal combinations, sample means and the central limit theorem

Target: I can find the distribution of aX + bY + c, of X1 − X2 and of Y1 + Y2 + Y3 for independent normal variables. (+4 more)

By the end of this section

Key points

Choosing the distribution

  • Ask first: one observation, a total of n observations, or a mean of n observations? They are \(N(\mu,\sigma^2)\), \(N(n\mu,n\sigma^2)\) and \(N(\mu,\sigma^2/n)\).
  • Two separate observations \(X_1,X_2\) of the same variable are independent. \(X_1-X_2\) has mean 0 and variance \(2\sigma^2\).
  • If the population is normal, the sample mean and the total are exactly normal. If it is not, you need n > 30 for the central limit theorem.
  • Always turn a standard deviation into a variance before you use the rule, and back into a standard deviation for the calculator.

Inequalities and two tails

  • To compare two combinations, move everything to one side: \(3X+1<5Y\) becomes \(3X-5Y+1<0\). Find the distribution of the left side and the probability below 0.
  • A modulus gives two tails: \(|\bar X-95|>10\) means \(\bar X>105\) or \(\bar X<85\). Find both probabilities and add them.
  • "Rejected if the mean is too low or too high" is also two tails. The probability of being accepted is the area between.

On the calculator

Normal probabilities for combinations, means and totals

  • Write the distribution first: N(mean, variance). Take the square root of the variance for the calculator.
  • For a probability between two values, enter both bounds. For a two-tailed probability, find each tail and add, or find the middle area and subtract it from 1.
  • Keep full accuracy in the calculator and round only the final answer to 3 significant figures.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Distribution of a combination

X and Y are independent, \(X\sim N(9,\,5)\) and \(Y\sim N(16,\,3)\). Find the distribution of \(W=5X+2Y-7\). Give the variance of W as your answer.

Working
  1. Mean

    E(W) = 5 × 9 + 2 × 16 − 7 = 70

    Coefficients and the constant, as they are.

  2. Variance

    Var(W) = 5² × 5 + 2² × 3 = 137

    Squared coefficients. The constant is left out.

  3. State

    \(W\sim N(70,\,137)\)

    Normal because X and Y are normal and independent.

Video walkthrough coming
An inequality between combinations

X and Y are independent, \(X\sim N(20,\,7)\) and \(Y\sim N(15,\,4)\). Find \(P(2X+3<4Y)\).

Working
  1. Move everything to one side

    \(2X+3<4Y \iff 2X-4Y+3<0\)

    Now it is a probability that one combination is below 0.

  2. Distribution of W

    W = 2X − 4Y + 3: mean 2 × 20 − 4 × 15 + 3 = −17, variance 2² × 7 + (−4)² × 4 = 92

    (−4)² = 16.

  3. From the calculator

    From the calculator: \(P(W<0)\) = 0.962 (normal cdf with μ = −17, σ = √92 = 9.59…)

    The mean of W is −17, well below 0, so the probability is large.

Video walkthrough coming
Independent observations of one variable

The mass of a loaf of bread is a random variable \(X\sim N(40,\,9)\) (in dekagrams). \(X_1\) and \(X_2\) are masses of two loaves chosen independently. Find the variance of \(X_1-X_2\). Then state the distribution of \(X_1+X_2+X_3\), the total mass of three loaves.

Working
  1. Difference of two loaves

    Var(X₁ − X₂) = 1² × 9 + (−1)² × 9 = 18

    Two separate observations, so each contributes its own variance. The mean is 40 − 40 = 0.

  2. Total of three loaves

    \(X_1+X_2+X_3\sim N(3\times 40,\,3\times 9)=N(120,\,27)\)

    Three observations: the variance is 3 × 9, not 3² × 9. That would be one loaf tripled.

Video walkthrough coming
A batch average is rejected

The length of a bolt is normally distributed with mean 52 mm and variance 0.18 mm². Bolts are checked in batches of 5. A batch is rejected if the mean length is less than 51.7 mm or more than 52.4 mm. Find the probability that a batch is rejected.

Working
  1. Distribution of the batch mean

    \(\bar{X}\sim N\left(52,\,\dfrac{0.18}{5}\right)=N(52,\,0.036)\)

    A normal population, so \(\bar X\) is normal.

  2. Lower tail

    From the calculator: \(P(\bar{X}<51.7)\) = 0.0569 (normal cdf with μ = 52, σ = √0.036 = 0.190…)

    σ = √0.036 = 0.190…

  3. Upper tail

    From the calculator: \(P(\bar{X}>52.4)\) = 0.0175 (normal cdf with μ = 52, σ = √0.036 = 0.190…)

    Rejected means one tail or the other, so add.

  4. Add

    0.05692 + 0.01751 = 0.0744

    About 1 batch in 8 is rejected.

Video walkthrough coming