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Combinations of normal random variables questions
Add independent normal variables and the result is normal again. Means add as you would expect. Variances add whether you are adding or subtracting, which is where the marks are lost.
6 original questions · 24 marks · the combinations of normal random variables notes · Further Statistics 2
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
State the distribution of aX + bY when X and Y are independent normal variables, and the one condition it needs.
Worked answer
It is normal, with mean aμx + bμy and variance a²σx² + b²σy². The variables must be independent; without that the variances cannot simply be added. B1 for the mean and variance, B1 for the independence condition.X is N(12, 9) and Y is N(7, 16), and X and Y are independent. Write down the distributions of X + Y and X − Y, and find the distribution of 2X − 3Y.
Worked answer
X + Y is N(19, 25) and X − Y is N(5, 25). The means follow the sign, but the variances add in both cases, so both have standard deviation 5.
For 2X − 3Y the mean is 2(12) − 3(7) = 24 − 21 = 3, and the variance is 2²(9) + 3²(16) = 36 + 144 = 180. So 2X − 3Y is N(3, 180), with standard deviation 13.4.
B1 for the sum, B1 for the difference, M1 for the mean and variance of the combination, A1 for its distribution.
Every coefficient squares, including the one attached to a subtraction. Writing 2(9) − 3(16) is the usual mistake, and it produces a negative variance.X is N(12, 9) and Y is N(7, 16), and X and Y are independent. Find P(X < Y).
Worked answer
P(X < Y) = P(X − Y < 0), and X − Y is N(5, 25) with standard deviation 5. Standardising: z = (0 − 5)/5 = −1, so the probability is P(Z < −1) = 0.1587. M1 for turning the comparison into a single variable, A1 for its distribution, M1 for standardising, A1 for 0.1587. Using variance 9 − 16 would give a negative variance, which is the check that the rule must be addition.X is N(12, 9). Find the distribution of X₁ + X₂ + X₃ + X₄ + X₅ for five independent readings, and of 5X for a single one. Explain the difference.
Worked answer
Five independent readings: mean 60, variance 5 × 9 = 45, so N(60, 45) with standard deviation 6.71. Five times one reading: mean 60, variance 5² × 9 = 225, so N(60, 225) with standard deviation 15. M1 A1 for the distribution of the sum, A1 for the distribution of the multiple, B1 for the explanation. The five separate readings vary independently and their errors partly cancel; scaling one reading magnifies its error fivefold.The weights of adults are modelled as N(70, 100) kg, independently. Eight adults get into a lift rated for 600 kg. Find the probability that the rating is exceeded.
Worked answer
The total for eight people is normal with mean 8(70) = 560 and variance 8(100) = 800, so the standard deviation is √800 = 28.28. The variance is multiplied by 8, not by 64; that is the distinction between eight people and one person weighed eight times.
Standardising: z = (600 − 560)/28.28 = 1.414, so P(total > 600) = 1 − 0.9214 = 0.0786, about one time in thirteen.
M1 A1 for the mean and variance of the total, M1 for standardising, A1 for 0.0786.
The independence assumption is doing real work here. A family travelling together is unlikely to satisfy it.The weights of adults are modelled as N(70, 100) kg, independently. A lift is rated for 600 kg. Find the greatest number of adults that can be allowed into the lift if the probability that the rating is exceeded is to be less than 0.01.
Worked answer
Let n be the number of adults. The total weight is N(70n, 100n), with standard deviation 10√n. Note the root; using 10n instead is the error that makes this question look easier than it is.
The requirement P(total > 600) < 0.01 means the value 600 must lie at least 2.3263 standard deviations above the mean, since Φ(2.3263) = 0.99. So
(600 − 70n)/(10√n) > 2.3263.
Substituting t = √n turns this into a quadratic: 600 − 70t² > 23.263t, that is 70t² + 23.263t − 600 < 0. Making that substitution is the method mark; there is no way through by inspection.
The positive root is t = 2.766, so n < 7.65 and the greatest whole number is n = 7.
M1 for the distribution of the total in terms of n, B1 for the value 2.3263, M1 for the inequality, M1 for the substitution t = √n, A1 for the positive root, A1 for the greatest whole number.
Check both sides. With n = 7 the z value is 4.16 and the probability is 0.000016, comfortably below 0.01. With n = 8 the z value is 1.41 and the probability is 0.0786, far above it. Rounding 7.65 up to 8 would break the condition, so the direction of the rounding carries a mark of its own.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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