Questions on combinations of normal random variables

Question 1

State the distribution of aX + bY when X and Y are independent normal variables, and the one condition it needs.

Accepted Answer

It is normal, with mean aμ_x + bμ_y and variance a²σ_x² + b²σ_y². The variables must be independent; without that the variance of the combination is not the sum a²σ_x² + b²σ_y². B1 for the mean and variance, B1 for the independence condition.

Question 2

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.

Accepted 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) would produce a negative variance, which no distribution has.

Question 3

X is N(40, 25). Write down the distribution of 3X − 10.

Accepted Answer

The mean follows the whole expression: 3(40) − 10 = 110. The variance takes only the multiplier, and squares it: 3²(25) = 225. So 3X − 10 is N(110, 225), with standard deviation 15. B1 for the mean, B1 for the variance. Subtracting 10 slides the whole distribution down the axis without altering its width, so 10 appears in the mean and nowhere in the variance.

Question 4

X is N(20, 16) and Y is N(14, 9), and X and Y are independent. Find P(X + Y < 40).

Accepted Answer

X + Y has mean 20 + 14 = 34 and variance 16 + 9 = 25, so it is N(34, 25) with standard deviation 5. M1 for the mean and variance of the sum. Standardising, z = (40 − 34)/5 = 1.2, so P(X + Y < 40) = P(Z < 1.2) = 0.8849. M1 for standardising, A1 for 0.8849. Standardise with the standard deviation 5 rather than the variance 25; dividing by 25 gives z = 0.24 and a probability near 0.595.

Question 5

X is N(12, 9) and Y is N(7, 16), and X and Y are independent. Find P(X < Y).

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

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.

Accepted 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.

Question 8

The masses of apples from an orchard are modelled as N(120, 64) grams, independently. Find the probability that one apple has mass greater than 124 g. Find the distribution of the mean mass of a random sample of 16 apples, and the probability that this mean is greater than 124 g. Explain why the two probabilities differ so much.

Accepted Answer

One apple: z = (124 − 120)/8 = 0.5, so P(X > 124) = 1 − 0.6915 = 0.3085.
The sample mean is X̄ = (X₁ + X₂ + … + X₁₆)/16. The sum of the sixteen has mean 16(120) = 1920 and variance 16(64) = 1024, sixteen variances added. Dividing by 16 divides the mean by 16 and the variance by 16² = 256, so X̄ is N(120, 4) with standard deviation 2. In general the mean of n independent readings is N(μ, σ²/n).
For the mean, z = (124 − 120)/2 = 2, so P(X̄ > 124) = 1 − 0.9772 = 0.0228.
M1 for standardising the single apple, A1 for 0.3085, M1 for the variance of the mean, A1 for N(120, 4), A1 for 0.0228, B1 for the explanation.
Averaging sixteen apples lets the heavy ones and the light ones cancel, so the mean is far less variable than one apple. The variance divides by n and the standard deviation by √n, so the spread falls by a factor of √16 = 4 rather than by 16.

Question 9

The mass of a bag of flour is modelled by N(1005, 40) grams and the mass of the box it is packed into by N(250, 9) grams, the two independent. Find the probability that a packed box has total mass below 1245 g.

Accepted Answer

The total T is the bag plus the box, so its mean is 1005 + 250 = 1255 and its variance is 40 + 9 = 49. M1 for the mean and the variance, A1 for T being N(1255, 49), with standard deviation 7.
Standardising: z = (1245 − 1255)/7 = −1.4286, so P(T < 1245) = P(Z < −1.4286) = 0.0766, about one packed box in thirteen. M1 for standardising, A1 for 0.0766.
The two masses are different quantities rather than repeated readings of one, so there is no coefficient to square anywhere: each variance enters once.

Question 10

The lengths of steel rods are modelled by N(250, 4) mm, independently. Two rods are chosen at random. Find the probability that their lengths differ by more than 5 mm.

Accepted Answer

Let D = X₁ − X₂ for two independent rods. Its mean is 250 − 250 = 0 and its variance is 4 + 4 = 8, the two variances adding, so D is N(0, 8) with standard deviation √8 = 2.8284. M1 for the distribution of the difference, A1 for N(0, 8).
A difference of more than 5 mm can fall either way, so the event is D > 5 or D < −5. Standardising, z = 5/2.8284 = 1.7678, and P(|D| > 5) = 2(1 − 0.9615) = 0.0771. M1 for doubling the tail, A1 for 0.0771.
Using variance 4 for D treats the second rod as fixed at 250 mm; both rods vary, so the spread of the gap between them is the wider √8.

Question 11

A machine fills packets with a mass modelled by N(502, 9) grams, the packets independent. A carton holds four packets. Find the probability that the total mass of the four packets is less than 2000 g, and the probability that their mean mass is less than 500 g.

Accepted Answer

The total of four packets has mean 4(502) = 2008 and variance 4(9) = 36, so it is N(2008, 36) with standard deviation 6. M1 for the mean and variance of the total, A1 for N(2008, 36).
Standardising: z = (2000 − 2008)/6 = −1.3333, so the probability is 0.0912. A1.
The mean of the four is N(502, 9/4), that is N(502, 2.25), with standard deviation 1.5. B1. Standardising: z = (500 − 502)/1.5 = −1.3333 again, so that probability is 0.0912 as well. B1 for the second probability with the reason.
The two z values agree because the two events are one event written twice: a total below 2000 g is exactly a mean below 500 g.

Question 12

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.

Accepted 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 an error.
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.
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.

Question 13

A carton has mass modelled by N(200, 25) grams, and each of the 12 bars packed into it has mass modelled by N(50, 4) grams, all 13 masses independent. Find the distribution of the mass of a full carton, and the probability that a full carton is heavier than 815 g. A second packer weighs one bar and multiplies by 12 to estimate the mass of the contents. State the distribution of that estimate and compare it with the distribution of the true contents.

Accepted Answer

The full carton is T = C + W₁ + W₂ + … + W₁₂. Its mean is 200 + 12(50) = 800. Its variance is 25 + 12(4) = 73, because twelve independent bars contribute twelve variances. So T is N(800, 73), with standard deviation √73 = 8.544.
Standardising: z = (815 − 800)/8.544 = 1.756, so P(T > 815) = 1 − 0.9604 = 0.0396.
The second packer's estimate is 12W for a single bar W. Here the coefficient squares, so the variance is 12²(4) = 576 and the estimate is N(600, 576) with standard deviation 24. The true contents W₁ + … + W₁₂ is N(600, 48) with standard deviation 6.93. The two have the same mean, but the estimate is more variable by a factor of √576/√48 = √12 = 3.46, because it magnifies the error of one bar instead of letting twelve errors cancel.
M1 for the mean of the total, M1 for the variance 25 + 12(4), A1 for N(800, 73), M1 for standardising, A1 for 0.0396, B1 for N(600, 576), B1 for the comparison.

Question 14

Machine A fills bottles with a volume modelled by N(505, 4) ml and machine B fills them with a volume modelled by N(500, 9) ml, all fills independent. Find the probability that a bottle from machine A contains at least 10 ml more than a bottle from machine B. Machine B is then reset so that its mean is μ ml, its variance unchanged. Find the value of μ for which the probability that a bottle from machine B holds more than one from machine A is 0.10.

Accepted Answer

Let D = A − B. Its mean is 505 − 500 = 5 and its variance is 4 + 9 = 13, the variances adding even though the means subtract, so D is N(5, 13) with standard deviation √13 = 3.606.
P(D ≥ 10) = P(Z ≥ (10 − 5)/3.606) = P(Z ≥ 1.3868) = 1 − 0.9172 = 0.0828.
After the reset, D = A − B is N(505 − μ, 13), with the same standard deviation 3.606 since the variance is unchanged. Machine B holding more is the event D < 0, and requiring P(D < 0) = 0.10 puts 0 at the bottom 10% point of D, which is z = −1.2816 because Φ(−1.2816) = 0.10. So (0 − (505 − μ))/3.606 = −1.2816, giving 505 − μ = 1.2816 × 3.606 = 4.621 and μ = 500.4 to one decimal place.
M1 for the distribution of the difference, A1 for N(5, 13), M1 for standardising, A1 for 0.0828, B1 for z = −1.2816, M1 for the equation in μ, A1 for 500.4.
Check the direction before writing the answer down. Raising the mean of machine B from 500 to 500.38 makes it more likely to beat machine A, and the probability rises from 0.0828 to 0.10, so the answer has moved the right way. A value below 500 would have moved it the wrong way, which is what taking z = +1.2816 produces.

Question 15

A component's length is modelled by N(20, 0.04) cm and the internal length of the socket it slides into by N(20.3, 0.05) cm, all lengths independent. The component fits when it is shorter than the socket. Find the probability that a randomly chosen component fits a randomly chosen socket. The socket maker can move the mean socket length without changing its variance. Find, to two decimal places, the smallest mean socket length for which the probability of a fit is at least 0.99.

Accepted Answer

Work with the clearance D = S − C, the socket length minus the component length. Its mean is 20.3 − 20 = 0.3 and its variance is 0.05 + 0.04 = 0.09, so D is N(0.3, 0.09) with standard deviation 0.3. M1 for turning the comparison into one variable, A1 for N(0.3, 0.09).
A fit is the event D > 0, so z = (0 − 0.3)/0.3 = −1, and P(D > 0) = P(Z > −1) = 0.8413. M1 for standardising, A1 for 0.8413.
Now let the mean socket length be μ. Then D is N(μ − 20, 0.09), still with standard deviation 0.3, and P(D > 0) ≥ 0.99 requires 0 to sit at least 2.3263 standard deviations below the mean, since Φ(2.3263) = 0.99. So μ − 20 ≥ 2.3263 × 0.3 = 0.6979, giving μ ≥ 20.6979, and the smallest such mean is 20.70 cm. M1 for the inequality with 2.3263, A1 for 20.70.
Round this limit upwards. Rounding 20.6979 down to 20.69 puts the probability of a fit at 0.9893, below the 0.99 asked for, so the direction of the inequality decides which way the last digit moves. Note also that widening the tolerance on either part raises the variance of D and pushes the required mean further out; only the two variances added matter, not which part carries them.