Questions on the normal distribution

Question 1

Describe the shape of a normal distribution in words: state its symmetry, and roughly what proportion of values fall within one and within two standard deviations of the mean.

Accepted Answer

A single symmetric peak at the mean, falling away identically on both sides, with the mean, median and mode all at the same place. About 68% of values lie within one standard deviation of the mean and about 95% within two. B1 for the symmetry about the mean, B1 for 68%, B1 for 95%.

Question 2

X ~ N(60, 5²). Find P(X < 70) to 3 significant figures.

Accepted Answer

Standardise: z = (70 − 60)/5 = 2, so P(X < 70) = P(Z < 2) = 0.977. M1 for standardising, A1 for 0.977. Two standard deviations above the mean leaves only about 2.3% in the upper tail, so an answer just below 1 is what the sketch predicts. Write the standardising line out even when the calculator will take μ and σ directly, so that the method is visible.

Question 3

X ~ N(70, 8²). Write down the values of x at which the normal curve for X has its points of inflection.

Accepted Answer

The points of inflection of a normal curve lie one standard deviation either side of the mean, at x = μ ± σ. B1 for that fact. Here that gives 70 − 8 = 62 and 70 + 8 = 78. B1 for both values. They mark where the curve stops falling away ever more steeply and starts to flatten out, which is what allows σ to be read off a sketch of the curve without any calculation.

Question 4

X ~ N(50, 12²). Find P(X > 62) and P(38 < X < 62), each to 3 significant figures.

Accepted Answer

Standardise both values: z = (62 − 50)/12 = 1 and z = (38 − 50)/12 = −1, so they sit exactly one standard deviation either side of the mean. M1 for standardising. P(X > 62) = P(Z > 1) = 0.159. A1. The middle region is what is left when both tails are taken away, and symmetry makes the two tails equal, so P(38 < X < 62) = 1 − 2 × 0.1587 = 0.683. A1 for 0.683.

Question 5

Masses are modelled by X ~ N(60, 10²) grams. Find P(65 < X < 75) to 3 significant figures.

Accepted Answer

Standardise both ends: z = (65 − 60)/10 = 0.5 and z = (75 − 60)/10 = 1.5. Then P(0.5 < Z < 1.5) = P(Z < 1.5) − P(Z < 0.5) = 0.9332 − 0.6915 = 0.242. M1 for standardising both ends, M1 for the subtraction, A1 for 0.242. Subtract the smaller cumulative probability from the larger; adding the probabilities, or subtracting the z values, gives a different quantity.

Question 6

X ~ N(40, σ²) and P(X > 46) = 0.05. Find σ to 3 significant figures.

Accepted Answer

The top 5% of the standard normal starts at z = 1.6449. So (46 − 40)/σ = 1.6449, giving σ = 6/1.6449 = 3.65. B1 for z = 1.6449, M1 for the standardising equation, M1 for rearranging, A1 for 3.65. Check forwards: 46 is then 1.64 standard deviations above the mean, which cuts off 5%, as stated. Quote the z value before forming the equation, and note that z here is positive because 46 lies above the mean.

Question 7

X ~ N(μ, 4²) and P(X < 50) = 0.975. Use z = 1.96 to find μ.

Accepted Answer

A probability of 0.975 below 50 puts 50 at z = +1.96, so (50 − μ)/4 = 1.96 and μ = 50 − 7.84 = 42.16. M1 for the standardising equation, M1 for rearranging, A1 for 42.16. The value 50 sits above the mean, not below it. Reading 0.975 as a lower-tail area is what keeps the sign of z honest, and a negative z here would give μ = 57.84, comfortably above a value that 97.5% of the distribution falls below.

Question 8

X ~ B(200, 0.5). Use a suitable approximation to estimate P(X ≥ 110), stating why the approximation is justified and showing the continuity correction you apply.

Accepted Answer

Match the mean and the variance: np = 100 and np(1 − p) = 50, so X is approximated by Y ~ N(100, 50) with σ = √50 = 7.071. B1 for the normal named with both parameters. The approximation is justified because n = 200 is large and p = 0.5 sits exactly where the binomial's bar chart is symmetric and bell-shaped. B1 for the conditions. A discrete count of at least 110 keeps the whole of the 110 bar, so with the continuity correction P(X ≥ 110) ≈ P(Y > 109.5). M1 for the correction. Standardising, z = (109.5 − 100)/7.071 = 1.343. M1. The tail beyond it is 0.0896 to three significant figures. A1. The exact binomial sum is 0.0895, so the corrected approximation is good to three figures; cutting at 110 instead of 109.5 loses the bar and understates the tail.

Question 9

The lengths, in centimetres, of manufactured components are modelled by N(24, 1.5²). Find the length exceeded by 20% of components, and the length below which 5% of components lie.

Accepted Answer

For the first, 20% lie above the value, so 80% lie below it and the inverse normal at 0.8 gives z = 0.8416. M1 for using the inverse normal at 0.8 rather than at 0.2. The length is 24 + 0.8416 × 1.5 = 25.3 cm. A1 for 25.3. For the second, 5% lie below the value, so z = −1.6449 and the length is 24 − 1.6449 × 1.5 = 21.5 cm. M1 for a negative z below the mean, A1 for 21.5. Check each answer against the mean before writing it down: a length exceeded by only 20% of components has to be above 24, and one below which only 5% lie has to be under it, so a positive z in the second part would show up immediately.

Question 10

X ~ N(180, σ²) and P(174 < X < 186) = 0.6. Find σ to 3 significant figures.

Accepted Answer

The interval is symmetric about the mean, since 180 − 174 and 186 − 180 are both 6, so the 0.4 outside it is split equally between the two tails at 0.2 each, which puts 0.8 below 186. M1 for using the symmetry to reach a single cumulative probability. The inverse normal at 0.8 gives z = 0.8416, so (186 − 180)/σ = 0.8416. B1 for z = 0.8416. Rearranging, σ = 6/0.8416 = 7.13. M1 for the rearrangement, A1 for 7.13. Taking 0.6 straight to the inverse normal, as though 186 stood at a cumulative probability of 0.6, gives z = 0.2533 and σ = 23.7; under that spread the stated interval would hold only a fifth of the distribution rather than the three fifths given.

Question 11

For each of the following, state whether a binomial model or a normal model is the more suitable, with a reason: (i) the number of heads obtained in 40 tosses of a fair coin; (ii) the height of a randomly chosen adult woman.

Accepted Answer

The number of heads counts successes in a fixed number of independent trials with a constant probability, so it takes the whole-number values 0 to 40 and B(40, 0.5) is the model that fits. B1 for the binomial with a reason. A height is a measurement rather than a count: it can take any value in a range, and heights cluster symmetrically about a central value with fewer women far from it, which is the shape a normal model describes. B1 for the normal with a reason. B1 for drawing the discrete-against-continuous distinction between the two.

Question 12

X ~ N(μ, σ²), with P(X < 20) = 0.10 and P(X > 35) = 0.05. Find μ and σ.

Accepted Answer

Turn each probability into a z value first. The bottom 10% ends at z = −1.2816, so (20 − μ)/σ = −1.2816. The top 5% begins at z = 1.6449, so (35 − μ)/σ = 1.6449. Rewrite them as 20 = μ − 1.2816σ and 35 = μ + 1.6449σ. Subtracting removes μ: 15 = 2.9265σ, so σ = 5.13. Then μ = 20 + 1.2816 × 5.1257 = 26.6. Check both statements before finishing: (35 − 26.57)/5.126 = 1.645, as required. B1 B1 for the two z values, M1 for the pair of simultaneous equations, M1 for the elimination, A1 for σ, A1 for μ. The 20 is below the mean, so its z must be negative. Writing +1.2816 there gives σ = 41.3 and a negative mean, which is plainly impossible for a variable whose values sit around 20 to 35.

Question 13

Daily rainfall totals at a weather station have mean 4.1 mm and standard deviation 5.3 mm. Explain why a normal model is unsuitable for this variable, using both the values a normal model allows and the shape the data must have.

Accepted Answer

A normal model allows values from minus infinity to infinity, and rainfall cannot be negative. B1 for the impossible values. Here the objection has a size: with σ = 5.3 above the mean of 4.1, zero sits at z = −0.77, so the model would put about 22% of days below zero rainfall. B1 for quantifying it. A normal curve is also symmetric about its mean, while rainfall piles up near zero with a long right tail of rare wet days, the very shape that pushes σ above the mean. B1 for the symmetry against the skew.

Question 14

A call centre records the number of calls arriving in each one-minute interval. Over a long period the mean is 3.2 calls a minute and the standard deviation is 1.76 calls. One student proposes a binomial model for the number of calls in a minute and another proposes a normal model. Explain why neither is suitable.

Accepted Answer

A binomial model counts successes among a fixed number of trials. B1 for a binomial needing a fixed n. Here there is no such number: any number of calls could in principle arrive in a minute, and there are no trials to count successes among. B1 for there being none in this context. A normal model is continuous and admits every real value, while a number of calls is a whole number that cannot be negative. B1 for the discrete count set against a continuous model. The second objection can be given a size: with mean 3.2 and standard deviation 1.76, zero sits at z = (0 − 3.2)/1.76 = −1.82, so a normal model would place about 3.5% of minutes below zero calls, which is one minute in thirty. M1 for standardising zero, A1 for about 3.5%.

Question 15

A packer fills bags with sugar so that the mass, in grams, of sugar in a bag is modelled by N(μ, 4²). The bags are labelled 1000 g, and the packer wants at most 1% of bags to hold less than the labelled mass. Find the smallest whole number of grams μ can be, and state how much sugar this gives away per bag on average.

Accepted Answer

At most 1% below 1000 g means P(X < 1000) ≤ 0.01, and the bottom 1% of the standard normal ends at z = −2.3263. B1 for that z value. So (1000 − μ)/4 ≤ −2.3263, giving 1000 − μ ≤ −9.305 and μ ≥ 1009.305. M1 for forming the inequality, M1 for rearranging it, A1 for 1009.305. The smallest whole number of grams satisfying that is μ = 1010 g. A1 for 1010 with the upward rounding justified: μ = 1009 gives z = (1000 − 1009)/4 = −2.25 and a tail of 0.0122, above the 1% allowed, while μ = 1010 gives z = −2.5 and a tail of 0.0062, inside it. The packer therefore aims 10 g above the labelled mass and gives away 10 g of sugar per bag on average. B1 for the 10 g.