Statistics: Distributions Depth
In A-level Statistics, Distributions Depth takes the binomial and normal distributions beyond a single direct probability calculation: finding the mean and variance of a linear transformation of a binomial or normal random variable (such as a total cost built from a fixed number of trials), finding an unknown mean or standard deviation of a normal distribution from a given probability, including solving simultaneously when BOTH are unknown, and applying the continuity correction precisely in all four directions when a normal distribution is used to approximate a binomial one. The examiner is testing algebraic control of E(aX+b) and Var(aX+b), fluency with the inverse normal function, and care with strict versus non-strict inequalities under a continuity correction.
Before you start
Make sure you're comfortable with these topics first:
Method
- Decide what the question needs. A new variable defined as Y = aX + b (or built from several independent trials): use E(aX+b) = aE(X) + b and Var(aX+b) = a^2 Var(X). One unknown parameter (mu or sigma) of a normal distribution with one given probability: standardise and use the inverse normal function. BOTH mu and sigma unknown, with two given probabilities: form two equations and solve simultaneously. Approximating a binomial with a normal distribution: check np>5 and n(1-p)>5, then apply a continuity correction.
- For X ~ B(n,p), E(X) = np and Var(X) = np(1-p). For any linear transformation Y = aX + b, E(Y) = aE(X) + b and Var(Y) = a^2 Var(X); adding a constant b never changes the variance, only multiplying by a does.
- To find one unknown parameter from a single probability statement about X ~ N(mu, sigma^2): standardise using z = (x-mu)/sigma, find the exact z-value that gives the stated probability using the inverse normal function, then substitute and rearrange for the unknown.
- When BOTH mu and sigma are unknown, form two equations of the form (x - mu)/sigma = z from two different given probabilities (each with its own x-value and its own z-value), then solve simultaneously, typically by subtracting one equation from the other to eliminate mu first.
- For a normal approximation Y ~ N(np, np(1-p)) to X ~ B(n,p), apply a continuity correction that always widens a strict inequality into the equivalent range for the continuous variable: P(X > r) becomes P(Y > r+0.5); P(X >= r) becomes P(Y > r-0.5); P(X < r) becomes P(Y < r-0.5); P(X <= r) becomes P(Y < r+0.5).
- Always check np > 5 and n(1-p) > 5 before using a normal approximation to a binomial distribution; if either condition fails, use the exact binomial distribution instead.
Worked example
The random variable X ~ N(mu, sigma^2). Given that P(X < 20) = 0.2 and P(X < 35) = 0.9, find the values of mu and sigma, each to 3 significant figures.
- Standardise both probability statements: (20 - mu)/sigma = z1, where P(Z < z1) = 0.2, and (35 - mu)/sigma = z2, where P(Z < z2) = 0.9.
- Find the z-values from the standard normal distribution: z1 = -0.8416 and z2 = 1.2816 (4dp each, using the inverse normal function).
- Write the two equations: 20 - mu = -0.8416 sigma ... (1); 35 - mu = 1.2816 sigma ... (2).
- Subtract equation (1) from equation (2) to eliminate mu: 15 = 2.1232 sigma.
- Solve for sigma: sigma = 15/2.1232 = 7.06 (3 s.f.).
- Substitute back into equation (1): mu = 20 + 0.8416(7.0649) = 25.9 (3 s.f.).
Practice questions
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Q1X ~ B(20, 0.3). Find E(4X + 5).Show answer
Answer: E(X) = 20(0.3) = 6, so E(4X+5) = 4(6)+5 = 29.
Q2X ~ B(20, 0.3). Find Var(4X + 5).Show answer
Answer: Var(X) = 20(0.3)(0.7) = 4.2, so Var(4X+5) = 4^2 x 4.2 = 67.2.
Q3X ~ N(mu, 6^2). Given that P(X > 40) = 0.15, find the value of mu, giving your answer to 1 decimal place.Show answer
Answer: P(X<40) = 0.85, so z = 1.0364 (4dp). (40-mu)/6 = 1.0364, giving mu = 40 - 6(1.0364) = 33.8.
Q4X ~ N(50, sigma^2). Given that P(X < 58) = 0.9, find sigma, giving your answer to 2 significant figures.Show answer
Answer: z = 1.2816 (4dp, since P(Z<1.2816)=0.9). (58-50)/sigma = 1.2816, so sigma = 8/1.2816 = 6.2.
Q5X ~ B(80, 0.4) is to be approximated by a normal distribution Y. State the continuity-corrected inequality used to approximate P(X >= 35).Show answer
Answer: P(Y > 34.5).
Q6X ~ B(15, 0.1). Explain, with a calculation, why a normal approximation would NOT be appropriate for X.Show answer
Answer: np = 15(0.1) = 1.5, which is not greater than 5, so the condition np>5 fails. A normal approximation should not be used here; the exact binomial distribution should be used instead.
Q7X ~ B(100, 0.5) is approximated by Y ~ N(50, 25). Use this approximation, with a continuity correction, to estimate P(45 <= X <= 55), giving your answer to 3 significant figures.Show answer
Answer: P(45<=X<=55) becomes P(44.5<Y<55.5). Standardising with mean 50 and standard deviation 5: z = -1.1 and z = 1.1. This gives a probability of 0.729 (3 s.f.).
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
A factory finds that 8% of items on a production line are defective. A random sample of 150 items is selected. Let X be the number of defective items in the sample. (a) State two conditions needed for a normal distribution to be a suitable approximation for X, and show that both are satisfied here. (2) (b) Using a suitable normal approximation, find P(X > 16), giving your answer to 3 significant figures. (5)
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A discrete random variable X ~ B(40, 0.25) models the number of faulty bulbs in a box of 40. Each faulty bulb costs the manufacturer 3 pounds to replace, plus a fixed inspection fee of 20 pounds per box regardless of the number of faults. The total cost, in pounds, of a box is modelled by C = 3X + 20. (a) Find E(C) and Var(C). (4) (b) A different box has 60 bulbs with the same fault probability 0.25, using the same cost model C = 3X + 20. Explain, without further calculation, whether E(C) would increase, decrease, or stay the same compared with the 40-bulb box. (1)
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X ~ B(200, 0.45) is approximated by a normal distribution. (a) State the mean and variance of the normal approximation. (2) (b) Write down the continuity-corrected inequality (in terms of the normal variable Y) that should be used to estimate each of: (i) P(X < 85) (ii) P(X >= 95). (2)
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Free printable worksheet
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