Statistics: Statistical Distributions (Binomial/Normal)
Statistical distributions describe how the probability of an outcome is spread across all its possible values. A Level Statistics focuses on the binomial distribution B(n,p) for independent trials and the normal distribution N(mu, sigma^2) for continuous data, including standardising values, finding unknown parameters, and approximating a binomial by a normal distribution with a continuity correction.
Method
- Check the binomial conditions apply: a fixed number of trials n, each with only two outcomes, a constant probability of success p, and independence between trials.
- Calculate a binomial probability using P(X = r) = nCr x p^r x (1-p)^(n-r), or sum several such terms for P(X <= r) type questions.
- For a normal distribution X ~ N(mu, sigma^2), standardise using z = (x - mu)/sigma, then use the standard normal distribution (tables or a calculator) to find the required probability.
- To find an unknown mean or standard deviation from a given probability, find the z-value matching that probability (using tables or an inverse normal function), then rearrange z = (x - mu)/sigma to solve for the unknown.
- For a normal approximation to a binomial X ~ B(n,p), check that np > 5 and n(1-p) > 5, then use Y ~ N(np, np(1-p)).
- When using a normal approximation to a binomial (a discrete distribution) with a continuous model, apply a continuity correction, e.g. P(X >= r) becomes P(Y > r - 0.5).
Worked example
The random variable X ~ B(12, 0.25). (a) Calculate P(X = 3), giving your answer to 3 significant figures. (b) Calculate P(X <= 1).
- Use P(X = r) = nCr x p^r x (1-p)^(n-r) with n=12, p=0.25, r=3: P(X=3) = 12C3 x 0.25^3 x 0.75^9.
- 12C3 = 220, 0.25^3 = 0.015625 and 0.75^9 = 0.075085 (6dp), so P(X=3) = 220 x 0.015625 x 0.075085 = 0.258 (3sf).
- For P(X <= 1), find P(X=0) = 0.75^12 = 0.031676 and P(X=1) = 12C1 x 0.25 x 0.75^11 = 0.126705.
- P(X <= 1) = P(X=0) + P(X=1) = 0.031676 + 0.126705 = 0.158 (3sf).
Practice questions
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Q1State two conditions required for a random variable to be modelled by a binomial distribution.Show answer
Answer: Fixed number of independent trials, each with only two outcomes and a constant probability of success (any two of these).
Q2X ~ B(8, 0.3). Write down E(X) and Var(X).Show answer
Answer: E(X) = 2.4, Var(X) = 1.68 (np and np(1-p)).
Q3X ~ N(50, 8^2). Find P(X > 58).Show answer
Answer: awrt 0.159 (z = (58-50)/8 = 1, P(Z>1) = 0.1587).
Q4X ~ B(15, 0.4). Calculate P(X = 5), giving your answer to 3 significant figures.Show answer
Answer: 0.186 (15C5 x 0.4^5 x 0.6^10).
Q5X ~ N(20, 5^2). Find the value of a such that P(X < a) = 0.9, using z = 1.2816.Show answer
Answer: a = 26.4 (3sf) (20 + 1.2816 x 5 = 26.408).
Q6The random variable X ~ B(n, 0.2) has Var(X) = 6.4. Find the value of n.Show answer
Answer: n = 40 (Var = np(1-p) = 0.16n = 6.4, so n = 40).
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
Zoe is a quality inspector at a factory producing glass bottles. From long-term records, 6% of bottles produced have a visible flaw. Zoe selects a random sample of 15 bottles. Let X be the number of flawed bottles in the sample. (a) State two conditions needed for X to be modelled by a binomial distribution. (2) (b) Given that X ~ B(15, 0.06), calculate P(X = 1). (2)
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The heights of adult men in a country, H cm, are modelled by H ~ N(178, 7^2). (a) Calculate P(H > 185). (3) (b) Calculate P(170 < H < 185). (2)
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A polling company surveys a random sample of 250 voters. National figures suggest 48% of voters support a proposed policy. Let X be the number, out of 250, who support the policy, so X ~ B(250, 0.48). (a) Give a reason why a normal approximation to X is appropriate here. (1) (b) Using a suitable approximation with a continuity correction, calculate P(X > 130). (5)
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Free printable worksheet
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