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A-Level Sampling
What the A-Level syllabus expects for Sampling, and how to practise it.
What the syllabus expects
- The ideas of a population and of a simple random sample
- Treating the sample mean X̄ as a random quantity for which E(X̄) = μ and Var(X̄) = σ²/n
- The distribution of the sample mean drawn from a normal population
- Applying the Central Limit Theorem so that a sample mean is treated as normally distributed once the sample is large enough (for instance n ≥ 30)
- Computing unbiased estimates of a population's mean and variance from a sample, including data summarised as ∑x and ∑x², or as ∑(x−a) and ∑(x−a)²
How it's examined
Questions on this topic most often ask you to find, estimate. About 2% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (5 marks)
The PE department also tests at the 3% level whether GJC female students run 2.4 km in under 14.5 minutes on average. A random sample of n female students (n large) has mean time 14.2 minutes and standard deviation 1.5 minutes. Given the department concludes the mean is below 14.5 minutes, find the range of possible n.
Show the worked answer
One-tailed test. H0: mu=14.5, H1: mu<14.5. n large so use z-test with s=1.5 as estimate of sigma. z = (xbar-14.5)/(s/sqrt(n)) = (14.2-14.5)/(1.5/sqrt(n)) = -0.3*sqrt(n)/1.5 = -0.2*sqrt(n). At 3% significance (one tail): reject H0 if z < -1.8808 (since P(Z<-1.8808)=0.03). Department concludes mean below 14.5 => reject H0: -0.2*sqrt(n) < -1.8808 => sqrt(n) > 9.404 => n > 88.4. Since n is a large integer, n >= 89.
Example 2 (3 marks)
A Large vessel's catch is normal with mean 540 kg and standard deviation 110 kg. Over n randomly chosen trips, the probability that a Large vessel's mean catch is below 552 kg is to be at least 0.7. Find the smallest possible n. [3]
Show the worked answer
Let X kg be one trip's catch, so X ~ N(540, 110²). Let X̄ be the mean catch over n randomly chosen trips. Since X is itself normal, X̄ is exactly normal (no need to appeal to the CLT): X̄ ~ N(540, 110²/n), so the standard error is 110/√n. The requirement is P(X̄ < 552) ≥ 0.7. Standardise, subtracting the mean and dividing by the standard error: P(Z < (552 − 540)/(110/√n)) ≥ 0.7 P(Z < 12√n/110) ≥ 0.7 From the normal distribution, P(Z < 0.5244005) = 0.7. Since the cumulative normal is an increasing function, the inequality holds precisely when 12√n/110 ≥ 0.5244005 √n ≥ 0.5244005 × 110/12 √n ≥ 4.80700... n ≥ 23.107... n must be a whole number of trips, and the inequality needs n at least 23.107, so round up: smallest n = 24. (Check: n = 23 gives 12√23/110 = 0.5233, and P(Z < 0.5233) = 0.6996 < 0.7, so 23 is too small.)
More A-Level H2 Maths topics
Functions · Graphs and their transformations · Equations and inequalities · Sequences and series · Vectors in two and three dimensions: basic properties · Scalar and vector products · all of A-Level H2 Maths