FS1-5 Central Limit Theorem — coverage pack

1 specification leaves · notes, questions, answers and worked methods

FS1-5.1 · Applications of the Central Limit Theorem to other distributions.

For a large independent random sample from a population with mean $\mu$ and variance $\sigma^2$ , $\overline X\mathrel{\dot\sim}N(\mu,\sigma^2/n)$ .
Standardise a sample mean with standard error $\sigma/\sqrt n$ : $Z=(\overline X-\mu)/(\sigma/\sqrt n)$ .
The Central Limit Theorem can be applied to populations from the A-level Mathematics and FS₁ distributions when the sample is sufficiently large.
A common error is to use population variance $\sigma^2$ for $\overline X$ instead of dividing it by $n$ .

1. Independent lifetimes have population mean $50$ hours and variance $100$ hours $^2$ . For a sample of $36$ lifetimes, use the Central Limit Theorem to estimate $P(\overline X<47.5)$ .[4 marks]
Answer
- $P(\overline X<47.5)\approx0.0668$
Method: $\overline X\mathrel{\dot\sim}N(50,100/36)$ , so its standard error is $10/6=5/3$ . Hence $P(\overline X<47.5)\approx P(Z<(47.5-50)/(5/3))=P(Z<-1.5)=0.0668072\ldots$ .

1. A geometric population has parameter $p=0.25$ . A random sample of $64$ observations is taken. Use the Central Limit Theorem to estimate $P(3.5<\overline X<4.4)$ .[6 marks]
Answer
- $P(3.5<\overline X<4.4)\approx0.6981$
Method: For the geometric population, $\mu=1/p=4$ and $\sigma^2=(1-p)/p^2=12$ . Thus $\overline X\mathrel{\dot\sim}N(4,12/64)$ with standard error $\sqrt{12/64}=0.433013\ldots$ . Therefore the required probability is $P(-1.15470<Z<0.92376)=0.6980879\ldots$ . The sample mean is modelled directly as continuous here.

1. A population has mean $\mu$ and variance $36$ . Using the Central Limit Theorem normal approximation, estimate the least sample size $n$ for which $P(|\overline X-\mu|<1.2)\geq0.95$ .[6 marks]
Answer
- $n=97$
Method: Under the CLT normal approximation, the standard error is $6/\sqrt n$ . For central probability at least $0.95$ , require $1.2/(6/\sqrt n)\geq z_{0.975}=1.959964$ . Hence $n\geq(1.959964\times6/1.2)^2=96.0365\ldots$ . The estimated least integer value is $97$ .