Further Mathematics · Pearson Edexcel 9FM0 · Option FS2 · Topic 5

Estimation, confidence intervals and tests using a normal distribution

Estimators and bias, confidence intervals for a normal mean, tests for the difference between two means, and large-sample methods.

5 sub-topics
Taught in Year 13
73 practice questions
5 animated lessons

Practice questions (73) Lessons Worksheets All Edexcel Further Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

5.1Estimators and biasY13
5.2Confidence intervalsY13
5.3Confidence limits for a normal mean (variance known)Y13
5.4Test for the difference of two normal means (variances known)Y13
5.5Large-sample methodsY13

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 73 questions in this topic, free, no account needed.

Estimation, confidence intervals and tests using a normal distribution

Not needed for Year 12Easy

Why large samples help · Year 13 · 1 mark

Which result lets the sample mean be treated as approximately Normal when the sample is large, whatever the distribution of the population?
(1)

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For large $n$ , $\bar{X}$ is approximately $N\!\left(\mu, \dfrac{\sigma^2}{n}\right)$ .A: The Central Limit Theorem

Total1

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Estimation, confidence intervals and tests using a normal distribution

Not needed for Year 12Medium

Confidence intervals · Year 13 · 5 marks

A measurement is normally distributed with standard deviation 8. A 99% confidence interval for the mean is to have width at most 3.

(a)
Find the smallest sample size needed.
(3)
(b)
A sample of this size has mean 52.4. Find the 99% confidence interval.
(2)

Check your answer before you look at the solution

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Part (a)3 marks

The width is twice $z \sigma /\sqrt{n}$ . $2 \times 2.5758 \times \frac{8}{\sqrt{n}} \leq 3$
$n \geq 188.72$
Round up. $n = 189$

Part (b)2 marks

$52.4 \pm 2.5758 \times \frac{8}{\sqrt{189}}$
$(50.901, 53.899)$

Total5

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Estimation, confidence intervals and tests using a normal distribution

Not needed for Year 12Hard

Estimators and bias · Year 13 · 6 marks

X_1, X_2, X_3

is a random sample from a population with mean

\mu

and variance

\sigma^2

. Two estimators of

\mu

are

T_1

{}= \tfrac{1}{3}(X_1 + X_2 + X_3)

and

T_2

{}= \frac{1}{8}X_1

{}+ \frac{3}{8}X_2

{}+ kX_3

.

(a)
Find the value of $k$ for which $T_2$ is unbiased.
(2)
(b)
Find the variances of $T_1$ and $T_2$ , and state which is the better estimator.
(4)

Check your answer before you look at the solution

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Part (a)2 marks

$\mathrm{E}(T_2)$ ${}= (\frac{1}{8} + \frac{3}{8} + k)\mu$ ${}= \mu$
$k = \frac{1}{2}$

Part (b)4 marks

$\operatorname{Var}(T_1)$ ${}= \tfrac{1}{3}\sigma^2$
$\operatorname{Var}(T_2)$ ${}= \left(\frac{1}{8}^2 + \frac{3}{8}^2 + \frac{1}{2}^2\right)\sigma^2$ ${}= \frac{13}{32}\sigma^2$
$T_1$ : both are unbiased and it has the smaller variance.

Total6

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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