S3 June 2013 (R) Q5

EdexcelOld spec9 marksConfidence Intervals

5. A manufacturer produces circular discs with diameter \(D\) mm, such that \(D \sim \mathrm{N}(\mu, \sigma^2)\). A random sample of discs is taken and, using tables of the normal distribution, a 90% confidence interval for \(\mu\) is found to be

\[(118.8,\ 121.2)\]

(a) Find a 98% confidence interval for \(\mu\). (6)

(b) Hence write down a 98% confidence interval for the circumference of the discs. (1)

Using three different random samples, three 98% confidence intervals for \(\mu\) are to be found.

Mark scheme (a)

Scheme	Marks
\(\bar{x} = \dfrac{1}{2}(118.8 + 121.2) = 120\)	B1
1.6449 (or better) “their 1.6449” \(\dfrac{\sigma}{\sqrt{n}} = 121.2 - 120\)	B1 M1
2.3263 (or better) “their 2.3263” \(\dfrac{\sigma}{\sqrt{n}} = 2.3263 \times \left(\dfrac{121.2 - 120}{1.6449}\right)\)	B1 dM1
So 98% C.I. = \(120 \pm 1.424\ldots = (118.3028\ldots,\ 121.699\ldots)\) awrt (118, 122)	A1
	(6)

Notes

NB in part (a) only lose one of the B1 marks for not using the percentage points table

1st B1 for \(\bar{x} = 120\)

2nd B1 for 1.6449 or better in an attempt (could be \(1.6449\sigma = k\) or even \(1.6449\ \sigma^2 = k\))
Condone strange notation for standard error (\(E\)) here if it is used correctly

1st M1 for an attempt to find “width” or “half-width” of a 90% CI ft their \(z\) value (\(|z| \gt 1.5\))
e.g. for \(zE = 121.2 - 120\) (o.e.) N.B. \(E = 0.7295\ldots\) Condone missing 2 here.

3rd B1 for 2.3263 or better in an attempt at CI.
If score 2nd B0 for using 1.64 or 1.645 allow 3rd B1 for 2.32 or 2.33 here

2nd dM1 for a correct attempt at “width” or “half-width” of a 98% CI ft their \(z\) value (\(|z| \gt 2\))
Dependent on 1st M1 and ft their value or expression for s.e.

A1 for lower limit in range [118, 118.35) and upper limit in range (121.65, 122]

Answer only of awrt (118, 122) with no incorrect working seen scores 6/6/ if 1.6449 and 2.3263 are seen and 5/6 (B1B1M1B0M1A1) otherwise.

Scheme	Marks
awrt \((118\pi,\ 122\pi)\) or (371/372, 382/383)	B1ft
	(1)

Scheme	Marks
P(All) = \((0.98)^3\)	M1
= 0.941	A1
	(2)
	(9 marks)