A2 June 2025 Q7

EdexcelCurrent spec11 marksConfidence Intervals

7. A random sample \(X_1, X_2, X_3, X_4\) and \(X_5\) is taken from a distribution \(X \sim \mathrm{N}\left(\mu, \sigma^2\right)\)

Nina and Jonah each use unbiased estimators to estimate \(\mu\)

Nina uses \(\displaystyle\bar{X} = \frac{1}{5}\sum_{r=1}^{5} X_r\) but Jonah uses \(\displaystyle Y = \frac{X_1 + X_5}{2}\)

(a) Find \(\mathrm{P}\left(\bar{X} - Y \gt \sigma\right)\) (7)

Nina and Jonah both calculated 95% confidence intervals for \(\mu\)

They used their estimates and the normal distribution with the same value of \(\sigma\)

Their intervals were

\[(21.823,\ 38.177) \quad\text{and}\quad (27.029,\ 37.371)\]

(b) State, giving a reason, which of these intervals Nina calculated. (1)

Mark scheme (a)

Scheme	Marks	AO
[Let \(D = \bar{X} - Y\)] \(\mathrm{E}(D) = 0\)	B1	1.1b
\((D =)\ \dfrac{X_1 + X_2 + X_3 + X_4 + X_5}{5} - \dfrac{X_1 + X_5}{2}\) oe	M1	3.1a
\(= \dfrac{1}{10}\Big[2\left(X_2 + X_3 + X_4\right) - 3\left(X_1 + X_5\right)\Big]\) oe	M1 A1	2.1 1.1b
\(\mathrm{Var}(D) = \dfrac{1}{100}\left(4 \times 3\sigma^2 + 9 \times 2\sigma^2\right)\)	dM1	1.1b
\(= \dfrac{3}{10}\sigma^2\)	A1	1.1b
So \(\mathrm{P}(D \gt \sigma) = \mathrm{P}\left(Z \gt \dfrac{\sigma - 0}{\sigma\sqrt{\frac{3}{10}}}\right) = 0.03394\ldots\) awrt 0.034	A1	3.2a
	(7)

Notes

B1: for \(\mathrm{E}(D) = 0\) (May be implied by their other working). May use other letters.

1^st M1: for an attempt at \(\bar{X} - Y\) (any expression using \(X_1, X_2, X_3, X_4\) and \(X_5\)). May be seen in attempt to find a probability or \(\mathrm{E}(D)\)

2^nd M1: for an attempt to eliminate the “repeats” – condone missing \(\frac{1}{10}\) or errors in “2” and “3”

1^st A1: for a correct expression for \(D\) with no “repeats”

3^rd dM1: for a correct application of \(\mathrm{Var}(aX - bY)\) dep on 2^nd M1 but can ft their expression

2^nd A1: for the correct variance (may be implied by a correct answer)

3^rd A1: for awrt 0.034

Mark scheme (b)

Scheme	Marks	AO
(27.029, 37.371) since this is the narrower interval or her sample was greater oe	B1	2.2a
	(1)

Notes

B1: for choosing the correct interval and giving a suitable reason based on width or sample size. Contradictory or incorrect statements is B0. Do not accept just referring to the standard deviation being smaller without going onto explain the effect on the width of the interval

Mark scheme (c)

Scheme	Marks	AO
e.g. \(37.371 - 27.029 = [2] \times \dfrac{\sigma}{\sqrt{\text{“}5\text{”}}} \times 1.96\) (or \(38.177 - 21.823 = [2] \times \dfrac{\sigma}{\sqrt{\text{“}2\text{”}}} \times 1.96\))	M1	3.4
\(\sigma = \dfrac{10.342 \times \sqrt{5}}{2 \times 1.96}\) \(\sigma = \dfrac{16.354 \times \sqrt{2}}{2 \times 1.96}\)	A1	1.1b
\(= 5.8993\ldots\) awrt 5.9 \(= 5.900\ldots\) awrt 5.9	A1	1.1b
	(3)
	(11 marks)

Notes

M1: for an attempt at a correct equation in \(\sigma\). Condone wrong \(n\) and missing \(\times 2\) must have 1.96.
Alternatively for an attempt to form two simultaneous equations in \(\bar{x}\) and \(\sigma\), condoning a wrong \(n\). (Condone labelling \(\bar{x}\) as \(\mu\)) Must have 1.96

1^st A1: for a correct expression for \(\sigma\). May be implied by awrt 5.9

2^nd A1: for awrt 5.9