June 2024 Paper 3 Q15

AQACurrent spec9 marksBinomial Distribution

15 It is given that

\[X \sim \mathrm{B}(48, 0.175)\]

(a) Find the mean of \(X\) [1 mark]

(b) Show that the variance of \(X\) is 6.93 [1 mark]

(d) Find \(\mathrm{P}(X \geqslant 6)\) [2 marks]

(e) Find \(\mathrm{P}(9 \leqslant X \leqslant 15)\) [2 marks]

(f) The aeroplanes used on a particular route have 48 seats.

The proportion of passengers who use this route to travel for business is known to be 17.5%

Make two comments on whether it would be appropriate to use \(X\) to model the number of passengers on an aeroplane who are travelling for business using this route. [2 marks]

Scheme	Marks	AO
Obtains 8.4	B1	1.1b
	(1)

Scheme	Marks	AO
Shows how to obtain 6.93 by using the correct formula for variance AG	B1	1.1b
	(1)

Scheme	Marks	AO
Obtains AWFW [0.674, 0.6742]	B1	1.1b
	(1)

Mark scheme (d)

Scheme	Marks	AO
Finds \(\mathrm{P}(X \leqslant 5)\) = [0.132, 0.133] or \(\mathrm{P}(X \leqslant 6)\) = [0.241, 0.242] or \(\mathrm{P}(X \gt 6)\) = [0.758, 0.759] or \(\mathrm{P}(X \geqslant 6)\) = [0.867, 0.868] PI by correct answer Condone incorrect or no labels	M1	1.1a
Obtains AWFW [0.867, 0.868]	A1	1.1b
	(2)

Typical solution

\[\mathrm{P}(X \geqslant 6) = 1 - \mathrm{P}(X \leqslant 5)\]\[= 1 - 0.1325\]\[= 0.8675\]

Mark scheme (e)

Scheme	Marks	AO
Finds \(\mathrm{P}(X \leqslant 15)\) = [0.994, 0.9941] or \(\mathrm{P}(X \leqslant 8)\) = [0.531, 0.532] PI by correct answer Condone incorrect or no labels	M1	1.1a
Obtains AWFW [0.462, 0.4631]	A1	1.1b
	(2)

Typical solution

\[\mathrm{P}(9 \leqslant X \leqslant 15)\]\[= 0.9940 - 0.5317\]\[= 0.4623\]

Mark scheme (f)

Scheme	Marks	AO
States that the model is appropriate because The 48 seats and 17.5% passengers travelling for business match the values of \(n\) and \(p\) in the model or Only business or non-business passenger Business/non-business passengers occur independently or states that the model is not appropriate because The plane may not be full so \(n\) might change or might not be 48 The \(p\) might change or is not fixed or might not be 0.175 due to eg peak or holiday periods Not independent because passengers might book seats in groups Do not allow probability being independent	E1	3.5b
Makes a second non-contradictory comment in context from the list above, about the suitability of the model	E1	3.5b
	(2)
	(9 marks)

Typical solution

Model is appropriate because 48 seats and 17.5% passengers travelling for business match the values of \(n\) and \(p\).

However the plane may not be full so \(n\) may not be 48.