Moderate -0.8 This is a straightforward binomial distribution question requiring only standard recall and calculator work. Parts (a)-(e) involve direct application of formulas (mean = np, variance = np(1-p)) and calculator functions for binomial probabilities. Part (f) requires stating standard conditions for binomial modeling (fixed n, independent trials, constant p), which is routine A-level statistics content with no problem-solving or novel insight required.
It is given that
$$X \sim \text{B}(48, 0.175)$$
\begin{enumerate}[label=(\alph*)]
\item Find the mean of \(X\)
[1 mark]
\item Show that the variance of \(X\) is 6.93
[1 mark]
\item Find P(\(X < 10\))
[1 mark]
\item Find P(\(X \geq 6\))
[2 marks]
\item Find P(\(9 \leq X \leq 15\))
[2 marks]
\item 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]
Question 15:
--- 15(a) ---
15(a) | Obtains 8.4 | 1.1b | B1 | 8.4
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 15(b) ---
15(b) | Shows how to obtain 6.93 by
using the correct formula for
variance AG | 1.1b | B1 | Variance = 48 × 0.175 × 0.825
= 6.93
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 15(c) ---
15(c) | Obtains AWFW [0.674, 0.6742] | 1.1b | B1 | 0.6742
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 15(d) ---
15(d) | Finds
P(X 5) = [0.132, 0.133] or
P(X 6) = [0.241, 0.242] or
P(X > 6) = [0.758, 0.759] or
P(X ≥ 6) = [0.867, 0.868]
PI by correct answer
Condone incorrect or no labels | 1.1a | M1 | P (X ≥ 6) = 1 – P (X ≤ 5)
= 1 – 0.1325
= 0.8675
Obtains AWFW [0.867, 0.868] | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 15(e) ---
15(e) | Finds
P(X 15) = [0.994, 0.9941] or
P(X 8) = [0.531, 0.532]
PI by correct answer
Condone incorrect or no labels | 1.1a | M1 | P (9 ≤ X ≤ 15)
= 0.9940 – 0.5317
= 0.4623
Obtains AWFW [0.462, 0.4631] | 1.1b | A1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 15(f) ---
15(f) | States that the model is
appropriate because
1. The 48 seats and 17.5%
passengers travelling for
business match the
values of n and p in the
model or
2. Only business or non-
business passenger
3. Business/non-business
passengers occur
independently
or
states that the model is not
appropriate because
1. The plane may not be
full so n might change or
might not be 48
2. The p might change or is
not fixed or might not be
0.175 due to eg peak or
holiday periods
3. Not independent
because passengers
might book seats in
groups
Do not allow probability being
independent | 3.5b | E1 | 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.
Makes a second
non-contradictory comment in
context from the list above,
about the suitability of the model | 3.5b | E1
Subtotal | 2
Question 15 Total | 9
Q | Marking instructions | AO | Marks | Typical solution
It is given that
$$X \sim \text{B}(48, 0.175)$$
\begin{enumerate}[label=(\alph*)]
\item Find the mean of $X$
[1 mark]
\item Show that the variance of $X$ is 6.93
[1 mark]
\item Find P($X < 10$)
[1 mark]
\item Find P($X \geq 6$)
[2 marks]
\item Find P($9 \leq X \leq 15$)
[2 marks]
\item 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]
</end{enumerate}
\hfill \mbox{\textit{AQA Paper 3 2024 Q15 [9]}}