| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Finding binomial parameters from properties |
| Difficulty | Moderate -0.8 This is a straightforward application of standard binomial distribution formulas. Part (a) requires recalling that mean of B(n,p) is np and variance is np(1-p), then solving a simple equation: 40p = 25(0.6)(0.4) gives p = 0.15. Part (b) is direct calculator work using the binomial probability formula. Both parts are routine recall and calculation with no problem-solving or conceptual challenge, making this easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks |
|---|---|
| 14(a) | State one correct expression for |
| Answer | Marks | Guidance |
|---|---|---|
| PI | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| variance of Y | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 1.1b | A1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 14(b) | Obtains correct probability | |
| AWRT 0.12 | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question 14 Total | 4 | |
| Q | Marking instructions | AO |
Question 14:
--- 14(a) ---
14(a) | State one correct expression for
the mean of X or the variance of
Y using given values of n and p
PI | 1.1b | B1 | Mean of X = 40p
Variance of Y = 25 × 0.6 × 0.4
= 6
So 40p = 6
3
p =
20
Equates their mean and
variance and solves to find a
value for their p provided
0 < p < 1
Do not accept mean of Y
3
(15 giving p = ) as their
8
variance of Y | 3.1a | M1
3
Obtains p =
2 0
ACF | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 14(b) ---
14(b) | Obtains correct probability
AWRT 0.12 | 1.1b | B1 | P(Y = 17) = 0.1199797
= 0.120
Subtotal | 1
Question 14 Total | 4
Q | Marking instructions | AO | Marks | Typical solutionThe discrete random variables $X$ and $Y$ can be modelled by the distributions
$$X \sim \text{B}(40, p)$$
$$Y \sim \text{B}(25, 0.6)$$
It is given that the mean of $X$ is equal to the variance of $Y$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $p$
[3 marks]
\item Find P($Y = 17$)
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2024 Q14 [4]}}