| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Probability of range of values |
| Difficulty | Standard +0.3 This is a straightforward S1 binomial distribution question requiring standard calculations: cumulative probabilities, mean/variance formulas, and basic data comparison. All parts use direct formula application with no novel problem-solving, though part (d)(ii) requires simple interpretation. Slightly easier than average due to routine nature. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02d Binomial: mean np and variance np(1-p) |
| Number of mornings | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Frequency | 10 | 8 | 7 | 7 | 5 | 5 | 4 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(p = 0.4\), attempted use of \(B(7, 0.4)\) | M1 | |
| \(P(X \leq 2) = 0.419\) to \(0.421\) | B1 | AWFW (0.4199) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(X > 1 \text{ and } X < 5) = P(2 \leq X \leq 4)\) | ||
| \(= P(X \leq 4)\) | M1 | Identification of at least 2, 3 and 4 |
| \(- P(X \leq 1)\) | M1 | Identification of exactly 2, 3 and 4 |
| \(= 0.9037 - 0.1586 = 0.744\) to \(0.746\) | A1 | AWFW (0.7451) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P(Y=7) = \binom{n}{7}(0.4)^7(0.6)^{n-7}\) | M1 | Correct expression for \(B(7; n, 0.4)\) with \(n \neq 7\) |
| \(= \binom{28}{7}(0.4)^7(0.6)^{21}\) | A1 | Fully correct expression, may be implied |
| \(= 0.0425\) to \(0.0427\) | A1 | AWFW (0.042556) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Mean \(= np = 2.8\) | B1 | CAO |
| \(SD = \sqrt{np(1-p)} = \sqrt{1.68} = 1.29\) to \(1.31\) | B1 | AWFW |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Mean \(= 2.8\) | B1 | CAO \(\Sigma fx = 140\) |
| \(SD = 2.24\) to \(2.27\) | B2 | AWFW \(\Sigma fx^2 = 644\); substitution of values into correct formula for variance or SD; \(SD = 5.03\) to \(5.15\) AWFW |
| \(s^2_{n-1} = 5.14\) to \(5.15\) and \(s^2_{n-1} = 5.04\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Means are the same | B1\(\checkmark\) | ft on (c) and (d)(i); must be \(s\) with \(\sigma\) or \(s^2\) with \(\sigma^2\) |
| SDs differ greatly | B1\(\checkmark\) | ft on (c) and (d)(i) |
| Thus answers do not support Aaron's belief | B1 | Dependent on B1 above CAO |
# Question 5:
## Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $p = 0.4$, attempted use of $B(7, 0.4)$ | M1 | |
| $P(X \leq 2) = 0.419$ to $0.421$ | B1 | AWFW (0.4199) |
## Part (a)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(X > 1 \text{ and } X < 5) = P(2 \leq X \leq 4)$ | | |
| $= P(X \leq 4)$ | M1 | Identification of **at least** 2, 3 and 4 |
| $- P(X \leq 1)$ | M1 | Identification of **exactly** 2, 3 and 4 |
| $= 0.9037 - 0.1586 = 0.744$ to $0.746$ | A1 | AWFW (0.7451) |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(Y=7) = \binom{n}{7}(0.4)^7(0.6)^{n-7}$ | M1 | Correct expression for $B(7; n, 0.4)$ with $n \neq 7$ |
| $= \binom{28}{7}(0.4)^7(0.6)^{21}$ | A1 | Fully correct expression, may be implied |
| $= 0.0425$ to $0.0427$ | A1 | AWFW (0.042556) |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Mean $= np = 2.8$ | B1 | CAO |
| $SD = \sqrt{np(1-p)} = \sqrt{1.68} = 1.29$ to $1.31$ | B1 | AWFW |
## Part (d)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| Mean $= 2.8$ | B1 | CAO $\Sigma fx = 140$ |
| $SD = 2.24$ to $2.27$ | B2 | AWFW $\Sigma fx^2 = 644$; substitution of values into correct formula for variance or SD; $SD = 5.03$ to $5.15$ AWFW | M1 |
| $s^2_{n-1} = 5.14$ to $5.15$ and $s^2_{n-1} = 5.04$ | | |
## Part (d)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Means are the same | B1$\checkmark$ | ft on (c) and (d)(i); must be $s$ with $\sigma$ or $s^2$ with $\sigma^2$ |
| SDs differ greatly | B1$\checkmark$ | ft on (c) and (d)(i) |
| Thus answers do not support Aaron's belief | B1 | Dependent on B1 above CAO |
---5 Each evening Aaron sets his alarm for 7 am. He believes that the probability that he wakes before his alarm rings each morning is 0.4 , and is independent from morning to morning.
\begin{enumerate}[label=(\alph*)]
\item Assuming that Aaron's belief is correct, determine the probability that, during a week (7 mornings), he wakes before his alarm rings:
\begin{enumerate}[label=(\roman*)]
\item on 2 or fewer mornings;
\item on more than 1 but fewer than 5 mornings.
\end{enumerate}\item Assuming that Aaron's belief is correct, calculate the probability that, during a 4 -week period, he wakes before his alarm rings on exactly 7 mornings.
\item Assuming that Aaron's belief is correct, calculate values for the mean and standard deviation of the number of mornings in a week when Aaron wakes before his alarm rings.\\
(2 marks)
\item During a 50-week period, Aaron records, each week, the number of mornings on which he wakes before his alarm rings. The results are as follows.
\begin{center}
\begin{tabular}{ | l | r | r | r | r | r | r | r | r | }
\hline
Number of mornings & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline
Frequency & 10 & 8 & 7 & 7 & 5 & 5 & 4 & 4 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Calculate the mean and standard deviation of these data.
\item State, giving reasons, whether your answers to part (d)(i) support Aaron's belief that the probability that he wakes before his alarm rings each morning is 0.4 , and is independent from morning to morning.\\
(3 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2005 Q5 [15]}}