AQA S1 2015 June — Question 6 12 marks

Exam BoardAQA
ModuleS1 (Statistics 1)
Year2015
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Distribution
TypeMultiple independent binomial calculations
DifficultyModerate -0.8 This is a straightforward application of standard binomial distribution formulas with no conceptual challenges. Part (a) is direct calculation of P(X=2), part (b) requires cumulative probability lookups from tables, and part (c) is simple recall of mean=np and variance=np(1-p). All are routine textbook exercises requiring only formula application and table reading.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities

6 Customers at a supermarket can pay at a checkout either by cash, debit card or credit card.
  1. The probability that a customer pays by cash is 0.22 . Calculate the probability that exactly 2 customers from a random sample of 24 customers pay by cash.
  2. The probability that a customer pays by debit card is 0.45 . Determine the probability that the number of customers who pay by debit card from a random sample of \(\mathbf { 4 0 }\) customers is:
    1. fewer than 20 ;
    2. more than 15 ;
    3. at least 12 but at most 24 .
  3. The random variable \(W\) denotes the number of customers who pay by credit card from a random sample of \(\mathbf { 2 0 0 }\) customers. Calculate values for the mean and the variance of \(W\).
    [0pt] [3 marks]

Question 6(a):
\(X \sim B(24, 0.22)\), find \(P(X = 2)\)
\(= \binom{24}{2}(0.22)^2(0.78)^{22}\)
Question 6(b):
\(X \sim B(40, 0.45)\)
- (i) \(P(X < 20)\)
- (ii) \(P(X > 15)\)
- (iii) \(P(12 \leq X \leq 24)\)
Question 6(c):
\(W \sim B(200, p)\) where \(p = 1 - 0.22 - 0.45 = 0.33\)
- Mean \(= 200 \times 0.33 = 66\)
- Variance \(= 200 \times 0.33 \times 0.67 = 44.22\)
I can see these are answer space pages (pages 22-24) from an AQA Statistics exam paper (P/Jun15/SS1B), but they are blank answer spaces — they do not contain a mark scheme. They are the pages where students write their answers.
To extract the mark scheme content, I would need the actual mark scheme document for this paper, which is a separate document published by AQA.
However, based on the questions shown on page 22, I can outline the expected working/answers:
**Question 6(a):**
$X \sim B(24, 0.22)$, find $P(X = 2)$
$= \binom{24}{2}(0.22)^2(0.78)^{22}$

**Question 6(b):**
$X \sim B(40, 0.45)$
- (i) $P(X < 20)$
- (ii) $P(X > 15)$
- (iii) $P(12 \leq X \leq 24)$

**Question 6(c):**
$W \sim B(200, p)$ where $p = 1 - 0.22 - 0.45 = 0.33$
- Mean $= 200 \times 0.33 = 66$
- Variance $= 200 \times 0.33 \times 0.67 = 44.22$

I can see these are answer space pages (pages 22-24) from an AQA Statistics exam paper (P/Jun15/SS1B), but they are **blank answer spaces** — they do not contain a mark scheme. They are the pages where students write their answers.

To extract the mark scheme content, I would need the actual **mark scheme document** for this paper, which is a separate document published by AQA.

However, based on the **questions shown on page 22**, I can outline the expected working/answers:

---
6 Customers at a supermarket can pay at a checkout either by cash, debit card or credit card.
\begin{enumerate}[label=(\alph*)]
\item The probability that a customer pays by cash is 0.22 .

Calculate the probability that exactly 2 customers from a random sample of 24 customers pay by cash.
\item The probability that a customer pays by debit card is 0.45 .

Determine the probability that the number of customers who pay by debit card from a random sample of $\mathbf { 4 0 }$ customers is:
\begin{enumerate}[label=(\roman*)]
\item fewer than 20 ;
\item more than 15 ;
\item at least 12 but at most 24 .
\end{enumerate}\item The random variable $W$ denotes the number of customers who pay by credit card from a random sample of $\mathbf { 2 0 0 }$ customers.

Calculate values for the mean and the variance of $W$.\\[0pt]
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA S1 2015 Q6 [12]}}