Question 4 - A-Level Maths

AQA S1 2012 January — Question 4 14 marks

Exam Board	AQA
Module	S1 (Statistics 1)
Year	2012
Session	January
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Binomial Distribution
Type	Probability of range of values
Difficulty	Moderate -0.3 This is a straightforward application of binomial distribution with standard calculations: direct use of binomial probability formula/tables for parts (a)(i)-(iii), routine mean/SD formulas for part (b), and basic comparison of sample statistics to theoretical values in part (c). All techniques are standard S1 content requiring no novel insight, though the multi-part structure and cumulative probability calculations place it slightly below average difficulty.
Spec	2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities 2.04d Normal approximation to binomial

4 The records at a passport office show that, on average, 15 per cent of photographs that accompany applications for passport renewals are unusable. Assume that exactly one photograph accompanies each application.

Determine the probability that in a random sample of 40 applications:
1. exactly 6 photographs are unusable;
2. at most 5 photographs are unusable;
3. more than 5 but fewer than 10 photographs are unusable.
Calculate the mean and the standard deviation for the number of photographs that are unusable in a random sample of $\mathbf { 3 2 }$ applications.
Mr Stickler processes 32 applications each day. His records for the previous 10 days show that the numbers of photographs that he deemed unusable were $$\begin{array} { l l l l l l l l l l } 8 & 6 & 10 & 7 & 9 & 7 & 8 & 9 & 6 & 7 \end{array}$$ By calculating the mean and the standard deviation of these values, comment, with reasons, on the suitability of the $\mathrm { B } ( 32,0.15 )$ model for the number of photographs deemed unusable each day by Mr Stickler.

Show mark scheme Show mark scheme source

Question 4:

Part (a)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$U \sim B(40, 0.15)$	M1	Used somewhere in (a)

Part (a)(i)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$P(U = 6) = 0.6067 - 0.4325$ or $= \binom{40}{6}(0.15)^6(0.85)^{34}$	M1	Accept 3 dp rounding or truncation. Can be implied by a correct answer
$= 0.174$	A1	AWRT $(0.1742)$

Part (a)(ii)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$P(U \leq 5) = 0.432$ to $0.433$	B1	AWFW $(0.4325)$

Part (a)(iii)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$P(5 < U < 10) = 0.9328$ or $0.9701$ $(p_1)$	M1	Accept 3 dp rounding or truncation but allow $0.97$. $p_2 - p_1 \Rightarrow$ M0 M0 A0; $(1-p_2)-p_1 \Rightarrow$ M0 M0 A0; $p_1-(1-p_2) \Rightarrow$ M1 M0 A0; $(1-p_2)-(1-p_1) \Rightarrow$ M1 M1 (A1) only providing result $> 0$
MINUS $0.4325$ or $0.2633$ $(p_2)$	M1	Accept 3 dp rounding or truncation
$= 0.5(00)$ to $0.501$	A1	AWFW $(0.5003)$

Part (b)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Mean or $\mu = 32 \times 0.15 = 4.8$	B1	CAO
$(V$ or $\sigma^2 =)\ 32 \times 0.15 \times 0.85$ or $(SD$ or $\sigma =)\ \sqrt{32 \times 0.15 \times 0.85}$	M1	Either numerical expression; ignore terminology. May be implied by $4.08$ CAO seen or $2.02$ AWRT seen
$(SD$ or $\sigma) = 2.02$	A1	AWRT $(2.0199)$. Do not award if labelled $V$ or $\sigma^2$

Part (c)

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Mean $= 7.7$	B1	CAO $\left(\sum x = 77\right)$
$SD = 1.26$ to $1.34$	B1	AWFW $\left(\sum x^2 = 609\right)$
(Sample) mean is bigger/greater/different or $7.7/32 = 0.24 > 0.15$ and (Sample) SD is smaller/less/different	Bdep1	Both; dependent on all previous 5 marks of B1 M1 A1 B1 B1. Can be scored for incorrect (b) re-done correctly in (c). Means & SDs different $\Rightarrow$ Bdep0
So model appears unsuitable	Bdep1	OE; dependent on Bdep1

Question 4:

Part (a)(iii):

Answer	Marks	Guidance
$B(40, 0.15)$ expressions stated for at least 3 terms within $5 \leq U \leq 10$ gives probability $= 0.5(00)$ to $0.501$	M2, A1	Can be implied by a correct answer; AWFW (0.5003)
$u$	(5)	6
$P(U=u)$	(0.1692)	0.1742

Total: 3 marks

# Question 4:

## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $U \sim B(40, 0.15)$ | M1 | Used somewhere in (a) |

## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(U = 6) = 0.6067 - 0.4325$ or $= \binom{40}{6}(0.15)^6(0.85)^{34}$ | M1 | Accept 3 dp rounding or truncation. Can be implied by a correct answer |
| $= 0.174$ | A1 | AWRT $(0.1742)$ |

## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(U \leq 5) = 0.432$ to $0.433$ | B1 | AWFW $(0.4325)$ |

## Part (a)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(5 < U < 10) = 0.9328$ or $0.9701$ $(p_1)$ | M1 | Accept 3 dp rounding or truncation but allow $0.97$. $p_2 - p_1 \Rightarrow$ M0 M0 A0; $(1-p_2)-p_1 \Rightarrow$ M0 M0 A0; $p_1-(1-p_2) \Rightarrow$ M1 M0 A0; $(1-p_2)-(1-p_1) \Rightarrow$ M1 M1 (A1) only providing result $> 0$ |
| MINUS $0.4325$ or $0.2633$ $(p_2)$ | M1 | Accept 3 dp rounding or truncation |
| $= 0.5(00)$ to $0.501$ | A1 | AWFW $(0.5003)$ |

## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean or $\mu = 32 \times 0.15 = 4.8$ | B1 | CAO |
| $(V$ or $\sigma^2 =)\ 32 \times 0.15 \times 0.85$ or $(SD$ or $\sigma =)\ \sqrt{32 \times 0.15 \times 0.85}$ | M1 | Either numerical expression; ignore terminology. May be implied by $4.08$ CAO seen or $2.02$ AWRT seen |
| $(SD$ or $\sigma) = 2.02$ | A1 | AWRT $(2.0199)$. Do not award if labelled $V$ or $\sigma^2$ |

## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean $= 7.7$ | B1 | CAO $\left(\sum x = 77\right)$ |
| $SD = 1.26$ to $1.34$ | B1 | AWFW $\left(\sum x^2 = 609\right)$ |
| (Sample) mean is bigger/greater/different or $7.7/32 = 0.24 > 0.15$ and (Sample) SD is smaller/less/different | Bdep1 | Both; dependent on all previous 5 marks of B1 M1 A1 B1 B1. Can be scored for incorrect (b) re-done correctly in (c). Means & SDs different $\Rightarrow$ Bdep0 |
| So model appears unsuitable | Bdep1 | OE; dependent on Bdep1 |

# Question 4:

## Part (a)(iii):
$B(40, 0.15)$ expressions stated for at least 3 terms within $5 \leq U \leq 10$ gives probability $= 0.5(00)$ to $0.501$ | M2, A1 | Can be implied by a correct answer; AWFW (0.5003)

| $u$ | (5) | 6 | 7 | 8 | 9 | (10) |
|---|---|---|---|---|---|---|
| $P(U=u)$ | (0.1692) | 0.1742 | 0.1492 | 0.1087 | 0.0682 | (0.0373) |

Total: 3 marks

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4 The records at a passport office show that, on average, 15 per cent of photographs that accompany applications for passport renewals are unusable.

Assume that exactly one photograph accompanies each application.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that in a random sample of 40 applications:
\begin{enumerate}[label=(\roman*)]
\item exactly 6 photographs are unusable;
\item at most 5 photographs are unusable;
\item more than 5 but fewer than 10 photographs are unusable.
\end{enumerate}\item Calculate the mean and the standard deviation for the number of photographs that are unusable in a random sample of $\mathbf { 3 2 }$ applications.
\item Mr Stickler processes 32 applications each day. His records for the previous 10 days show that the numbers of photographs that he deemed unusable were

$$\begin{array} { l l l l l l l l l l } 
8 & 6 & 10 & 7 & 9 & 7 & 8 & 9 & 6 & 7
\end{array}$$

By calculating the mean and the standard deviation of these values, comment, with reasons, on the suitability of the $\mathrm { B } ( 32,0.15 )$ model for the number of photographs deemed unusable each day by Mr Stickler.
\end{enumerate}

\hfill \mbox{\textit{AQA S1 2012 Q4 [14]}}

This paper (7 questions)

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Q1 4 Q2 3 Q3 12 Q4 14 Q5 17 Q6 11 Q7 14

Answer	Marks	Guidance
\(B(40, 0.15)\) expressions stated for at least 3 terms within \(5 \leq U \leq 10\) gives probability \(= 0.5(00)\) to \(0.501\)	M2, A1	Can be implied by a correct answer; AWFW (0.5003)
\(u\)	(5)	6
\(P(U=u)\)	(0.1692)	0.1742