AQA S3 2015 June — Question 5 16 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2015
SessionJune
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating the Binomial to the Poisson distribution
TypeProve binomial mean or variance
DifficultyStandard +0.3 This is a standard S3 multi-part question testing routine bookwork (proving binomial mean from first principles, deriving variance using given result) and straightforward applications (finding parameters, checking distribution properties, Poisson approximation). All parts follow textbook methods with no novel insight required, making it slightly easier than average.
Spec2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial5.02b Expectation and variance: discrete random variables5.02d Binomial: mean np and variance np(1-p)5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!
5
  1. The random variable \(X\) has a binomial distribution with parameters \(n\) and \(p\).
    1. Prove, from first principles, that \(\mathrm { E } ( X ) = n p\).
    2. Given that \(\mathrm { E } ( X ( X - 1 ) ) = n ( n - 1 ) p ^ { 2 }\), find an expression for \(\operatorname { Var } ( X )\).
    1. The random variable \(Y\) has a binomial distribution with \(\mathrm { E } ( Y ) = 3\) and \(\operatorname { Var } ( Y ) = 2.985\). Find values for \(n\) and \(p\).
    2. The random variable \(U\) has \(\mathrm { E } ( U ) = 5\) and \(\operatorname { Var } ( U ) = 6.25\). Show that \(U\) does not have a binomial distribution.
  3. The random variable \(V\) has the distribution \(\operatorname { Po } ( 5 )\) and \(W = 2 V + 10\). Show that \(\mathrm { E } ( W ) = \operatorname { Var } ( W )\) but that \(W\) does not have a Poisson distribution.
  4. The probability that, in a particular country, a person has blood group AB negative is 0.2 per cent. A sample of 5000 people is selected. Given that the sample may be assumed to be random, use a distributional approximation to estimate the probability that at least 6 people but at most 12 people have blood group AB negative.
    [0pt] [3 marks]
Question 5:
Part 5(a)(i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E(X) = \sum_{x=0}^{n} x \binom{n}{x} p^x (1-p)^{n-x}\)M1 Writing out sum with binomial probabilities
\(= \sum_{x=1}^{n} x \cdot \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x}\) x=0 term vanishes
\(= \sum_{x=1}^{n} \frac{n!}{(x-1)!(n-x)!} p^x (1-p)^{n-x}\)M1 Cancelling x with x!
\(= np \sum_{x=1}^{n} \binom{n-1}{x-1} p^{x-1}(1-p)^{n-x}\) Factoring out np
\(= np(p+(1-p))^{n-1} = np\)A1 Recognising binomial sum = 1
Part 5(a)(ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\text{Var}(X) = E(X(X-1)) + E(X) - [E(X)]^2\)M1 Using \(\text{Var}(X)=E(X^2)-[E(X)]^2\) and \(E(X^2)=E(X(X-1))+E(X)\)
\(= n(n-1)p^2 + np - n^2p^2 = np(1-p)\)A1 Correct simplification
Part 5(b)(i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(np = 3\) and \(np(1-p) = 2.985\)M1 Setting up two equations
\(1-p = \frac{2.985}{3} = 0.995\), so \(p = 0.005\)A1 Correct value of p
\(n = \frac{3}{0.005} = 600\)A1 Correct value of n
Part 5(b)(ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
For binomial: \(\text{Var}(U) < E(U)\), i.e. \(np(1-p) < np\)M1 Correct condition stated
But \(6.25 > 5\), so \(\text{Var}(U) > E(U)\), contradictionA1 Correct conclusion
Part 5(c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E(W) = E(2V+10) = 2E(V)+10 = 2(5)+10 = 20\)B1 Correct E(W)
\(\text{Var}(W) = 4\text{Var}(V) = 4(5) = 20\)B1 Correct Var(W)
So \(E(W) = \text{Var}(W) = 20\)
For Poisson, must take only non-negative integer values; W takes values \(10, 12, 14,...\) only (even integers \(\geq 10\)), so not PoissonB1 Valid reason e.g. W cannot equal 0,1,2...
Part 5(d)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\lambda = 5000 \times 0.002 = 10\)B1 Correct Poisson approximation \(X \sim \text{Po}(10)\)
\(P(6 \leq X \leq 12) = P(X \leq 12) - P(X \leq 5)\)M1 Correct probability expression
\(= 0.7916 - 0.0671 = 0.7245\)A1 Correct answer (from tables) 
# Question 5:

## Part 5(a)(i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = \sum_{x=0}^{n} x \binom{n}{x} p^x (1-p)^{n-x}$ | M1 | Writing out sum with binomial probabilities |
| $= \sum_{x=1}^{n} x \cdot \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x}$ | | x=0 term vanishes |
| $= \sum_{x=1}^{n} \frac{n!}{(x-1)!(n-x)!} p^x (1-p)^{n-x}$ | M1 | Cancelling x with x! |
| $= np \sum_{x=1}^{n} \binom{n-1}{x-1} p^{x-1}(1-p)^{n-x}$ | | Factoring out np |
| $= np(p+(1-p))^{n-1} = np$ | A1 | Recognising binomial sum = 1 |

## Part 5(a)(ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(X) = E(X(X-1)) + E(X) - [E(X)]^2$ | M1 | Using $\text{Var}(X)=E(X^2)-[E(X)]^2$ and $E(X^2)=E(X(X-1))+E(X)$ |
| $= n(n-1)p^2 + np - n^2p^2 = np(1-p)$ | A1 | Correct simplification |

## Part 5(b)(i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $np = 3$ and $np(1-p) = 2.985$ | M1 | Setting up two equations |
| $1-p = \frac{2.985}{3} = 0.995$, so $p = 0.005$ | A1 | Correct value of p |
| $n = \frac{3}{0.005} = 600$ | A1 | Correct value of n |

## Part 5(b)(ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| For binomial: $\text{Var}(U) < E(U)$, i.e. $np(1-p) < np$ | M1 | Correct condition stated |
| But $6.25 > 5$, so $\text{Var}(U) > E(U)$, contradiction | A1 | Correct conclusion |

## Part 5(c)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(W) = E(2V+10) = 2E(V)+10 = 2(5)+10 = 20$ | B1 | Correct E(W) |
| $\text{Var}(W) = 4\text{Var}(V) = 4(5) = 20$ | B1 | Correct Var(W) |
| So $E(W) = \text{Var}(W) = 20$ | | |
| For Poisson, must take only non-negative integer values; W takes values $10, 12, 14,...$ only (even integers $\geq 10$), so not Poisson | B1 | Valid reason e.g. W cannot equal 0,1,2... |

## Part 5(d)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\lambda = 5000 \times 0.002 = 10$ | B1 | Correct Poisson approximation $X \sim \text{Po}(10)$ |
| $P(6 \leq X \leq 12) = P(X \leq 12) - P(X \leq 5)$ | M1 | Correct probability expression |
| $= 0.7916 - 0.0671 = 0.7245$ | A1 | Correct answer (from tables) |
5
\begin{enumerate}[label=(\alph*)]
\item The random variable $X$ has a binomial distribution with parameters $n$ and $p$.
\begin{enumerate}[label=(\roman*)]
\item Prove, from first principles, that $\mathrm { E } ( X ) = n p$.
\item Given that $\mathrm { E } ( X ( X - 1 ) ) = n ( n - 1 ) p ^ { 2 }$, find an expression for $\operatorname { Var } ( X )$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item The random variable $Y$ has a binomial distribution with $\mathrm { E } ( Y ) = 3$ and $\operatorname { Var } ( Y ) = 2.985$.

Find values for $n$ and $p$.
\item The random variable $U$ has $\mathrm { E } ( U ) = 5$ and $\operatorname { Var } ( U ) = 6.25$.

Show that $U$ does not have a binomial distribution.
\end{enumerate}\item The random variable $V$ has the distribution $\operatorname { Po } ( 5 )$ and $W = 2 V + 10$.

Show that $\mathrm { E } ( W ) = \operatorname { Var } ( W )$ but that $W$ does not have a Poisson distribution.
\item The probability that, in a particular country, a person has blood group AB negative is 0.2 per cent. A sample of 5000 people is selected.

Given that the sample may be assumed to be random, use a distributional approximation to estimate the probability that at least 6 people but at most 12 people have blood group AB negative.\\[0pt]
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA S3 2015 Q5 [16]}}
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