Edexcel S2 2022 June — Question 1 10 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypePoisson with binomial combination
DifficultyModerate -0.8 This is a straightforward S2 question testing basic recall of Poisson/binomial properties (parts a,b), independence for joint probability (part c), and enumeration of cases (part d). All parts use standard techniques with no novel insight required, making it easier than average A-level questions.
Spec5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^25.02m Poisson: mean = variance = lambda

  1. The independent random variables \(W\) and \(X\) have the following distributions.
$$W \sim \operatorname { Po } ( 4 ) \quad X \sim \mathrm {~B} ( 3,0.8 )$$
  1. Write down the value of the variance of \(W\)
  2. Determine the mode of \(X\) Show your working clearly. One observation from each distribution is recorded as \(W _ { 1 }\) and \(X _ { 1 }\) respectively.
  3. Find \(\mathrm { P } \left( W _ { 1 } = 2 \right.\) and \(\left. X _ { 1 } = 2 \right)\)
  4. Find \(\mathrm { P } \left( X _ { 1 } < W _ { 1 } \right)\)

AnswerMarks Guidance
(a) \(4\)B1 cao
(b) \(P(X = 2) = 3x0.2×0.8^2 = \frac{-48}{125} = 0.384\) or \(P(X = 3) = 0.8^3 = \frac{64}{125} = 0.512\)M1, A1 M1 valid attempt at either probability. A1 3 (M1 must be scored). NB answer only with no method is M0A0
\([X =] 3\) is the modeA1
(c) \(P(W_1 = 2) = \frac{e^{-4}4^2}{2} = [0.1465]\) and \(P(X_1 = 2) = 3 × 0.2 × 0.8^2 = [\frac{48}{125} = 0.384]\)M1, M1 1st M1 both \(P(W_1 = 2)\) Allow \((0.2381 - 0.0916)\) and \(P(X_1 = 2)\). 2nd M1 Poisson probability × binomial probability. If no working shown these probabilities must be correct
\(P(W_1 \text{ and } X_1 = 2) = \frac{e^{-4}4^2}{2} × (3 × 0.2 × 0.8^2) = [0.1465 × 0.384]\)M1
\(= 0.05626564…\) awrt 0.0563A1 A1 awrt 0.0563
(d) \(X_1 = 0 \text{ and } W_1 > 0, X_1 = 1 \text{ and } W_1 > 1, X_1 = 2 \text{ and } W_1 > 2, X_1 = 3 \text{ and } W_1 > 3\)M1
\(0.008×(1 - 0.0183) + 0.096×(1 - 0.0916) + 0.384×(1 - 0.2381) + 0.512×(1 - 0.4335)\)M1M1
\(= 0.677677…\) awrt 0.678A1 1st M1 for listing at least 3 combinations. Implied by 2nd M1. 2nd M1 for sum of at least 3 correct products. Condone consistent use of the tables for 3.5 or 4.5 rather than 4. 3rd M1 for a fully correct expression eg \(0.008×(0.9817) + 0.096×(0.9084) + 0.384×(0.7619) + 0.512×(0.5665)\) condone 0.9816 and 0.7618 Allow figures to 3sf for method or awrt 0.00785 + awrt 0.0872 + awrt 0.293 + awrt 0.290 (allow 0.29). A1 awrt 0.678
[10 marks]
**(a)** $4$ | B1 | cao

**(b)** $P(X = 2) = 3x0.2×0.8^2 = \frac{-48}{125} = 0.384$ or $P(X = 3) = 0.8^3 = \frac{64}{125} = 0.512$ | M1, A1 | M1 valid attempt at either probability. A1 3 (M1 must be scored). NB answer only with no method is M0A0

$[X =] 3$ is the mode | A1 |

**(c)** $P(W_1 = 2) = \frac{e^{-4}4^2}{2} = [0.1465]$ and $P(X_1 = 2) = 3 × 0.2 × 0.8^2 = [\frac{48}{125} = 0.384]$ | M1, M1 | 1st M1 both $P(W_1 = 2)$ Allow $(0.2381 - 0.0916)$ and $P(X_1 = 2)$. 2nd M1 Poisson probability × binomial probability. If no working shown these probabilities must be correct

$P(W_1 \text{ and } X_1 = 2) = \frac{e^{-4}4^2}{2} × (3 × 0.2 × 0.8^2) = [0.1465 × 0.384]$ | M1 |

$= 0.05626564…$ awrt **0.0563** | A1 | A1 awrt 0.0563

**(d)** $X_1 = 0 \text{ and } W_1 > 0, X_1 = 1 \text{ and } W_1 > 1, X_1 = 2 \text{ and } W_1 > 2, X_1 = 3 \text{ and } W_1 > 3$ | M1 |

$0.008×(1 - 0.0183) + 0.096×(1 - 0.0916) + 0.384×(1 - 0.2381) + 0.512×(1 - 0.4335)$ | M1M1 |

$= 0.677677…$ awrt **0.678** | A1 | 1st M1 for listing at least 3 combinations. Implied by 2nd M1. 2nd M1 for sum of at least 3 correct products. Condone consistent use of the tables for 3.5 or 4.5 rather than 4. 3rd M1 for a fully correct expression eg $0.008×(0.9817) + 0.096×(0.9084) + 0.384×(0.7619) + 0.512×(0.5665)$ condone 0.9816 and 0.7618 Allow figures to 3sf for method or awrt 0.00785 + awrt 0.0872 + awrt 0.293 + awrt 0.290 (allow 0.29). A1 awrt 0.678

**[10 marks]**

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\begin{enumerate}
  \item The independent random variables $W$ and $X$ have the following distributions.
\end{enumerate}

$$W \sim \operatorname { Po } ( 4 ) \quad X \sim \mathrm {~B} ( 3,0.8 )$$

(a) Write down the value of the variance of $W$\\
(b) Determine the mode of $X$

Show your working clearly.

One observation from each distribution is recorded as $W _ { 1 }$ and $X _ { 1 }$ respectively.\\
(c) Find $\mathrm { P } \left( W _ { 1 } = 2 \right.$ and $\left. X _ { 1 } = 2 \right)$\\
(d) Find $\mathrm { P } \left( X _ { 1 } < W _ { 1 } \right)$

\hfill \mbox{\textit{Edexcel S2 2022 Q1 [10]}}