Edexcel S1 2007 January — Question 3 13 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2007
SessionJanuary
Marks13
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Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeConstruct probability distribution from scenario
DifficultyModerate -0.8 This is a straightforward S1 question requiring routine application of standard formulas for discrete probability distributions. Students substitute values into the given probability function, sum probabilities over an interval, calculate E(X) and Var(X) using standard formulas, then apply the variance transformation rule. All steps are mechanical with no problem-solving or insight required—easier than average A-level.
Spec2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance
  1. The random variable \(X\) has probability function
$$\mathrm { P } ( X = x ) = \frac { ( 2 x - 1 ) } { 36 } \quad x = 1,2,3,4,5,6$$
  1. Construct a table giving the probability distribution of \(X\). Find
  2. \(\mathrm { P } ( 2 < X \leqslant 5 )\),
  3. the exact value of \(\mathrm { E } ( X )\).
  4. Show that \(\operatorname { Var } ( X ) = 1.97\) to 3 significant figures.
  5. Find \(\operatorname { Var } ( 2 - 3 X )\).
AnswerMarks Guidance
(a) N.B. Part (a) doesn't have to be in a table, could be a list \(P(X = 1) = \ldots\) etcB1, B1, B1
\(x\)1 2
\(P(X = x)\)\(\frac{1}{36}\) \(\frac{3}{36}\)
0.0278, 0.0833, 0.139, 0.194, 0.25, 0.306(Accept awrt 3 s.f) (3 marks)
(b) \(P(3) + P(4) + P(5) = \frac{21}{36}\) or \(\frac{7}{12}\) or awrt 0.583M1, A1 (2 marks)
(c) \(E(X) = 1 \times \frac{1}{36} + 2 \times \frac{3}{36} + ... = \frac{161}{36}\) or 4.472 or \(4\frac{17}{36}\)M1, A1 (2 marks)
(d) \(E(X^2) = \frac{1}{36} + 2^2 \times \frac{3}{36} + ... = \frac{791}{36}\) or full expression or \(21\frac{35}{36}\) or awrt 21.97M1, A1
\(\text{Var}(X) = \frac{791}{36} - \left(\frac{161}{36}\right)^2 = \mathbf{1.9714...} *\)M1, A1 c.s.o. (4 marks)
(e) \(\text{Var}(2 - 3X) = 9 \times 1.97\) or \((-3)^2 \times 1.97 = 17.73\)M1, A1 awrt 17.7 or \(\frac{2555}{144}\)
Total: 13 marks
**(a)** N.B. Part (a) doesn't have to be in a table, could be a list $P(X = 1) = \ldots$ etc | B1, B1, B1 |

| $x$ | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| $P(X = x)$ | $\frac{1}{36}$ | $\frac{3}{36}$ | $\frac{5}{36}$ | $\frac{7}{36}$ | $\frac{9}{36}$ | $\frac{11}{36}$ |

0.0278, 0.0833, 0.139, 0.194, 0.25, 0.306 | (Accept awrt 3 s.f) | (3 marks)

**(b)** $P(3) + P(4) + P(5) = \frac{21}{36}$ or $\frac{7}{12}$ or awrt 0.583 | M1, A1 | (2 marks)

**(c)** $E(X) = 1 \times \frac{1}{36} + 2 \times \frac{3}{36} + ... = \frac{161}{36}$ or 4.472 or $4\frac{17}{36}$ | M1, A1 | (2 marks)

**(d)** $E(X^2) = \frac{1}{36} + 2^2 \times \frac{3}{36} + ... = \frac{791}{36}$ or full expression or $21\frac{35}{36}$ or awrt 21.97 | M1, A1 |
$\text{Var}(X) = \frac{791}{36} - \left(\frac{161}{36}\right)^2 = \mathbf{1.9714...} *$ | M1, A1 c.s.o. | (4 marks)

**(e)** $\text{Var}(2 - 3X) = 9 \times 1.97$ or $(-3)^2 \times 1.97 = 17.73$ | M1, A1 | awrt **17.7** or $\frac{2555}{144}$ | (2 marks)

**Total: 13 marks**

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\begin{enumerate}
  \item The random variable $X$ has probability function
\end{enumerate}

$$\mathrm { P } ( X = x ) = \frac { ( 2 x - 1 ) } { 36 } \quad x = 1,2,3,4,5,6$$

(a) Construct a table giving the probability distribution of $X$.

Find\\
(b) $\mathrm { P } ( 2 < X \leqslant 5 )$,\\
(c) the exact value of $\mathrm { E } ( X )$.\\
(d) Show that $\operatorname { Var } ( X ) = 1.97$ to 3 significant figures.\\
(e) Find $\operatorname { Var } ( 2 - 3 X )$.

\hfill \mbox{\textit{Edexcel S1 2007 Q3 [13]}}
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