OCR MEI S1 2010 June — Question 2 7 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Year2010
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeSimple algebraic expression for P(X=x)
DifficultyModerate -0.8 This is a straightforward discrete probability distribution question requiring only routine techniques: summing probabilities to find k, then calculating E(X) and Var(X) using standard formulas. The algebra is simple with only 4 values, and the methods are direct textbook applications with no problem-solving insight needed.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables

2 The probability distribution of the random variable \(X\) is given by the formula $$\mathrm { P } ( X = r ) = k r ( 5 - r ) \text { for } r = 1,2,3,4 .$$
  1. Show that \(k = 0.05\).
  2. Find \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).

(i) \(4k + 6k + 6k + 4k = 1\)
\(20k = 1\)
AnswerMarks Guidance
\(k = 0.05\)M1, A1 NB Answer given 2 marks
(ii) \(E(X) = 1 \times 0.2 + 2 \times 0.3 + 3 \times 0.3 + 4 \times 0.2 = 2.5\) (or by inspection)M1 for \(\sum rp\) (at least 3 terms correct), A1 CAO
\(E(X^2) = 1 \times 0.2 + 4 \times 0.3 + 9 \times 0.3 + 16 \times 0.2 = 7.3\)M1 for \(\sum r^2p\) (at least 3 terms correct), M1 dep for – their \(E(X)^2\)
\(\text{Var}(X) = 7.3 - 2.5^2 = 1.05\)A1 FT their \(E(X)\) provided Var\((X) > 0\) 5 marks
TOTAL: 7 marks
**(i)** $4k + 6k + 6k + 4k = 1$
$20k = 1$
$k = 0.05$ | M1, A1 NB Answer given | 2 marks

**(ii)** $E(X) = 1 \times 0.2 + 2 \times 0.3 + 3 \times 0.3 + 4 \times 0.2 = 2.5$ (or by inspection) | M1 for $\sum rp$ (at least 3 terms correct), A1 CAO |

$E(X^2) = 1 \times 0.2 + 4 \times 0.3 + 9 \times 0.3 + 16 \times 0.2 = 7.3$ | M1 for $\sum r^2p$ (at least 3 terms correct), M1 dep for – their $E(X)^2$ |

$\text{Var}(X) = 7.3 - 2.5^2 = 1.05$ | A1 FT their $E(X)$ provided Var$(X) > 0$ | 5 marks

**TOTAL: 7 marks**

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2 The probability distribution of the random variable $X$ is given by the formula

$$\mathrm { P } ( X = r ) = k r ( 5 - r ) \text { for } r = 1,2,3,4 .$$

(i) Show that $k = 0.05$.\\
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.

\hfill \mbox{\textit{OCR MEI S1 2010 Q2 [7]}}