Question 1 - A-Level Maths

OCR S2 2006 June — Question 1 6 marks

Exam Board	OCR
Module	S2 (Statistics 2)
Year	2006
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Continuous Probability Distributions and Random Variables
Type	Direct variance calculation from pdf
Difficulty	Moderate -0.5 This is a straightforward application of the variance formula for continuous distributions requiring calculation of E(X²) and E(X)² using standard integration of polynomial functions. While it involves multiple steps and careful arithmetic, it's a routine textbook exercise with no conceptual challenges beyond knowing the formula Var(X) = E(X²) - [E(X)]².
Spec	5.03c Calculate mean/variance: by integration

1 Calculate the variance of the continuous random variable with probability density function given by $$f ( x ) = \begin{cases} \frac { 3 } { 37 } x ^ { 2 } & 3 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$

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Answer	Marks	Guidance
$\mu = \frac{1}{3}\int_3^4 x \cdot x \, dx = \frac{1}{3}\left[\frac{x^4}{4}\right]_3^4 = [3, 3\frac{1}{4}]$	M1	Integrate $xf(x)$, limits 3 & 4 [can be implied]
$\frac{1}{3}\int_3^4 x^3 \, dx = \left[\frac{x^4}{3}\right]_3^4 = 12\frac{163}{3}$ or $12.665$	M1, A1	[range 12.6, 12.7] [can be implied]
$\sigma^2 = 12\frac{163}{3} - 3\frac{81}{148}^2 = 0.0815$	M1, A1	Attempt to integrate $x^2f(x)$, limits 3 & 4. Correct indefinite integral, any form. Subtract their $\mu^2$. Answer, in range [0.0575, 0.084]

$\mu = \frac{1}{3}\int_3^4 x \cdot x \, dx = \frac{1}{3}\left[\frac{x^4}{4}\right]_3^4 = [3, 3\frac{1}{4}]$ | M1 | Integrate $xf(x)$, limits 3 & 4 [can be implied]

$\frac{1}{3}\int_3^4 x^3 \, dx = \left[\frac{x^4}{3}\right]_3^4 = 12\frac{163}{3}$ or $12.665$ | M1, A1 | [range 12.6, 12.7] [can be implied]

$\sigma^2 = 12\frac{163}{3} - 3\frac{81}{148}^2 = 0.0815$ | M1, A1 | Attempt to integrate $x^2f(x)$, limits 3 & 4. Correct indefinite integral, any form. Subtract their $\mu^2$. Answer, in range [0.0575, 0.084]

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1 Calculate the variance of the continuous random variable with probability density function given by

$$f ( x ) = \begin{cases} \frac { 3 } { 37 } x ^ { 2 } & 3 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$

\hfill \mbox{\textit{OCR S2 2006 Q1 [6]}}

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Q1 6 Q2 7 Q3 8 Q4 10 Q5 9 Q6 14 Q7 18

Answer	Marks	Guidance
\(\mu = \frac{1}{3}\int_3^4 x \cdot x \, dx = \frac{1}{3}\left[\frac{x^4}{4}\right]_3^4 = [3, 3\frac{1}{4}]\)	M1	Integrate \(xf(x)\), limits 3 & 4 [can be implied]
\(\frac{1}{3}\int_3^4 x^3 \, dx = \left[\frac{x^4}{3}\right]_3^4 = 12\frac{163}{3}\) or \(12.665\)	M1, A1	[range 12.6, 12.7] [can be implied]
\(\sigma^2 = 12\frac{163}{3} - 3\frac{81}{148}^2 = 0.0815\)	M1, A1	Attempt to integrate \(x^2f(x)\), limits 3 & 4. Correct indefinite integral, any form. Subtract their \(\mu^2\). Answer, in range [0.0575, 0.084]