OCR S2 2005 June — Question 7 13 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Year2005
SessionJune
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeGeometric/graphical PDF with k
DifficultyStandard +0.3 This is a straightforward S2 question covering standard PDF concepts: finding k using area=1, calculating mean by symmetry and variance by integration, applying independence for repeated observations, and stating the Central Limit Theorem result. All techniques are routine textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.05a Sample mean distribution: central limit theorem

7 The continuous random variable \(X\) has the probability density function shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{b69b1fe8-790d-4727-a892-8ab2ade08962-3_364_766_1229_699}
  1. Find the value of the constant \(k\).
  2. Write down the mean of \(X\), and use integration to find the variance of \(X\).
  3. Three observations of \(X\) are made. Find the probability that \(X < 9\) for all three observations.
  4. The mean of 32 observations of \(X\) is denoted by \(\bar { X }\). State the approximate distribution of \(\bar { X }\), giving its mean and variance. \section*{[Question 8 is printed overleaf.]}

AnswerMarks Guidance
(i) \((11 - 3)k = 1\)M1 Use area = 1 [e.g. \(\int kdx = 1\) with limits 3, 11]
\(k = 1/8\)A1 2 Answer 1/8 or 0.125 only
(ii) \(\mu = \frac{1}{2}(3 + 11) = 7\)B1 Mean 7, cwd
\(\int_3^{11} x \cdot f(x)dx = [\frac{x^4}{24}]_{= 54 \frac{1}{4}}\)M1 Attempt \(\int x^2f(x)dx\), correct limits
A1Indefinite integral \(\frac{x^3}{3k}\), their k
\(\sigma^2 = 54\frac{1}{4} - 7^2 = 5\frac{1}{4}\)M1 Subtract their μ²
A1 5Correct answer, \(5\frac{1}{4}\) or a.r.t. 5.33
(iii) \(P(X < 9) = 6k \quad [= \frac{3}{4}]\)B1V Correct p for their k
\((\frac{3}{4})^3 = \frac{27}{64}\) or 0.421875M1 Work out their \(p^3\), \(0 < p < 1\)
A1 3Answer \(\frac{27}{64}\) or a.r.t. 0.422
(iv) NormalB1 "Normal" distribution stated
Mean is 7B1V Mean same as in (ii) √
Variance is \(5\frac{1}{4} + 32 (= \frac{k}{3})\)B1V 3 Variance is (iii) + 32 √ [not V errors]
**(i)** $(11 - 3)k = 1$ | M1 | Use area = 1 [e.g. $\int kdx = 1$ with limits 3, 11]
$k = 1/8$ | A1 2 | Answer 1/8 or 0.125 only

**(ii)** $\mu = \frac{1}{2}(3 + 11) = 7$ | B1 | Mean 7, cwd
$\int_3^{11} x \cdot f(x)dx = [\frac{x^4}{24}]_{= 54 \frac{1}{4}}$ | M1 | Attempt $\int x^2f(x)dx$, correct limits
| A1 | Indefinite integral $\frac{x^3}{3k}$, their k
$\sigma^2 = 54\frac{1}{4} - 7^2 = 5\frac{1}{4}$ | M1 | Subtract their μ²
| A1 5 | Correct answer, $5\frac{1}{4}$ or a.r.t. 5.33

**(iii)** $P(X < 9) = 6k \quad [= \frac{3}{4}]$ | B1V | Correct p for their k
$(\frac{3}{4})^3 = \frac{27}{64}$ or 0.421875 | M1 | Work out their $p^3$, $0 < p < 1$
| A1 3 | Answer $\frac{27}{64}$ or a.r.t. 0.422

**(iv)** Normal | B1 | "Normal" distribution stated
Mean is 7 | B1V | Mean same as in (ii) √
Variance is $5\frac{1}{4} + 32 (= \frac{k}{3})$ | B1V 3 | Variance is (iii) + 32 √ [not V errors]
7 The continuous random variable $X$ has the probability density function shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{b69b1fe8-790d-4727-a892-8ab2ade08962-3_364_766_1229_699}\\
(i) Find the value of the constant $k$.\\
(ii) Write down the mean of $X$, and use integration to find the variance of $X$.\\
(iii) Three observations of $X$ are made. Find the probability that $X < 9$ for all three observations.\\
(iv) The mean of 32 observations of $X$ is denoted by $\bar { X }$. State the approximate distribution of $\bar { X }$, giving its mean and variance.

\section*{[Question 8 is printed overleaf.]}

\hfill \mbox{\textit{OCR S2 2005 Q7 [13]}}