OCR S2 — Question 8 15 marks

Exam BoardOCR
ModuleS2 (Statistics 2)
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeDirect variance calculation from pdf
DifficultyModerate -0.3 This is a standard S2 question testing routine application of pdf properties and variance formulas. Parts (i)-(iii) involve straightforward integration and formula application (finding k, calculating E(X), then Var(X)). Part (iv) tests CLT knowledge (direct recall), and part (v) requires recognizing a linear transformation. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec5.01b Selection/arrangement: probability problems5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration
8 A continuous random variable \(X\) has probability density function given by $$f ( x ) = \begin{cases} k x ^ { n } & 0 \leqslant x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$ where \(n\) and \(k\) are positive constants.
  1. Find \(k\) in terms of \(n\).
  2. Show that \(\mathrm { E } ( X ) = \frac { n + 1 } { n + 2 }\). It is given that \(n = 3\).
  3. Find the variance of \(X\).
  4. One hundred observations of \(X\) are taken, and the mean of the observations is denoted by \(\bar { X }\). Write down the approximate distribution of \(\bar { X }\), giving the values of any parameters.
  5. Write down the mean and the variance of the random variable \(Y\) with probability density function given by $$g ( y ) = \begin{cases} 4 \left( y + \frac { 4 } { 5 } \right) ^ { 3 } & - \frac { 4 } { 5 } \leqslant y \leqslant \frac { 1 } { 5 } \\ 0 & \text { otherwise } \end{cases}$$
(i) Answer: \(\int_0^1 x^n \, dx = \left[\frac{x^{n+1}}{n+1}\right]_0^1 = \frac{1}{n+1}\)
AnswerMarks Guidance
\(k/(n+1) = 1\) so \(k = n+1\)Marks: M1 M1 A1 Guidance: Integrate \(x^n\), limits 0 and 1. Equate to 1 and solve for \(k\). Answer \(n + 1\), not \(1^{n+1}\), e.w.o.
(ii) Answer: \(\int_0^1 x^{n+1} \, dx = \left[\frac{x^{n+2}}{n+2}\right]_0^1 = \frac{1}{n+2}\)
AnswerMarks Guidance
\(\mu = \frac{k}{n+2} = \frac{n+1}{n+2}\) AGMarks: M1 M1 A1 A1 Guidance: Integrate \(x^{n+1}\), limits 0 and 1, not just \(x^n\). Answer \(\frac{1}{n+2}\). Correctly obtain given answer.
(iii) Answer: \(\int_0^1 x^3 \, dx = \left[\frac{x^6}{6}\right]_0^1 = \frac{1}{6}\)
AnswerMarks Guidance
\(\sigma^2 = \frac{4}{3} - \left(\frac{3}{4}\right)^2 = \frac{7}{75}\)Marks: A1 A1 Guidance: Integrate \(x^3\), limits 0 and 1, allow with \(n\). Subtract \((\frac{3}{4})^2\). Answer \(\frac{7}{75}\) or a.r.t. 0.027
(iv) Answer: \(\text{N}\left(3, \frac{7}{500}\right)\)Marks: B1 B1 B1 Guidance: Normal stated. Mean \(\frac{3}{4}\) or \(\frac{n+1}{n+2}\). Variance their (iii)/100, a.e.f., allow √
(v) Answer: Same distribution, translated
Mean 0
AnswerMarks Guidance
Variance \(\frac{7}{75}\)Marks: M1 A1 B1 B1 Guidance: Can be negative translation; or integration, must include correct method for integral. (Their mean) – \(\frac{3}{4}\), c.w.d. Variance same as their (iii), or \(\frac{7}{75}\) by integration
*Note: This extraction covers the first complete mark scheme provided. Additional questions and variations from different exam sessions would follow the same format.*
**(i)** **Answer:** $\int_0^1 x^n \, dx = \left[\frac{x^{n+1}}{n+1}\right]_0^1 = \frac{1}{n+1}$
$k/(n+1) = 1$ so $k = n+1$ | **Marks:** M1 M1 A1 | **Guidance:** Integrate $x^n$, limits 0 and 1. Equate to 1 and solve for $k$. Answer $n + 1$, not $1^{n+1}$, e.w.o.

**(ii)** **Answer:** $\int_0^1 x^{n+1} \, dx = \left[\frac{x^{n+2}}{n+2}\right]_0^1 = \frac{1}{n+2}$
$\mu = \frac{k}{n+2} = \frac{n+1}{n+2}$ AG | **Marks:** M1 M1 A1 A1 | **Guidance:** Integrate $x^{n+1}$, limits 0 and 1, not just $x^n$. Answer $\frac{1}{n+2}$. Correctly obtain given answer.

**(iii)** **Answer:** $\int_0^1 x^3 \, dx = \left[\frac{x^6}{6}\right]_0^1 = \frac{1}{6}$
$\sigma^2 = \frac{4}{3} - \left(\frac{3}{4}\right)^2 = \frac{7}{75}$ | **Marks:** A1 A1 | **Guidance:** Integrate $x^3$, limits 0 and 1, allow with $n$. Subtract $(\frac{3}{4})^2$. Answer $\frac{7}{75}$ or a.r.t. 0.027

**(iv)** **Answer:** $\text{N}\left(3, \frac{7}{500}\right)$ | **Marks:** B1 B1 B1 | **Guidance:** Normal stated. Mean $\frac{3}{4}$ or $\frac{n+1}{n+2}$. Variance their (iii)/100, a.e.f., allow √

**(v)** **Answer:** Same distribution, translated
Mean 0
Variance $\frac{7}{75}$ | **Marks:** M1 A1 B1 B1 | **Guidance:** Can be negative translation; or integration, must include correct method for integral. (Their mean) – $\frac{3}{4}$, c.w.d. Variance same as their (iii), or $\frac{7}{75}$ by integration

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*Note: This extraction covers the first complete mark scheme provided. Additional questions and variations from different exam sessions would follow the same format.*
8 A continuous random variable $X$ has probability density function given by

$$f ( x ) = \begin{cases} k x ^ { n } & 0 \leqslant x \leqslant 1 \\ 0 & \text { otherwise } \end{cases}$$

where $n$ and $k$ are positive constants.\\
(i) Find $k$ in terms of $n$.\\
(ii) Show that $\mathrm { E } ( X ) = \frac { n + 1 } { n + 2 }$.

It is given that $n = 3$.\\
(iii) Find the variance of $X$.\\
(iv) One hundred observations of $X$ are taken, and the mean of the observations is denoted by $\bar { X }$. Write down the approximate distribution of $\bar { X }$, giving the values of any parameters.\\
(v) Write down the mean and the variance of the random variable $Y$ with probability density function given by

$$g ( y ) = \begin{cases} 4 \left( y + \frac { 4 } { 5 } \right) ^ { 3 } & - \frac { 4 } { 5 } \leqslant y \leqslant \frac { 1 } { 5 } \\ 0 & \text { otherwise } \end{cases}$$

\hfill \mbox{\textit{OCR S2  Q8 [15]}}
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