CAIE FP2 2014 June — Question 9

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeValidity of normal model
DifficultyEasy -3.0 The question text is completely corrupted and unreadable, making it impossible to assess the mathematical content or difficulty. This appears to be a data encoding/OCR error rather than an actual exam question.
Spec5.03a Continuous random variables: pdf and cdf5.03c Calculate mean/variance: by integration5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05a Sample mean distribution: central limit theorem
9 The continuous random variable \(X\) has distribution function F given by $$\mathrm { F } ( x ) = \begin{cases} 0 & x < 2 , \\ \frac { 1 } { 8 } x - \frac { 1 } { 4 } & 2 \leqslant x \leqslant 10 , \\ 1 & x > 10 . \end{cases}$$ Find the value of \(k\) for which \(\mathrm { P } ( X \geqslant k ) = 0.6\). The random variable \(Y\) is defined by \(Y = 2 \ln X\). Find the distribution function of \(Y\). Find the probability density function of \(Y\) and sketch its graph.
Question 9:
Find k for which P(X ≥ k) = 0.6:
AnswerMarks Guidance
\(0.6 = 1 - F(k)\)M1
\(= 1 - (k/8 - \frac{1}{4})\)M1
\(k = 26/5\) or \(5.2\)A1 3 marks total
Find G(y) from Y = 2 ln X for \(2 \leq x \leq 10\):
AnswerMarks Guidance
\(G(y) = P(Y < y) = P(2\ln X < y)\)
\(= P(X < e^{y/2}) = F(e^{y/2})\) allow < or ≤ throughout
\(= e^{y/2}/8 - \frac{1}{4}\) for \(2\ln 2 \leq y \leq 2\ln 10\)M1 A1 result may be stated; or \((\ln 4 \leq y \leq \ln 100)\) or \((1.39 \leq y \leq 4.61)\); 3 marks total
State G(y) for other values of x:
AnswerMarks Guidance
\(0\ (y < 2\ln 2)\) and \(1\ (y > 2\ln 10)\)B1 3 marks total
Find g(y) for \(2\ln 2 \leq y \leq 2\ln 10\):
AnswerMarks
\(g(y) = e^{y/2}/16\)M1 A1
Sketch:
AnswerMarks Guidance
Sketch positive exponential for \(2\ln 2 \leq y \leq 2\ln 10\)B1
Show \(g(y) = 0\) on either side of this intervalB1 4 marks total; 10 marks overall 
## Question 9:

**Find k for which P(X ≥ k) = 0.6:**

| $0.6 = 1 - F(k)$ | M1 | |
|---|---|---|
| $= 1 - (k/8 - \frac{1}{4})$ | M1 | |
| $k = 26/5$ or $5.2$ | A1 | 3 marks total |

**Find G(y) from Y = 2 ln X for $2 \leq x \leq 10$:**

| $G(y) = P(Y < y) = P(2\ln X < y)$ | | |
|---|---|---|
| $= P(X < e^{y/2}) = F(e^{y/2})$ | | allow < or ≤ throughout |
| $= e^{y/2}/8 - \frac{1}{4}$ for $2\ln 2 \leq y \leq 2\ln 10$ | M1 A1 | result may be stated; or $(\ln 4 \leq y \leq \ln 100)$ or $(1.39 \leq y \leq 4.61)$; 3 marks total |

**State G(y) for other values of x:**

| $0\ (y < 2\ln 2)$ and $1\ (y > 2\ln 10)$ | B1 | 3 marks total |
|---|---|---|

**Find g(y) for $2\ln 2 \leq y \leq 2\ln 10$:**

| $g(y) = e^{y/2}/16$ | M1 A1 | |
|---|---|---|

**Sketch:**

| Sketch positive exponential for $2\ln 2 \leq y \leq 2\ln 10$ | B1 | |
|---|---|---|
| Show $g(y) = 0$ on either side of this interval | B1 | 4 marks total; **10 marks overall** |

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9 The continuous random variable $X$ has distribution function F given by

$$\mathrm { F } ( x ) = \begin{cases} 0 & x < 2 , \\ \frac { 1 } { 8 } x - \frac { 1 } { 4 } & 2 \leqslant x \leqslant 10 , \\ 1 & x > 10 . \end{cases}$$

Find the value of $k$ for which $\mathrm { P } ( X \geqslant k ) = 0.6$.

The random variable $Y$ is defined by $Y = 2 \ln X$. Find the distribution function of $Y$.

Find the probability density function of $Y$ and sketch its graph.

\hfill \mbox{\textit{CAIE FP2 2014 Q9}}
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