S2 June 2011 Q7

EdexcelOld spec17 marksContinuous Random Variables

7. The continuous random variable \(X\) has probability density function given by

\[\mathrm{f}(x) = \begin{cases} \dfrac{3}{32}\left(x - 1\right)\left(5 - x\right) & 1 \leqslant x \leqslant 5 \\ 0 & \text{otherwise} \end{cases}\]

(a) Sketch \(\mathrm{f}(x)\) showing clearly the points where it meets the \(x\)-axis. (2)

(b) Write down the value of the mean, \(\mu\), of \(X\). (1)

(d) Find the standard deviation, \(\sigma\), of \(X\). (2)

The cumulative distribution function of \(X\) is given by

\[\mathrm{F}(x) = \begin{cases} 0 & x \lt 1 \\ \dfrac{1}{32}\left(a - 15x + 9x^2 - x^3\right) & 1 \leqslant x \leqslant 5 \\ 1 & x \gt 5 \end{cases}\]

where \(a\) is a constant.

(e) Find the value of \(a\). (2)

(f) Show that the lower quartile of \(X\), \(q_1\), lies between 2.29 and 2.31 (3)

(g) Hence find the upper quartile of \(X\), giving your answer to 1 decimal place. (1)

(h) Find, to 2 decimal places, the value of \(k\) so that\[\mathrm{P}(\mu - k\sigma \lt X \lt \mu + k\sigma) = 0.5\] (2)

Scheme	Marks
\(\cap\) shape which does not go below the \(x\)-axis [condone missing patios] Graph must end at the points (1,0) and (5,0) and the points labelled at 1 and 5	B1 B1
	(2)

Scheme	Marks
\(\mathrm{E}(X) = 3\) (by symmetry)	B1
	(1)

Mark scheme (c)

Scheme	Marks
\(\left[E(X^2)\right] = \displaystyle\int x^2\mathrm{f}(x)\,\mathrm{d}x = \frac{3}{32}\int\left(6x^3 - x^4 - 5x^2\right)\mathrm{d}x\)	M1
\(= \tfrac{3}{32}\left[\dfrac{6x^4}{4} - \dfrac{x^5}{5} - \dfrac{5x^3}{3}\right]_1^5\)	A1
\(= \tfrac{3}{32}\left(\left[\dfrac{6 \times 625}{4} - 625 - \dfrac{625}{3}\right] - \left[\dfrac{6}{4} - \dfrac{1}{5} - \dfrac{5}{3}\right]\right) = 9.8\) (*)	M1 A1 cso
	(4)

Notes

This part is a “show that” therefore we need to see all the steps in the working
1st M1 for showing intention of doing \(\displaystyle\int x^2\mathrm{f}(x)\) and attempt to multiply out bracket
1st A1 for correct integration, cao, ignore limits for this mark.
2nd M1 for use of correct limits. Need to see evidence of subst both 5 and 1.
2nd A1 for cso leading to 9.8. Do not ignore subsequent working for this final A mark.

Mark scheme (d)

Scheme	Marks
\(\text{s.d.} = \sqrt{9.8 - \mathrm{E}(X)^2}\),	M1
\(= 0.8944\ldots\) awrt 0.894	A1
	(2)

Notes

M1 for a correct expression for standard deviation, must include \(\sqrt{\ldots}\)
A1 allow awrt 0.894, \(\sqrt{0.8}\), \(\dfrac{2\sqrt{5}}{5}\) oe

Mark scheme (e)

Scheme	Marks
\(\mathrm{F}(1) = 0 \Rightarrow \tfrac{1}{32}\left(a - 15 + 9 - 1\right) = 0\), leading to \(\underline{a = 7}\)	M1 A1
	(2)

Notes

M1 for a correct method to find \(a\). e.g \(\mathrm{F}(5) = 1\) or \(\displaystyle\int_1^5 f(x) = 1\)

Mark scheme (f)

Scheme	Marks
\(\mathrm{F}(2.29) = 0.2449\ldots,\ \mathrm{F}(2.31) = 0.2515\ldots\)	M1 A1
Since \(\mathrm{F}(q_1) = 0.25\) and these values are either side of 0.25 then \(2.29 \lt q_1 \lt 2.31\)	A1
	(3)

Notes

M1 for an attempt at F(2.29) or F(2.31)    or    put \(\mathrm{F}(x) = 0.25\) (ft their value of \(a\))
1st A1 for both values seen. awrt 0.245 and 0.252    or    find 3 solutions awrt 6.76/6.75, 2.305, -0.064
2nd A1 for comparison with 0.25 and stating Q₁ lies between 2.29 and 2.31    or    state only 2.30 in range and stating Q₁ lies between 2.29 and 2.31

Scheme	Marks
Since the distribution is symmetric \(q_3 = 5 - 1.3 = \underline{3.7}\) cao	B1
	(1)

Mark scheme (h)

Scheme	Marks
We know \(\mathrm{P}(q_1 = 2.3 \lt X \lt 3.7 = q_3) = 0.5\) so \(k\sigma = 0.7\)	M1
so \(k = \dfrac{0.7}{0.894\ldots} = 0.7826.. = \textbf{awrt 0.78}\)	A1
	(2)
	(17 marks)

Notes

M1 For \(k\sigma\) = awrt 0.7
A1 Allow awrt 0.78
NB a correct awrt 0.78 gains M1 A1