OCR MEI Further Statistics Major 2023 June — Question 10 15 marks

Exam BoardOCR MEI
ModuleFurther Statistics Major (Further Statistics Major)
Year2023
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Probability Distributions and Random Variables
TypeFind parameter from median
DifficultyChallenging +1.2 This is a structured multi-part question on continuous probability distributions requiring integration to find the CDF, using the normalization condition to find parameter a, setting up and verifying a median equation, and finding expectation and mode. While it involves several standard techniques (integration, solving equations, differentiation), each part follows directly from the previous with clear signposting. The algebra is moderately involved but routine for Further Maths students. This is slightly above average difficulty due to the length and algebraic manipulation required, but doesn't require novel insights.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration5.03e Find cdf: by integration
10 The continuous random variable \(X\) has probability density function given by \(f ( x ) = \begin{cases} \frac { 4 } { 15 } \left( \frac { a } { x ^ { 2 } } + 3 x ^ { 2 } - \frac { 7 } { 2 } \right) & 1 \leqslant x \leqslant 2 , \\ 0 & \text { otherwise, } \end{cases}\) where \(a\) is a positive constant.
  1. Find the cumulative distribution function of \(X\) in terms of \(a\).
  2. Hence or otherwise determine the value of \(a\).
  3. Show that the median value \(m\) of \(X\) satisfies the equation $$8 m ^ { 4 } - 28 m ^ { 2 } + 9 m - 4 = 0 .$$
  4. Verify that the median value of \(X\) is 1.74, correct to \(\mathbf { 2 }\) decimal places.
  5. Find \(\mathrm { E } ( X )\).
  6. Determine the mode of \(X\).
Question 10:
AnswerMarks Guidance
10(a) 𝑥 4 𝑎 7
F(𝑥) = ∫ ( +3𝑢2− )d𝑢
15 𝑢2 2
1
4 𝑎 7 𝑥
= [− +𝑢3− 𝑢]
15 𝑢 2
1
= 4 (− 𝑎 +𝑥3− 7 𝑥)− 4 (−𝑎− 5 )
15 𝑥 2 15 2
F(𝑥)= 0 for 𝑥 < 1
F(𝑥) = 4 (− 𝑎 +𝑥3− 7 𝑥+𝑎+ 5 ) for 1 ≤ 𝑥 ≤ 2
15 𝑥 2 2
AnswerMarks
F(𝑥) =1 for 𝑥 > 2M1
M1
A1
A1
AnswerMarks
[4]2.1
1.1a
1.1
AnswerMarks
1.1For attempt to integrate (Condone ito x). No need for limits.
For correct integral (Condone ito x) No need for limits.
A1 for correct answer AEF
Fully correct AEF
4 1 7 5
[= (𝑎(1− )+𝑥3− 𝑥+ )]
15 𝑥 2 2
Or using F(2) = 1 leads to 4 (− 𝑎 +𝑥3− 7 𝑥+ 11 + 1 𝑎)
15 𝑥 2 4 2
AnswerMarks Guidance
10(b) 4 𝑎 4 5
F(2) = 1 ⇒ (− +8−7)− (−𝑎− ) = 1
15 2 15 2
2 14
⇒ 𝑎+ = 1
15 15
AnswerMarks
𝑎 = 0.5M1
A1
AnswerMarks
[2]3.1a
1.1For setting their F(2) = 1.. Their F(x) must be i.t.o. a
[⇒ F(𝑥) = 4 (− 0.5 +𝑥3− 7 𝑥+3) = 1]
15 𝑥 2
NB Alternative method F(2)−F(1) = 1 leads to same
equation if correct but FT their F(𝑥) for M1. Condone a
wrong constant term in F(𝑥) which leads to a correct value
of a
AnswerMarks Guidance
10(c) 4 (− 𝑎 +𝑚3− 7 𝑚)+0.8 = 0.5
15 𝑚 2
Or
4 𝑎 7
(− +𝑚3− 𝑚+3) = 0.5
15 𝑚 2
0.5 7
8(− +𝑚3− 𝑚) = −9
𝑚 2
AnswerMarks
⇒ 8𝑚4−28𝑚2+9𝑚−4 = 0M1
A1
AnswerMarks
[2]2.1
1.1For setting their F(m) = 0.5 Condone F(x) = 0.5
Or ∫ 𝑚 4 ( 𝑎 +3𝑢2− 7 ) = 0.5 followed by attempt at
1 15 𝑢2 2
integration
AG Must show at least one line of correct working
AnswerMarks Guidance
10(d) F(1.735) = 0.4965, F(1.745) = 0.5119
So median is 1.74 to 2 dpB1
E1
AnswerMarks
[2]3.4
1.1For either
OR evaluation of 8𝑚4−28𝑚2+9𝑚−4 for 1.735
(= −0.1797) and 1.745 (= 0.6217), so change of sign
No marks for calculator answer
AnswerMarks Guidance
10(e) 2 4 0.5 7
E(𝑋) = ∫ 𝑥( +3𝑥2− )d𝑥
15 𝑥2 2
1
AnswerMarks
= 1.69M1
A1
AnswerMarks
[2]1.1
1.1Allow with wrong value of a or ito a. Correct limits required.
2 8
BC (1.692419…) or ln2+
15 5
AnswerMarks Guidance
10(f) 4 1
f′(𝑥) = (− +6𝑥)
15 𝑥3
4 3
f′′(𝑥) = ( +6)
15 𝑥4
AnswerMarks
hence f is increasing so mode = 2M1
M1
A1
AnswerMarks
[3]3.1a
1.1
AnswerMarks
1.1Allow with wrong value of a
Allow with wrong value of a
Alternatives for second mark:
4 1
When 𝑥 = √ the curve has a minimum;
6
For 1  x  2, 6x is clearly greater than 1 so f′ is positive
𝑥3
Allow third mark if full answer based on the above.
Allow SCB1 for mode = 2 without proper justification
Question 10:
10 | (a) | 𝑥 4 𝑎 7
F(𝑥) = ∫ ( +3𝑢2− )d𝑢
15 𝑢2 2
1
4 𝑎 7 𝑥
= [− +𝑢3− 𝑢]
15 𝑢 2
1
= 4 (− 𝑎 +𝑥3− 7 𝑥)− 4 (−𝑎− 5 )
15 𝑥 2 15 2
F(𝑥)= 0 for 𝑥 < 1
F(𝑥) = 4 (− 𝑎 +𝑥3− 7 𝑥+𝑎+ 5 ) for 1 ≤ 𝑥 ≤ 2
15 𝑥 2 2
F(𝑥) =1 for 𝑥 > 2 | M1
M1
A1
A1
[4] | 2.1
1.1a
1.1
1.1 | For attempt to integrate (Condone ito x). No need for limits.
For correct integral (Condone ito x) No need for limits.
A1 for correct answer AEF
Fully correct AEF
4 1 7 5
[= (𝑎(1− )+𝑥3− 𝑥+ )]
15 𝑥 2 2
Or using F(2) = 1 leads to 4 (− 𝑎 +𝑥3− 7 𝑥+ 11 + 1 𝑎)
15 𝑥 2 4 2
10 | (b) | 4 𝑎 4 5
F(2) = 1 ⇒ (− +8−7)− (−𝑎− ) = 1
15 2 15 2
2 14
⇒ 𝑎+ = 1
15 15
𝑎 = 0.5 | M1
A1
[2] | 3.1a
1.1 | For setting their F(2) = 1.. Their F(x) must be i.t.o. a
[⇒ F(𝑥) = 4 (− 0.5 +𝑥3− 7 𝑥+3) = 1]
15 𝑥 2
NB Alternative method F(2)−F(1) = 1 leads to same
equation if correct but FT their F(𝑥) for M1. Condone a
wrong constant term in F(𝑥) which leads to a correct value
of a
10 | (c) | 4 (− 𝑎 +𝑚3− 7 𝑚)+0.8 = 0.5
15 𝑚 2
Or
4 𝑎 7
(− +𝑚3− 𝑚+3) = 0.5
15 𝑚 2
0.5 7
8(− +𝑚3− 𝑚) = −9
𝑚 2
⇒ 8𝑚4−28𝑚2+9𝑚−4 = 0 | M1
A1
[2] | 2.1
1.1 | For setting their F(m) = 0.5 Condone F(x) = 0.5
Or ∫ 𝑚 4 ( 𝑎 +3𝑢2− 7 ) = 0.5 followed by attempt at
1 15 𝑢2 2
integration
AG Must show at least one line of correct working
10 | (d) | F(1.735) = 0.4965, F(1.745) = 0.5119
So median is 1.74 to 2 dp | B1
E1
[2] | 3.4
1.1 | For either
OR evaluation of 8𝑚4−28𝑚2+9𝑚−4 for 1.735
(= −0.1797) and 1.745 (= 0.6217), so change of sign
No marks for calculator answer
10 | (e) | 2 4 0.5 7
E(𝑋) = ∫ 𝑥( +3𝑥2− )d𝑥
15 𝑥2 2
1
= 1.69 | M1
A1
[2] | 1.1
1.1 | Allow with wrong value of a or ito a. Correct limits required.
2 8
BC (1.692419…) or ln2+
15 5
10 | (f) | 4 1
f′(𝑥) = (− +6𝑥)
15 𝑥3
4 3
f′′(𝑥) = ( +6)
15 𝑥4
hence f is increasing so mode = 2 | M1
M1
A1
[3] | 3.1a
1.1
1.1 | Allow with wrong value of a
Allow with wrong value of a
Alternatives for second mark:
4 1
When 𝑥 = √ the curve has a minimum;
6
For 1  x  2, 6x is clearly greater than 1 so f′ is positive
𝑥3
Allow third mark if full answer based on the above.
Allow SCB1 for mode = 2 without proper justification
10 The continuous random variable $X$ has probability density function given by\\
$f ( x ) = \begin{cases} \frac { 4 } { 15 } \left( \frac { a } { x ^ { 2 } } + 3 x ^ { 2 } - \frac { 7 } { 2 } \right) & 1 \leqslant x \leqslant 2 , \\ 0 & \text { otherwise, } \end{cases}$\\
where $a$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Find the cumulative distribution function of $X$ in terms of $a$.
\item Hence or otherwise determine the value of $a$.
\item Show that the median value $m$ of $X$ satisfies the equation

$$8 m ^ { 4 } - 28 m ^ { 2 } + 9 m - 4 = 0 .$$
\item Verify that the median value of $X$ is 1.74, correct to $\mathbf { 2 }$ decimal places.
\item Find $\mathrm { E } ( X )$.
\item Determine the mode of $X$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Major 2023 Q10 [15]}}
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