OCR MEI Further Statistics B AS Specimen — Question 2 7 marks

Exam BoardOCR MEI
ModuleFurther Statistics B AS (Further Statistics B AS)
SessionSpecimen
Marks7
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Mark schemeDownload PDF ↗
TopicCumulative distribution functions
TypeCalculate probabilities from CDF
DifficultyStandard +0.3 This is a straightforward CDF question requiring direct substitution for part (i), simple evaluation at two points for part (ii), and differentiation for part (iii). All techniques are routine for Further Statistics students with no problem-solving insight needed, making it slightly easier than average.
Spec5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03e Find cdf: by integration5.03f Relate pdf-cdf: medians and percentiles
2 The cumulative distribution function of the continuous random variable, \(Y\), is given below. $$\mathrm { F } ( y ) = \left\{ \begin{array} { c c } 0 & y < 0 \\ \frac { y ^ { 3 } - y ^ { 2 } } { 4 } & 1 \leq y \leq 2 \\ 1 & y > 2 \end{array} \right.$$
  1. Find \(\mathrm { P } ( Y \leq 1.5 )\)
  2. Verify that the median of \(Y\) lies between 1.6 and 1.7.
  3. Find the probability density function of \(Y\).
Question 2:
AnswerMarks Guidance
2(i) 1.53 (cid:16)1.52
4
9
(cid:32)0.28125
AnswerMarks
32M1
A1
AnswerMarks
[2]1.1a
1.1
AnswerMarks Guidance
2(ii) F(1.6) = 0.384
F(1.7) = 0.50575
At the median, m, F(m) = 0.5 so the median must be
AnswerMarks
between 1.6 and 1.7M1
M1
A1
AnswerMarks
[3]1.1a
1.1
AnswerMarks
2.2bN
E
M
AnswerMarks Guidance
2(iii) 3y2 (cid:16)2y
f(y)(cid:32) , 1(cid:100) y(cid:100)2
AnswerMarks
4M1
A1
AnswerMarks
[2]I
1.1a
1.1
Question 2:
2 | (i) | 1.53 (cid:16)1.52
4
9
(cid:32)0.28125
32 | M1
A1
[2] | 1.1a
1.1
2 | (ii) | F(1.6) = 0.384
F(1.7) = 0.50575
At the median, m, F(m) = 0.5 so the median must be
between 1.6 and 1.7 | M1
M1
A1
[3] | 1.1a
1.1
2.2b | N
E
M
2 | (iii) | 3y2 (cid:16)2y
f(y)(cid:32) , 1(cid:100) y(cid:100)2
4 | M1
A1
[2] | I
1.1a
1.1
2 The cumulative distribution function of the continuous random variable, $Y$, is given below.

$$\mathrm { F } ( y ) = \left\{ \begin{array} { c c } 
0 & y < 0 \\
\frac { y ^ { 3 } - y ^ { 2 } } { 4 } & 1 \leq y \leq 2 \\
1 & y > 2
\end{array} \right.$$

(i) Find $\mathrm { P } ( Y \leq 1.5 )$\\
(ii) Verify that the median of $Y$ lies between 1.6 and 1.7.\\
(iii) Find the probability density function of $Y$.

\hfill \mbox{\textit{OCR MEI Further Statistics B AS  Q2 [7]}}
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Q1 9 Q2 7 Q3 11 Q4 8 Q5 11 Q6 8 Q7 6