OCR Further Statistics 2019 June — Question 7 10 marks

Exam BoardOCR
ModuleFurther Statistics (Further Statistics)
Year2019
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeVariance of geometric distribution
DifficultyStandard +0.3 This question tests standard properties of variance and expectation with linear transformations, plus basic geometric distribution formulas. Part (a) uses Var(aX+b)=a²Var(X), part (b) requires finding p from the variance formula then using E(D)=1/p, and part (c) involves straightforward probability calculation. All steps are routine applications of known formulas with no novel problem-solving required, making it slightly easier than average.
Spec5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2
7 The random variable \(D\) has the distribution \(\operatorname { Geo } ( p )\). It is given that \(\operatorname { Var } ( D ) = \frac { 40 } { 9 }\).
Determine
  1. \(\operatorname { Var } ( 3 D + 5 )\),
  2. \(\mathrm { E } ( 3 \mathrm { D } + 5 )\),
  3. \(\mathrm { P } ( D > \mathrm { E } ( D ) )\).
Question 7:
AnswerMarks Guidance
7(a) 9  40 = 40
9B1
[1]1.1 40 or awrt 40.0 only
(b)1 p
 40
p2 9
 p 3 or 3
8 5
Reject 3
5
E(D) = 1/p [ 8]
3
E(3D + 5) 385 [= 13]
AnswerMarks
3M1
M1
A1
B1ft
M1
A1ft
AnswerMarks
[6]2.1
3.1b
2.2a
2.3
1.1
AnswerMarks
1.1Use correct formula for variance
Solve quadratic, can be implied BC
3 stated
8
Explicitly reject second solution, e.g.
“”, no reason needed but not just
written down and then ignored
Use formula for E(D)
AnswerMarks
3(their E(D)) + 5SC: insufficient working, 3
8
only: M0B1 for 3 , then B0
8
Allow for explicit rejection
of a solution even if both are
wrong
p doesn’t need to be between
0 and 1 for either of these
marks
1 p 1 161 1 161
SC: 40(their 40), p , reject negative solution, E(D) = 6.844, E(3D + 5) = 25.53: M1, M1A0, B1, B2 total 5/6
p2 80 2
AnswerMarks
(c)P(D > E(D)) = P(D  3)
= (1 – p)2
 25 or 0.390625
AnswerMarks
64M1ft
M1
A1
AnswerMarks
[3]3.1a
1.1a
AnswerMarks
1.1Convert inequality to integer,
their [1/p] + 1, allow >
(1 – p)r, ft on their p, r, e.g. 8/3 or 13
Allow (1 – p)3 = 125/512 or 0.244
AnswerMarks
Answer, exact or art 0.391, wwwNot their 13
(1 – p)8/3 [0.286]: M0M1A0
Need 0 < p < 1 here
Allow (1 – p)6 = 0.3876
from SC above
Question 7:
7 | (a) | 9  40 = 40
9 | B1
[1] | 1.1 | 40 or awrt 40.0 only
(b) | 1 p
 40
p2 9
 p 3 or 3
8 5
Reject 3
5
E(D) = 1/p [ 8]
3
E(3D + 5) 385 [= 13]
3 | M1
M1
A1
B1ft
M1
A1ft
[6] | 2.1
3.1b
2.2a
2.3
1.1
1.1 | Use correct formula for variance
Solve quadratic, can be implied BC
3 stated
8
Explicitly reject second solution, e.g.
“”, no reason needed but not just
written down and then ignored
Use formula for E(D)
3(their E(D)) + 5 | SC: insufficient working, 3
8
only: M0B1 for 3 , then B0
8
Allow for explicit rejection
of a solution even if both are
wrong
p doesn’t need to be between
0 and 1 for either of these
marks
1 p 1 161 1 161
SC: 40(their 40), p , reject negative solution, E(D) = 6.844, E(3D + 5) = 25.53: M1, M1A0, B1, B2 total 5/6
p2 80 2
(c) | P(D > E(D)) = P(D  3)
= (1 – p)2
 25 or 0.390625
64 | M1ft
M1
A1
[3] | 3.1a
1.1a
1.1 | Convert inequality to integer,
their [1/p] + 1, allow >
(1 – p)r, ft on their p, r, e.g. 8/3 or 13
Allow (1 – p)3 = 125/512 or 0.244
Answer, exact or art 0.391, www | Not their 13
(1 – p)8/3 [0.286]: M0M1A0
Need 0 < p < 1 here
Allow (1 – p)6 = 0.3876
from SC above
7 The random variable $D$ has the distribution $\operatorname { Geo } ( p )$. It is given that $\operatorname { Var } ( D ) = \frac { 40 } { 9 }$.\\
Determine
\begin{enumerate}[label=(\alph*)]
\item $\operatorname { Var } ( 3 D + 5 )$,
\item $\mathrm { E } ( 3 \mathrm { D } + 5 )$,
\item $\mathrm { P } ( D > \mathrm { E } ( D ) )$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Statistics 2019 Q7 [10]}}
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