OCR MEI Further Statistics Minor 2020 November — Question 4 10 marks

Exam BoardOCR MEI
ModuleFurther Statistics Minor (Further Statistics Minor)
Year2020
SessionNovember
Marks10
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Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeName geometric distribution and parameter
DifficultyModerate -0.8 This is a straightforward application of the geometric distribution with standard bookwork for parts (a)-(c): identifying the distribution, calculating a single probability, and recalling formulas for mean and variance. Part (d) requires solving a quadratic equation from P(Y=2) but involves only basic algebraic manipulation. All steps are routine for Further Statistics students with no novel problem-solving required.
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
4 Cards are drawn at random from a standard pack of 52 cards, one at a time, until one of the 4 aces is drawn. After each card is drawn, it is replaced in the pack before the next one is drawn. The random variable \(X\) represents the number of draws required to draw the first ace.
  1. State fully the distribution of \(X\).
  2. Find \(\mathrm { P } ( X = 10 )\).
  3. Find each of the following.
    A further \(k\) aces are added to the full pack and the process described above is repeated. The random variable \(Y\) represents the number of draws required to draw the first ace.
  4. In this question you must show detailed reasoning. Given that \(\mathrm { P } ( Y = 2 ) = \frac { 8 } { 81 }\), find the two possible values of \(k\).
Question 4:
AnswerMarks Guidance
4(a) Distribution is Geo(metric)
1
AnswerMarks Guidance
�13�B1
[1]1.2 B0 if any additional
parameters given
AnswerMarks Guidance
4(b) P(X = 10) =
9
12 1
�13� ×13
AnswerMarks
= 0.037(428…)M1
A1
AnswerMarks
[2]3.3
1.1
AnswerMarks Guidance
4(c) E(X) = 13
Var(X) = 156B1
B1
AnswerMarks
[2]1.2
1.1
AnswerMarks Guidance
4(d) DR
p(Ace) = soi
𝑘𝑘+4
𝑘𝑘+52
(1 – p)p =
8
81p2 – 8811p + 8 = 0
⇒
p = or p =
1 8
9 9
or used to reach k =
𝑘𝑘+4 1 𝑘𝑘+4 8
𝑘𝑘 +52=9 𝑘𝑘+52 =9
AnswerMarks
⇒k = 2 okr = 380B1
M1
A1
M1
A1
AnswerMarks
[5]3.1a
1.1a
1.1
3.1a
AnswerMarks
1.1At any point
𝑘𝑘+4 𝑘𝑘+4 8
� 1−𝑘𝑘+52��𝑘𝑘+52�= 81
2
8 𝑘𝑘 − 3056𝑘𝑘 + 6080 = 0
Method to solve quadratic from
above approach
AnswerMarks
Both values needed2
𝑘𝑘 −382𝑘𝑘+760=0
Question 4:
4 | (a) | Distribution is Geo(metric)
1
�13� | B1
[1] | 1.2 | B0 if any additional
parameters given
4 | (b) | P(X = 10) =
9
12 1
�13� ×13
= 0.037(428…) | M1
A1
[2] | 3.3
1.1
4 | (c) | E(X) = 13
Var(X) = 156 | B1
B1
[2] | 1.2
1.1
4 | (d) | DR
p(Ace) = soi
𝑘𝑘+4
𝑘𝑘+52
(1 – p)p =
8
81p2 – 8811p + 8 = 0
⇒
p = or p =
1 8
9 9
or used to reach k =
𝑘𝑘+4 1 𝑘𝑘+4 8
𝑘𝑘 +52=9 𝑘𝑘+52 =9
⇒k = 2 okr = 380 | B1
M1
A1
M1
A1
[5] | 3.1a
1.1a
1.1
3.1a
1.1 | At any point
𝑘𝑘+4 𝑘𝑘+4 8
� 1−𝑘𝑘+52��𝑘𝑘+52�= 81
2
8 𝑘𝑘 − 3056𝑘𝑘 + 6080 = 0
Method to solve quadratic from
above approach
Both values needed | 2
𝑘𝑘 −382𝑘𝑘+760=0
4 Cards are drawn at random from a standard pack of 52 cards, one at a time, until one of the 4 aces is drawn. After each card is drawn, it is replaced in the pack before the next one is drawn. The random variable $X$ represents the number of draws required to draw the first ace.
\begin{enumerate}[label=(\alph*)]
\item State fully the distribution of $X$.
\item Find $\mathrm { P } ( X = 10 )$.
\item Find each of the following.

\begin{itemize}
  \item $\mathrm { E } ( X )$
  \item $\operatorname { Var } ( X )$
\end{itemize}

A further $k$ aces are added to the full pack and the process described above is repeated. The random variable $Y$ represents the number of draws required to draw the first ace.
\item In this question you must show detailed reasoning.

Given that $\mathrm { P } ( Y = 2 ) = \frac { 8 } { 81 }$, find the two possible values of $k$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2020 Q4 [10]}}
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