OCR MEI Further Statistics Minor Specimen — Question 3 10 marks

Exam BoardOCR MEI
ModuleFurther Statistics Minor (Further Statistics Minor)
SessionSpecimen
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicUniform Distribution
TypeVariance of sum of independent values
DifficultyStandard +0.3 This is a straightforward discrete uniform distribution question requiring standard formulas for mean and variance, basic properties of sums of independent random variables, and enumeration of outcomes for a convolution. While part (iii) requires some careful counting of the most likely sum, all techniques are routine for Further Statistics students with no novel problem-solving required.
Spec5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02e Discrete uniform distribution

A website awards a random number of loyalty points each time a shopper buys from it. The shopper gets a whole number of points between \(0\) and \(10\) (inclusive). Each possibility is equally likely, each time the shopper buys from the website. Awards of points are independent of each other.
  1. Let \(X\) be the number of points gained after shopping once. Find
  2. Let \(Y\) be the number of points gained after shopping twice. Find
  3. Find the probability of the most likely number of points gained after shopping twice. Justify your answer. [4]

Question 3:
AnswerMarks Guidance
3(i) E(X) = 5
10(cid:117)12
Var (cid:11)X(cid:12) (cid:32)
12
AnswerMarks
(cid:32)10B1
M1
A1
AnswerMarks
[3]1.1
3.3
1.1
AnswerMarks Guidance
3(ii) For 2 visits mean = 10
Variance = 10 + 10 = 20B1FT
M1
A1
AnswerMarks
[3]1.1
3.4
AnswerMarks
1.1N
FT their (i)
AnswerMarks Guidance
3(iii) [Working leading to] 121 possibilities
Most likely total is 10.
E.g. This is the one on the main diagonal of the sample
space so occurs most often.
11
Probability = oe
AnswerMarks
121M1
I
A1
C
E1
B1FT
AnswerMarks
[4]3.1b
M
1.1
2.4
AnswerMarks
3.2aE
E.g. starting list (0, 10), (0, 9)
or as sample space or
calculation 11 × 11. Implied
by probability out of 121
Justification
FT correct probability for their
most likely total
Question 3:
3 | (i) | E(X) = 5
10(cid:117)12
Var (cid:11)X(cid:12) (cid:32)
12
(cid:32)10 | B1
M1
A1
[3] | 1.1
3.3
1.1
3 | (ii) | For 2 visits mean = 10
Variance = 10 + 10 = 20 | B1FT
M1
A1
[3] | 1.1
3.4
1.1 | N
FT their (i)
3 | (iii) | [Working leading to] 121 possibilities
Most likely total is 10.
E.g. This is the one on the main diagonal of the sample
space so occurs most often.
11
Probability = oe
121 | M1
I
A1
C
E1
B1FT
[4] | 3.1b
M
1.1
2.4
3.2a | E
E.g. starting list (0, 10), (0, 9)
or as sample space or
calculation 11 × 11. Implied
by probability out of 121
Justification
FT correct probability for their
most likely total
A website awards a random number of loyalty points each time a shopper buys from it. The shopper gets a whole number of points between $0$ and $10$ (inclusive). Each possibility is equally likely, each time the shopper buys from the website. Awards of points are independent of each other.

\begin{enumerate}[label=(\roman*)]
\item Let $X$ be the number of points gained after shopping once.
Find
\begin{itemize}
\item the mean of $X$
\item the variance of $X$. [3]
\end{itemize}

\item Let $Y$ be the number of points gained after shopping twice.
Find
\begin{itemize}
\item the mean of $Y$
\item the variance of $Y$. [3]
\end{itemize}

\item Find the probability of the most likely number of points gained after shopping twice. Justify your answer. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor  Q3 [10]}}