OCR MEI Further Statistics Major 2023 June — Question 8 12 marks

Exam BoardOCR MEI
ModuleFurther Statistics Major (Further Statistics Major)
Year2023
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Uniform Random Variables
TypeMultiple observations or trials
DifficultyModerate -0.5 Part (a) is a straightforward application of independence with uniform distribution probabilities (P(X<3)=0.3, P(X>3)=0.7, multiply and account for order). Part (b) requires recognizing symmetry of the sum distribution around its mean (E[T]=25), making P(T≤25)=0.5. Both parts test standard concepts without requiring novel insight, though part (b) needs some probabilistic reasoning about symmetry, making this slightly easier than average overall.
Spec5.03a Continuous random variables: pdf and cdf5.04a Linear combinations: E(aX+bY), Var(aX+bY)
8 The random variable \(X\) has a continuous uniform distribution over [0,10].
  1. Find the probability that, if two independent values of \(X\) are taken, one is less than 3 and the other is greater than 3 . The random variable \(T\) denotes the sum of 5 independent values of \(X\).
  2. State the value of \(\mathrm { P } ( T \leqslant 25 )\). The spreadsheet below shows the heading row and the first 20 data rows from a total of 100 data rows of a simulation of the distribution of \(X\). Each of the 100 rows shows a simulation of 5 independent values of \(X\), together with \(T\), the sum of the 5 values. All of the values have been rounded to 2 decimal places. In column I the spreadsheet shows the number of values of \(T\) that are less than or equal to the corresponding values in column H . For example, there are 75 simulated values of \(T\) that are less than or equal to 30 .
    ABcDEFGHI
    1\(\mathrm { X } _ { 1 }\)\(\mathrm { X } _ { 2 }\)\(\mathrm { X } _ { 3 }\)\(\mathrm { X } _ { 4 }\)\(\mathrm { X } _ { 5 }\)TtNumber \(\leqslant \mathrm { t }\)
    23.736.654.930.419.3325.0600
    34.956.584.482.517.2625.7950
    48.104.874.263.830.7921.85101
    56.704.105.101.826.7624.48154
    63.738.388.499.871.3131.792023
    73.224.360.121.349.4918.532548
    89.177.135.474.352.4428.553075
    93.421.936.042.998.8523.243593
    100.980.689.829.837.2828.584099
    115.861.677.774.087.1426.5245100
    129.200.315.825.316.4527.1050100
    137.044.302.060.064.1617.62
    140.315.021.485.371.7713.94
    153.776.041.217.675.0123.69
    161.215.541.901.436.9117.00
    179.271.985.809.379.3435.76
    184.305.662.801.561.1915.51
    197.153.196.895.412.1824.82
    206.186.323.016.499.1231.13
    215.035.995.196.973.5526.73
  3. Use the spreadsheet output to estimate each of the following.
    The random variable \(Y\) is the mean of 100 independent values of \(T\). Determine an estimate of \(\mathrm { P } ( Y > 26 )\).
Question 8:
AnswerMarks Guidance
8(a) 3 7
2× ×
10 10
42 21
= or or 0.42
AnswerMarks
100 50M1
A1
AnswerMarks
[2]1.1
1.13 7
For × or 0.3×0.7
10 10
AnswerMarks Guidance
8(b) P(T ≤ 25) = 0.5
[1]1.1
8(c) 48 12
P(𝑇 ≤ 25) = or or 0.48
100 25
7
P(𝑇 > 35) = or 0.07
AnswerMarks
100B1
B1
AnswerMarks
[2]1.1
1.1
AnswerMarks Guidance
8(d) DR
E(X) = 5
Var(X) = 1 (10−0)2
12
25
=
3
125
E(T) = 25 Var(T) =
3
125 5
[E(Y) = 25] Var(Y) = [= ]
300 12
5
By CLT distribution is approx N(25, )
12
AnswerMarks
P(Y > 26) = 0.0607B1
M1
A1
M1
M1
M1
A1
AnswerMarks
[7]3.1a
1.2
1.1
1.1
1.1
2.2a
AnswerMarks
1.1s.o.i.
25
Allow M0A0 SCB1 if used below but not explicitly
3
found without full explanation.
Allow Var(T) = 5× their Var(X)
Allow Var(Y) = their Var(T)/100
BC (0.060667…) Do not allow a continuity correction
125
Allow equivalent method M1 for Var(T) = , M1 for
3
12500 12500
Var(total of 100 values) = , M1 for N(2500, ),
3 3
A1 for P(Total > 2600) = 0.0607
Question 8:
8 | (a) | 3 7
2× ×
10 10
42 21
= or or 0.42
100 50 | M1
A1
[2] | 1.1
1.1 | 3 7
For × or 0.3×0.7
10 10
8 | (b) | P(T ≤ 25) = 0.5 | B1
[1] | 1.1
8 | (c) | 48 12
P(𝑇 ≤ 25) = or or 0.48
100 25
7
P(𝑇 > 35) = or 0.07
100 | B1
B1
[2] | 1.1
1.1
8 | (d) | DR
E(X) = 5
Var(X) = 1 (10−0)2
12
25
=
3
125
E(T) = 25 Var(T) =
3
125 5
[E(Y) = 25] Var(Y) = [= ]
300 12
5
By CLT distribution is approx N(25, )
12
P(Y > 26) = 0.0607 | B1
M1
A1
M1
M1
M1
A1
[7] | 3.1a
1.2
1.1
1.1
1.1
2.2a
1.1 | s.o.i.
25
Allow M0A0 SCB1 if used below but not explicitly
3
found without full explanation.
Allow Var(T) = 5× their Var(X)
Allow Var(Y) = their Var(T)/100
BC (0.060667…) Do not allow a continuity correction
125
Allow equivalent method M1 for Var(T) = , M1 for
3
12500 12500
Var(total of 100 values) = , M1 for N(2500, ),
3 3
A1 for P(Total > 2600) = 0.0607
8 The random variable $X$ has a continuous uniform distribution over [0,10].
\begin{enumerate}[label=(\alph*)]
\item Find the probability that, if two independent values of $X$ are taken, one is less than 3 and the other is greater than 3 .

The random variable $T$ denotes the sum of 5 independent values of $X$.
\item State the value of $\mathrm { P } ( T \leqslant 25 )$.

The spreadsheet below shows the heading row and the first 20 data rows from a total of 100 data rows of a simulation of the distribution of $X$. Each of the 100 rows shows a simulation of 5 independent values of $X$, together with $T$, the sum of the 5 values. All of the values have been rounded to 2 decimal places.

In column I the spreadsheet shows the number of values of $T$ that are less than or equal to the corresponding values in column H . For example, there are 75 simulated values of $T$ that are less than or equal to 30 .

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline
 & A & B & c & D & E & F & G & H & I \\
\hline
1 & $\mathrm { X } _ { 1 }$ & $\mathrm { X } _ { 2 }$ & $\mathrm { X } _ { 3 }$ & $\mathrm { X } _ { 4 }$ & $\mathrm { X } _ { 5 }$ & T &  & t & Number $\leqslant \mathrm { t }$ \\
\hline
2 & 3.73 & 6.65 & 4.93 & 0.41 & 9.33 & 25.06 &  & 0 & 0 \\
\hline
3 & 4.95 & 6.58 & 4.48 & 2.51 & 7.26 & 25.79 &  & 5 & 0 \\
\hline
4 & 8.10 & 4.87 & 4.26 & 3.83 & 0.79 & 21.85 &  & 10 & 1 \\
\hline
5 & 6.70 & 4.10 & 5.10 & 1.82 & 6.76 & 24.48 &  & 15 & 4 \\
\hline
6 & 3.73 & 8.38 & 8.49 & 9.87 & 1.31 & 31.79 &  & 20 & 23 \\
\hline
7 & 3.22 & 4.36 & 0.12 & 1.34 & 9.49 & 18.53 &  & 25 & 48 \\
\hline
8 & 9.17 & 7.13 & 5.47 & 4.35 & 2.44 & 28.55 &  & 30 & 75 \\
\hline
9 & 3.42 & 1.93 & 6.04 & 2.99 & 8.85 & 23.24 &  & 35 & 93 \\
\hline
10 & 0.98 & 0.68 & 9.82 & 9.83 & 7.28 & 28.58 &  & 40 & 99 \\
\hline
11 & 5.86 & 1.67 & 7.77 & 4.08 & 7.14 & 26.52 &  & 45 & 100 \\
\hline
12 & 9.20 & 0.31 & 5.82 & 5.31 & 6.45 & 27.10 &  & 50 & 100 \\
\hline
13 & 7.04 & 4.30 & 2.06 & 0.06 & 4.16 & 17.62 &  &  &  \\
\hline
14 & 0.31 & 5.02 & 1.48 & 5.37 & 1.77 & 13.94 &  &  &  \\
\hline
15 & 3.77 & 6.04 & 1.21 & 7.67 & 5.01 & 23.69 &  &  &  \\
\hline
16 & 1.21 & 5.54 & 1.90 & 1.43 & 6.91 & 17.00 &  &  &  \\
\hline
17 & 9.27 & 1.98 & 5.80 & 9.37 & 9.34 & 35.76 &  &  &  \\
\hline
18 & 4.30 & 5.66 & 2.80 & 1.56 & 1.19 & 15.51 &  &  &  \\
\hline
19 & 7.15 & 3.19 & 6.89 & 5.41 & 2.18 & 24.82 &  &  &  \\
\hline
20 & 6.18 & 6.32 & 3.01 & 6.49 & 9.12 & 31.13 &  &  &  \\
\hline
21 & 5.03 & 5.99 & 5.19 & 6.97 & 3.55 & 26.73 &  &  &  \\
\hline
\end{tabular}
\end{center}
\item Use the spreadsheet output to estimate each of the following.

\begin{itemize}
  \item $\mathrm { P } ( T \leqslant 25 )$
  \item $\mathrm { P } ( T > 35 )$
\item In this question you must show detailed reasoning.
\end{itemize}

The random variable $Y$ is the mean of 100 independent values of $T$.

Determine an estimate of $\mathrm { P } ( Y > 26 )$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Major 2023 Q8 [12]}}
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