OCR MEI Further Statistics Minor 2023 June — Question 1 8 marks

Exam BoardOCR MEI
ModuleFurther Statistics Minor (Further Statistics Minor)
Year2023
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeFirst success on specific trial
DifficultyStandard +0.3 This is a straightforward application of the geometric distribution with clearly defined success probability (0.3). Parts (a) and (b) require direct formula substitution, while part (c) involves calculating mean and standard deviation then finding probabilities within that range—all standard textbook exercises requiring recall and basic computation rather than problem-solving insight.
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

1 A fair spinner has ten sectors, labelled \(1,2 , \ldots , 10\). In order to start a game, Kofi has to obtain an 8,9 or 10 on the spinner.
  1. Find the probability that Kofi starts the game on the third spin.
  2. Find the probability that Kofi takes at least 5 spins to start the game.
  3. Determine the probability that the number of spins required to start the game is within 1 standard deviation of its mean.

Question 1:
AnswerMarks Guidance
1(a) 0.72×0.3
= 0.147M1
A1
AnswerMarks
[2]3.3
1.1For geometric
Accept 0.15
AnswerMarks Guidance
1(b) P(At least 5) = 0.24(01)
[1]1.1 0.74 or 1−0.3(1+0.7+0.72+0.73) oe
1(c) 10
Mean = soi
3
1−0.3
Variance =
0.32
√70
Standard deviation =
3
10 √70
Evaluate 𝑡ℎ𝑒𝑖𝑟 ( ± )
3 3
and identify their correct integer range
AnswerMarks
(1−0.76) = 0.88(2351)B1
M1
A1
M1
A1 FT
AnswerMarks
[5]3.1b
1.1a
1.1
3.4
AnswerMarks
1.11
Accept 3.3
0.3
(2.78886…)
P(0.54 < 𝑋 < 6.12)
= P(𝑋 ≤ 6)
FT their 𝐸(𝑋) and 𝑆𝐷(𝑋) provided more than one positive
integer included in their range.
Question 1:
1 | (a) | 0.72×0.3
= 0.147 | M1
A1
[2] | 3.3
1.1 | For geometric
Accept 0.15
1 | (b) | P(At least 5) = 0.24(01) | B1
[1] | 1.1 | 0.74 or 1−0.3(1+0.7+0.72+0.73) oe
1 | (c) | 10
Mean = soi
3
1−0.3
Variance =
0.32
√70
Standard deviation =
3
10 √70
Evaluate 𝑡ℎ𝑒𝑖𝑟 ( ± )
3 3
and identify their correct integer range
(1−0.76) = 0.88(2351) | B1
M1
A1
M1
A1 FT
[5] | 3.1b
1.1a
1.1
3.4
1.1 | 1
Accept 3.3
0.3
(2.78886…)
P(0.54 < 𝑋 < 6.12)
= P(𝑋 ≤ 6)
FT their 𝐸(𝑋) and 𝑆𝐷(𝑋) provided more than one positive
integer included in their range.
1 A fair spinner has ten sectors, labelled $1,2 , \ldots , 10$. In order to start a game, Kofi has to obtain an 8,9 or 10 on the spinner.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Kofi starts the game on the third spin.
\item Find the probability that Kofi takes at least 5 spins to start the game.
\item Determine the probability that the number of spins required to start the game is within 1 standard deviation of its mean.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2023 Q1 [8]}}