OCR MEI Further Statistics Minor 2020 November — Question 2 11 marks

Exam BoardOCR MEI
ModuleFurther Statistics Minor (Further Statistics Minor)
Year2020
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypePoisson approximation justification or comparison
DifficultyStandard +0.3 This is a straightforward application of standard Poisson approximation to binomial with routine calculations. Part (a) requires recognizing the conditions (large n, small p), parts (b) and (d) involve direct Poisson probability calculations, and part (c) uses binomial distribution with logarithms to solve for n. All techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec5.02b Expectation and variance: discrete random variables5.02d Binomial: mean np and variance np(1-p)5.02e Discrete uniform distribution5.02f 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^25.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda
2 On computer monitor screens there are often one or more tiny dots which are permanently dark and do not display any of the image. Such dots are known as 'dead pixels'. Dead pixels occur on screens randomly and independently of each other. A company manufactures three types of monitor, Types A, B and C. For a monitor of Type A, the screen has a total of 2304000 pixels. For this type of monitor, the probability of a randomly chosen pixel being dead is 1 in 500000 . Let \(X\) represent the number of dead pixels on a monitor screen of this type.
  1. Explain why you could use either a binomial distribution or a Poisson distribution to model the distribution of \(X\).
  2. Use a Poisson distribution to calculate estimates of each of the following probabilities.
    For a monitor of Type B, the probability of a randomly chosen pixel being dead is also 1 in 500 000. The screen of a monitor of Type B has a total of \(n\) pixels. Use a binomial distribution to find the least value of \(n\) for which the probability of finding at least 1 dead pixel is greater than 0.99 . Give your answer in millions correct to 3 significant figures. For a monitor of Type C, the number of dead pixels on the screen is modelled by a Poisson distribution with mean \(\lambda\).
  3. Given that the probability of finding at least one dead pixel is 0.8 , find \(\lambda\).
Question 2:
AnswerMarks Guidance
2(a) Dead pixels (occur randomly and) occurrences are
independent with constant probability .
1
The number of β€˜successes’ in 2304000 pixels are
500000
being counted, so a binomial distribution is
appropriate.
Because n (2304000) is large and p ( ) is small
1
a Poisson distribution is also appropriate
AnswerMarks
500000E1
E1
E1
AnswerMarks
[3]2.4
2.4
AnswerMarks
2.4For explanation of binomial citing at
least two conditions
For full explanation including context
For explanation of Poisson
AnswerMarks
approximationNeed not give specific
values but if given they
should be correct
AnswerMarks Guidance
2(b) Poisson soi
(4.608)
P(X = 4) = 0.187 (0.18733...)
P(X > 4) = 1 – 0.5117
AnswerMarks
= 0.488 (0.48826...)M1
A1
A1
AnswerMarks
[3]3.3
1.1
AnswerMarks
1.1BC
BC576
125
AnswerMarks Guidance
2(c) DR
1 – (>) 0.99 so <) 0.01
𝑛𝑛 𝑛𝑛
499999 499999
οΏ½500000οΏ½ οΏ½500000οΏ½ (
n (>) (= 2302582.79) oe
log0.01
499999
log500000
AnswerMarks
Sample size is (at least) 2.31 million (2302583)M1*
M1dep*
A1
AnswerMarks
[3]3.1b
1.1
AnswerMarks
3.2a𝑙𝑙𝑙𝑙𝑙𝑙0.999998(0.01)
Allow (at least) 2.30 million
AnswerMarks Guidance
2(d) P( X β‰₯ 1) = 0.8 so P(X = 0) = 0.2
eβˆ’Ξ» = 0.2 oe
AnswerMarks
Ξ» = 1.6(094…) or orM1
A1
AnswerMarks
[2]3.3
1.1
Question 2:
2 | (a) | Dead pixels (occur randomly and) occurrences are
independent with constant probability .
1
The number of β€˜successes’ in 2304000 pixels are
500000
being counted, so a binomial distribution is
appropriate.
Because n (2304000) is large and p ( ) is small
1
a Poisson distribution is also appropriate
500000 | E1
E1
E1
[3] | 2.4
2.4
2.4 | For explanation of binomial citing at
least two conditions
For full explanation including context
For explanation of Poisson
approximation | Need not give specific
values but if given they
should be correct
2 | (b) | Poisson soi
(4.608)
P(X = 4) = 0.187 (0.18733...)
P(X > 4) = 1 – 0.5117
= 0.488 (0.48826...) | M1
A1
A1
[3] | 3.3
1.1
1.1 | BC
BC | 576
125
2 | (c) | DR
1 – (>) 0.99 so <) 0.01
𝑛𝑛 𝑛𝑛
499999 499999
οΏ½500000οΏ½ οΏ½500000οΏ½ (
n (>) (= 2302582.79) oe
log0.01
499999
log500000
Sample size is (at least) 2.31 million (2302583) | M1*
M1dep*
A1
[3] | 3.1b
1.1
3.2a | 𝑙𝑙𝑙𝑙𝑙𝑙0.999998(0.01)
Allow (at least) 2.30 million
2 | (d) | P( X β‰₯ 1) = 0.8 so P(X = 0) = 0.2
eβˆ’Ξ» = 0.2 oe
Ξ» = 1.6(094…) or or | M1
A1
[2] | 3.3
1.1
2 On computer monitor screens there are often one or more tiny dots which are permanently dark and do not display any of the image. Such dots are known as 'dead pixels'. Dead pixels occur on screens randomly and independently of each other.

A company manufactures three types of monitor, Types A, B and C. For a monitor of Type A, the screen has a total of 2304000 pixels. For this type of monitor, the probability of a randomly chosen pixel being dead is 1 in 500000 . Let $X$ represent the number of dead pixels on a monitor screen of this type.
\begin{enumerate}[label=(\alph*)]
\item Explain why you could use either a binomial distribution or a Poisson distribution to model the distribution of $X$.
\item Use a Poisson distribution to calculate estimates of each of the following probabilities.

\begin{itemize}
  \item $\mathrm { P } ( X = 4 )$
  \item $\mathrm { P } ( X > 4 )$
\item In this question you must show detailed reasoning.
\end{itemize}

For a monitor of Type B, the probability of a randomly chosen pixel being dead is also 1 in 500 000. The screen of a monitor of Type B has a total of $n$ pixels. Use a binomial distribution to find the least value of $n$ for which the probability of finding at least 1 dead pixel is greater than 0.99 . Give your answer in millions correct to 3 significant figures.

For a monitor of Type C, the number of dead pixels on the screen is modelled by a Poisson distribution with mean $\lambda$.
\item Given that the probability of finding at least one dead pixel is 0.8 , find $\lambda$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Statistics Minor 2020 Q2 [11]}}
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