OCR S1 2012 January — Question 6 5 marks

Exam BoardOCR
ModuleS1 (Statistics 1)
Year2012
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGeometric Distribution
TypeIdentify distribution from diagram/table
DifficultyModerate -0.8 This is a straightforward pattern recognition question requiring students to identify geometric and binomial distributions from their graphical representations. Part (i) requires recognizing the characteristic exponentially decreasing shape of a geometric distribution, while part (ii) requires identifying the symmetric bell shape of B(4, 0.5). Both are standard S1 content with minimal calculation—primarily visual recognition and brief justification.
Spec5.02c Linear coding: effects on mean and variance5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)

6 The diagrams illustrate all or part of the probability distributions of the discrete random variables \(V , W , X , Y\) and \(Z\). \includegraphics[max width=\textwidth, alt={}, center]{56ca7462-d061-48d3-bc5f-274d925e4e34-4_419_365_370_296} \includegraphics[max width=\textwidth, alt={}, center]{56ca7462-d061-48d3-bc5f-274d925e4e34-4_419_376_370_838} \includegraphics[max width=\textwidth, alt={}, center]{56ca7462-d061-48d3-bc5f-274d925e4e34-4_419_362_370_1400} \includegraphics[max width=\textwidth, alt={}, center]{56ca7462-d061-48d3-bc5f-274d925e4e34-4_421_359_879_580} \includegraphics[max width=\textwidth, alt={}, center]{56ca7462-d061-48d3-bc5f-274d925e4e34-4_419_355_881_1142}
  1. One of these variables has the distribution \(\operatorname { Geo } \left( \frac { 1 } { 2 } \right)\). State, with a reason, which variable this is.
  2. One of these variables has the distribution \(\mathrm { B } \left( 4 , \frac { 1 } { 2 } \right)\). State, with reasons, which variable this is. \(760 \%\) of the voters at a certain polling station are women. Voters enter the polling station one at a time. The number of voters who enter, up to and including the first woman, is denoted by \(X\).

Question 6:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
VB1 \(X\) because mode \(= 1\) oe, or Highest prob is \(P(1)\) oe
because [probs or values or geometric or etc] decreasing or halving, or Highest prob is 1st. Allow "decreasing" or "halving" or "sloping downwards" or equivalentB1 \(Z\) because \(P(0) = 0\) or variable can't be 0; oe Allow "Geo distr'n cannot be zero"
NOT "Positive skew" "None of them": Ignore any reason given
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
Y. Peaks at 2; Like normal, peak at 2; Highest prob is middle one (or is at 2); \(P(X=2)\) is max; Increase to 2 then decr; \(1\ 4\ 6\ 4\ 1\) alone or with \(0.5^4 \times\); \(0.0625, 0.25, 0.375, 0.25, 0.0625\); \(P(1)=P(3)\) and \(P(2)\) is greater/different; or equiv of any of the aboveB1B1B1 If values of some probs listed: 2 to 4 values: B1, Y: B1. For \(3^{rd}\) B1 must link list with Y diag
Any implication that values not all equal, e.g. Not uniform, or values increase (then decrease), or there is a peakB1 \({}^4C_0,\ {}^4C_1,\ {}^4C_2\), etc; indep
Symmetrical or mirror image oe, or \({}^4C_0 = {}^4C_4\) or 2nd \(=\) 4th or similar, or mean \(= 2\), or \(E(X) = 2\), or 2 is highest prob, or peak at 2, or peak is middle valueB1 indep
YB1 indep
## Question 6:

### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| V | B1 | $X$ because mode $= 1$ oe, or Highest prob is $P(1)$ oe |
| because [probs or values or geometric or etc] decreasing or halving, or Highest prob is 1st. Allow "decreasing" or "halving" or "sloping downwards" or equivalent | B1 | $Z$ because $P(0) = 0$ or variable can't be 0; oe Allow "Geo distr'n cannot be zero" |
| NOT "Positive skew" | | "None of them": Ignore any reason given |

### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Y. Peaks at 2; Like normal, peak at 2; Highest prob is middle one (or is at 2); $P(X=2)$ is max; Increase to 2 then decr; $1\ 4\ 6\ 4\ 1$ alone or with $0.5^4 \times$; $0.0625, 0.25, 0.375, 0.25, 0.0625$; $P(1)=P(3)$ and $P(2)$ is greater/different; or equiv of any of the above | B1B1B1 | If values of some probs listed: 2 to 4 values: B1, Y: B1. For $3^{rd}$ B1 must link list with Y diag |
| Any implication that values not all equal, e.g. Not uniform, or values increase (then decrease), or there is a peak | B1 | ${}^4C_0,\ {}^4C_1,\ {}^4C_2$, etc; indep |
| Symmetrical or mirror image oe, or ${}^4C_0 = {}^4C_4$ or 2nd $=$ 4th or similar, or mean $= 2$, or $E(X) = 2$, or 2 is highest prob, or peak at 2, or peak is middle value | B1 | indep |
| Y | B1 | indep |

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6 The diagrams illustrate all or part of the probability distributions of the discrete random variables $V , W , X , Y$ and $Z$.\\
\includegraphics[max width=\textwidth, alt={}, center]{56ca7462-d061-48d3-bc5f-274d925e4e34-4_419_365_370_296}\\
\includegraphics[max width=\textwidth, alt={}, center]{56ca7462-d061-48d3-bc5f-274d925e4e34-4_419_376_370_838}\\
\includegraphics[max width=\textwidth, alt={}, center]{56ca7462-d061-48d3-bc5f-274d925e4e34-4_419_362_370_1400}\\
\includegraphics[max width=\textwidth, alt={}, center]{56ca7462-d061-48d3-bc5f-274d925e4e34-4_421_359_879_580}\\
\includegraphics[max width=\textwidth, alt={}, center]{56ca7462-d061-48d3-bc5f-274d925e4e34-4_419_355_881_1142}\\
(i) One of these variables has the distribution $\operatorname { Geo } \left( \frac { 1 } { 2 } \right)$. State, with a reason, which variable this is.\\
(ii) One of these variables has the distribution $\mathrm { B } \left( 4 , \frac { 1 } { 2 } \right)$. State, with reasons, which variable this is.\\
$760 \%$ of the voters at a certain polling station are women. Voters enter the polling station one at a time. The number of voters who enter, up to and including the first woman, is denoted by $X$.\\

\hfill \mbox{\textit{OCR S1 2012 Q6 [5]}}