| Exam Board | OCR MEI |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Sorting and searching algorithms |
| Difficulty | Easy -1.2 This is a straightforward simulation question testing basic probability concepts and random number assignment. Students need only to allocate digit ranges proportionally to given probabilities and apply the rules mechanically—no problem-solving, proof, or novel insight required. The calculations are arithmetic and the concepts are routine for D1. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| number of occupants | 0 | 1 | 2 | 3 | 4 |
| probability | 0.1 | 0.2 | 0.3 | 0.2 | 0.2 |
| number of occupants | 0 | 1 | 2 | 3 | 4 |
| probability | \(\frac { 1 } { 13 }\) | \(\frac { 1 } { 13 }\) | \(\frac { 3 } { 13 }\) | \(\frac { 3 } { 13 }\) | \(\frac { 5 } { 13 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 \to 0\); \(1,2 \to 1\); \(3,4,5 \to 2\); \(6,7 \to 3\); \(8,9 \to 4\) | M1 A1 | either 0.2 for 1 or 0.3 for 2; all proportions correct |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| random numbers: 5, 3, 0, 2, 4, 7, 9, 1, 1, 8; number of occupants: 2, 2, 0, 1, 2, 3, 4, 1, 1, 4 | M1 A1 | 8 outcomes correct; all correct |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0, 1 \to \text{child}\); \(2\text{--}9 \to \text{adult}\) | B1 | must use all 10 digits; cao |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Simulation table with 8 chairs showing child (C) or adult (A) assignments | M1 | 8 chairs OK |
| Number of children \(= 5\), number of adults \(= 15\) | A1 | all OK |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 40 children and 120 adults | B1 | \(FT \times 8\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(00\text{--}06 \to 0\); \(07\text{--}13 \to 1\); \(14\text{--}34 \to 2\); \(35\text{--}55 \to 3\); \(56\text{--}90 \to 4\); \(91\text{--}99\) ignore and "redraw" | M1 A1 A1 | ignore some; proportions correct; efficient |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| random number: 23, 65, 07, 99, 37, 45 | M1 | 3 OK |
| number of occupants: 2, 4, 1, –, 3, 3 | A1 | all correct FT |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| chair table with occupants: occ1: 1,C / 9,A / 6,A / 8,A / 1,C; occ2: 2,A / 2,A / 8 / 0,C / 8,A; occ3: 6 / 3,A / 2 / 2,A / 1,C; occ4: 4 / 6,A / 1 / 9 / 4 | ||
| number of children \(= 4\), number of adults \(= 9\) | B1 | FT ... all correct |
| 64 children and 144 adults | B1 | FT ... \(\times\) by 16 |
| [2] |
| Answer | Marks |
|---|---|
| greater reliability or more representative | B1 |
| [1] |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 \to 0$; $1,2 \to 1$; $3,4,5 \to 2$; $6,7 \to 3$; $8,9 \to 4$ | M1 A1 | either 0.2 for 1 or 0.3 for 2; all proportions correct |
| **[2]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| random numbers: 5, 3, 0, 2, 4, 7, 9, 1, 1, 8; number of occupants: 2, 2, 0, 1, 2, 3, 4, 1, 1, 4 | M1 A1 | 8 outcomes correct; all correct |
| **[2]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0, 1 \to \text{child}$; $2\text{--}9 \to \text{adult}$ | B1 | must use all 10 digits; cao |
| **[1]** | | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Simulation table with 8 chairs showing child (C) or adult (A) assignments | M1 | 8 chairs OK |
| Number of children $= 5$, number of adults $= 15$ | A1 | all OK |
| **[2]** | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 40 children and 120 adults | B1 | $FT \times 8$ |
| **[1]** | | |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $00\text{--}06 \to 0$; $07\text{--}13 \to 1$; $14\text{--}34 \to 2$; $35\text{--}55 \to 3$; $56\text{--}90 \to 4$; $91\text{--}99$ ignore and "redraw" | M1 A1 A1 | ignore some; proportions correct; efficient |
| **[3]** | | |
# Question 5:
## Part (vii):
| random number: 23, 65, 07, 99, 37, 45 | M1 | 3 OK |
| number of occupants: 2, 4, 1, –, 3, 3 | A1 | all correct FT |
| | **[2]** | |
## Part (viii):
| chair table with occupants: occ1: 1,C / 9,A / 6,A / 8,A / 1,C; occ2: 2,A / 2,A / 8 / 0,C / 8,A; occ3: 6 / 3,A / 2 / 2,A / 1,C; occ4: 4 / 6,A / 1 / 9 / 4 | | |
| number of children $= 4$, number of adults $= 9$ | B1 | FT ... all correct |
| 64 children and 144 adults | B1 | FT ... $\times$ by 16 |
| | **[2]** | |
## Part (ix):
| greater reliability or more representative | B1 | |
| | **[1]** | |
---5 A chairlift for a ski slope has 160 4-person chairs. At any one time half of the chairs are going up and half are coming down empty. An observer watches the loading of the chairs during a moderately busy period, and concludes that the number of occupants per 'up' chair has the following probability distribution.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
number of occupants & 0 & 1 & 2 & 3 & 4 \\
\hline
probability & 0.1 & 0.2 & 0.3 & 0.2 & 0.2 \\
\hline
\end{tabular}
\end{center}
(i) Give a rule for using 1-digit random numbers to simulate the number of occupants of an up chair in a moderately busy period.\\
(ii) Use the 10 random digits provided to simulate the number of occupants in 10 up chairs.
The observer estimates that, at all times, on average $20 \%$ of chairlift users are children.\\
(iii) Give an efficient rule for using 1-digit random numbers to simulate whether an occupant of an up chair is a child or an adult.\\
(iv) Use the random digits provided to simulate how many of the occupants of the 10 up chairs are children, and how many are adults. There are more random digits than you will need.\\
(v) Use your results from part (iv) to estimate how many children and how many adults are on the chairlift (ie on the 80 up chairs) at any instant during a moderately busy period.
In a very busy period the number of occupants of an up chair has the following probability distribution.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | }
\hline
number of occupants & 0 & 1 & 2 & 3 & 4 \\
\hline
probability & $\frac { 1 } { 13 }$ & $\frac { 1 } { 13 }$ & $\frac { 3 } { 13 }$ & $\frac { 3 } { 13 }$ & $\frac { 5 } { 13 }$ \\
\hline
\end{tabular}
\end{center}
(vi) Give an efficient rule for using 2-digit random numbers to simulate the number of occupants of an up chair in a very busy period.\\
(vii) Use the 2-digit random numbers provided to simulate the number of occupants in 5 up chairs. There are more random numbers provided than you will need.\\
(viii) Simulate how many of the occupants of the 5 up chairs are children and how many are adults, and thus estimate how many children and how many adults are on the chairlift at any instant during a very busy period.\\
(ix) Discuss the relative merits of simulating using a sample of 10 chairs as against simulating using a sample of 5 chairs.
\hfill \mbox{\textit{OCR MEI D1 2013 Q5 [16]}}