| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Single batch expected count |
| Difficulty | Moderate -0.8 This is a straightforward application of binomial distribution with clearly stated parameters (n=30, p=0.85). Parts (i) and (ii) require direct calculation of P(X=29) and P(X≥29), while part (iii) simply multiplies the probability from (ii) by 10. All steps are routine with no conceptual challenges beyond recognizing the binomial model and performing standard calculations. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X \sim B(30, 0.85)\) | M1 | For \(0.85^{29} \times 0.15^1 = 0.0013466\) |
| \(P(X=29) = \binom{30}{29} \times 0.85^{29} \times 0.15^1 = 30 \times 0.0013466 = 0.0404\) | M1 | For \(\binom{30}{29} \times p^{29} \times q^1\) with \(p+q=1\) |
| A1 [3] | CAO; allow 0.04 www; if further working (e.g. \(P(X=29) - P(X=28)\)) give M2A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=30) = 0.85^{30} = 0.0076\) | M1 | For \(0.85^{30}\) |
| \(P(X \geq 29) = 0.0404 + 0.0076 = 0.0480\) | M1 | For \(P(X=29) + P(X=30)\); both attempts at binomial including coefficient in (i) |
| A1 [3] | CAO; allow e.g. \(0.04+0.0076=0.0476\); allow 0.05 with working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Expected number \(= 10 \times 0.0480 = 0.480\) | M1 | For \(10 \times\) their (ii); provided (ii) between 0 and 1 |
| A1 [2] | FT their (ii); if answer to (ii) leads to whole number for (iii) give M1A0; do not allow answer rounded to 0 or 1 |
# Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X \sim B(30, 0.85)$ | M1 | For $0.85^{29} \times 0.15^1 = 0.0013466$ |
| $P(X=29) = \binom{30}{29} \times 0.85^{29} \times 0.15^1 = 30 \times 0.0013466 = 0.0404$ | M1 | For $\binom{30}{29} \times p^{29} \times q^1$ with $p+q=1$ |
| | A1 [3] | CAO; allow 0.04 www; if further working (e.g. $P(X=29) - P(X=28)$) give M2A0 |
# Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=30) = 0.85^{30} = 0.0076$ | M1 | For $0.85^{30}$ |
| $P(X \geq 29) = 0.0404 + 0.0076 = 0.0480$ | M1 | For $P(X=29) + P(X=30)$; both attempts at binomial including coefficient in (i) |
| | A1 [3] | CAO; allow e.g. $0.04+0.0076=0.0476$; allow 0.05 with working |
# Question 3(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expected number $= 10 \times 0.0480 = 0.480$ | M1 | For $10 \times$ their (ii); provided (ii) between 0 and 1 |
| | A1 [2] | FT their (ii); if answer to (ii) leads to whole number for (iii) give M1A0; do not allow answer rounded to 0 or 1 |3 At a call centre, $85 \%$ of callers are put on hold before being connected to an operator. A random sample of 30 callers is selected.\\
(i) Find the probability that exactly 29 of these callers are put on hold.\\
(ii) Find the probability that at least 29 of these callers are put on hold.\\
(iii) If 10 random samples, each of 30 callers, are selected, find the expected number of samples in which at least 29 callers are put on hold.
\hfill \mbox{\textit{OCR MEI S1 2012 Q3 [8]}}