OCR S1 2013 June — Question 7 11 marks

Exam BoardOCR
ModuleS1 (Statistics 1)
Year2013
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicBinomial Distribution
TypeTwo-stage binomial problems
DifficultyStandard +0.3 This is a straightforward application of binomial probability with clearly defined scenarios. Students need to calculate P(X>2) for part (i)(a), then add P(X=2)×P(Y≥1) for part (i)(b), and apply geometric distribution for part (ii). All steps are routine S1 techniques with no novel insight required, making it slightly easier than average.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)
In a factory, an inspector checks a random sample of 30 mugs from a large batch and notes the number, \(X\), which are defective. He then deals with the batch as follows. • If \(X < 2\), the batch is accepted. • If \(X > 2\), the batch is rejected. • If \(X = 2\), the inspector selects another random sample of only 15 mugs from the batch. If this second sample contains 1 or more defective mugs, the batch is rejected. Otherwise the batch is accepted. It is given that 5\% of mugs are defective.
    1. Find the probability that the batch is rejected after just the first sample is checked. [3]
    2. Show that the probability that the batch is rejected is 0.327, correct to 3 significant figures. [5]
  2. Batches are checked one after another. Find the probability that the first batch to be rejected is either the 4th or the 5th batch that is checked. [3]
(i) (a)
Answer:
- \(X \sim B(30, 0.05)\) seen or implied
- \(P(X > 2) = 1 - 0.8122\) or \(1 - 0.5535\) or \(0.95^x \times 0.05^z\) (r.s.>1)
- \(= 0.1878\) or \(0.188\) (3 sfs)
Marks: B1, M1, A1[3]
Guidance:
- eg by 0.8122 or 1–0.5535 or \(0.95^x \times 0.05^z\) (r.s.>1)
- Allow \(B(30, 0.95)\) or \(B(30, 0.5)\) for B1
- 30 × 0.05 alone insufficient for B1
- \(^{\text{0}}C_x\times 0.95^{\text{0}}\times 0.05^z\) insufficient for B1
- If \(n = 15\): B(15, 0.05) B1
- If \(n = 15\): B(15, 0.05) B1 then: eg by 0.8122 or 1–0.5535 or \(0.95^x \times 0.05^z\) (r.s.>1) B1
- \(= 0.1878\) or \(0.188\) (3 sfs) B1
- Allow \(B(30, 0.95)\) or \(B(30, 0.5)\) for B1
- 30 × 0.05 alone insufficient for B1
- \(^{\text{0}}C_x\times 0.95^{\text{0}}\times 0.05^z\) insufficient for B1
- If \(n = 15\): \(B(15, 0.05)\) \(B1\)
- 1–\((0.95^{15}+15\times 0.95^{14}\times 0.05 +^{15}C_x\times 0.95^{13}\times 0.05^2)\) M1
- \(= 0.0362\) A0
(i) (b)
Answer:
- Addition method: \(X \sim B(30, 0.05)\) & \(Y \sim B(15, 0.05)\) stated or implied
- \(P(X = 2) = (0.8122 - 0.5535)\) or \(^{\text{0}}C_x\times 0.95^{\text{0}}\times 0.05^z\) or \(0.2587/6\)
- OR \(P(X \geq 1) = (1 - 0.95^{15})\) or 0.5367
- fully correct method for \(P(X=2)\times P(Y=0)\) "0.2587" × "0.5367" or 0.1388 M1
- \(P(X > 2) + P(X = 2) \times P(Y \geq 1) = "0.1878" + "0.1388"\) alone M1
- \(= 0.327\) (3 sf) AG
Marks: B1, M1, M1, M1, A1[5]
Guidance:
- Subtraction methods:
- \(P(X = 2) = (0.8122 - 0.5535)\) or \(^{\text{0}}C_x\times 0.95^{\text{0}}\times 0.05^z\) or \(0.2587/6\)
- OR \(P(X = 0) = 0.95^{15}\) or 0.4633 M1
- \(P(X=2)\times P(Y=0)\) "0.2587/6" × "0.4633" or 0.1388 M1
- fully correct method for \(P(X=2)\times P(Y=0)\) "0.2587" × "0.5367" or 0.1388 M1
- If \(n = 15\) for both distr's, see next page
- NB eg 0.0362 implies B(15, 0.05) see below
- If \(n = 15\) for both distr's:
- \(B(15, 0.05)\) B0
- \(P(X = 2) = ^{15}C_x\times 0.05^2\times 0.95^{13}\) or 0.1348 or \(P(X \geq 1) = 1 - 0.95^{15}\) or 0.5367 M1
- "0.1348"×"0.5367" or 0.0723 correct method M1
- their (i)(a) + "0.0732" Dep 1st M1 M1
- \(= 0.1085\) A0
- NB Also mark subtraction methods if seen.
(i) (b) (alternative)
Answer: Alternative scheme for the case where \(n = 15\) is used for both distr's
Marks: B0, M1, M1, M1, A0
Guidance:
- If \(n = 15\) for both distr's:
- \(B(15, 0.05)\) B0
- \(P(X = 2) = ^{15}C_x\times 0.05^2\times 0.95^{13}\) or 0.1348 or \(P(X \geq 1) = 1 - 0.95^{15}\) or 0.5367 M1
- "0.1348"×"0.5367" or 0.0723 correct method M1
- their (i)(a) + "0.0732" Dep 1st M1 M1
- \(= 0.1085\) A0
- NB Also mark subtraction methods if seen.
(ii)
Answer:
- Any use of 0.327 or their (i)(b) for 1st M1
- \((1 - 0.327)^3\times 0.327 + (1 - 0.327)^4\times 0.327\)
- Allow "correct" use of their (i)(a) or (i)(b) for 2nd M1
- \(= 0.167\) (3 sf)
Marks: M1, M1, A1[3]
Guidance:
- Allow any use of their (i)(b) for 1st M1 then if "correct" use, also 2nd M1
- Allow use of their (i)(a) in "correct" method for M0M1A0
- No marks for use of 0.95 & 0.05
### (i) (a)
**Answer:**
- $X \sim B(30, 0.05)$ seen or implied
- $P(X > 2) = 1 - 0.8122$ or $1 - 0.5535$ or $0.95^x \times 0.05^z$ (r.s.>1)
- $= 0.1878$ or $0.188$ (3 sfs)

**Marks:** B1, M1, A1[3]

**Guidance:**
- eg by 0.8122 or 1–0.5535 or $0.95^x \times 0.05^z$ (r.s.>1)
- Allow $B(30, 0.95)$ or $B(30, 0.5)$ for B1
- 30 × 0.05 alone insufficient for B1
- $^{\text{0}}C_x\times 0.95^{\text{0}}\times 0.05^z$ insufficient for B1
- If $n = 15$: B(15, 0.05) B1
- If $n = 15$: B(15, 0.05) B1 then: eg by 0.8122 or 1–0.5535 or $0.95^x \times 0.05^z$ (r.s.>1) B1
  - $= 0.1878$ or $0.188$ (3 sfs) B1
  - Allow $B(30, 0.95)$ or $B(30, 0.5)$ for B1
  - 30 × 0.05 alone insufficient for B1
  - $^{\text{0}}C_x\times 0.95^{\text{0}}\times 0.05^z$ insufficient for B1
  - If $n = 15$: $B(15, 0.05)$ $B1$
  - 1–$(0.95^{15}+15\times 0.95^{14}\times 0.05 +^{15}C_x\times 0.95^{13}\times 0.05^2)$ M1
  - $= 0.0362$ A0

### (i) (b)
**Answer:**
- Addition method: $X \sim B(30, 0.05)$ & $Y \sim B(15, 0.05)$ stated or implied
- $P(X = 2) = (0.8122 - 0.5535)$ or $^{\text{0}}C_x\times 0.95^{\text{0}}\times 0.05^z$ or $0.2587/6$
- OR $P(X \geq 1) = (1 - 0.95^{15})$ or 0.5367
- fully correct method for $P(X=2)\times P(Y=0)$ "0.2587" × "0.5367" or 0.1388 M1
- $P(X > 2) + P(X = 2) \times P(Y \geq 1) = "0.1878" + "0.1388"$ alone M1
- $= 0.327$ (3 sf) AG

**Marks:** B1, M1, M1, M1, A1[5]

**Guidance:**
- Subtraction methods:
  - $P(X = 2) = (0.8122 - 0.5535)$ or $^{\text{0}}C_x\times 0.95^{\text{0}}\times 0.05^z$ or $0.2587/6$
  - OR $P(X = 0) = 0.95^{15}$ or 0.4633 M1
  - $P(X=2)\times P(Y=0)$ "0.2587/6" × "0.4633" or 0.1388 M1
  - fully correct method for $P(X=2)\times P(Y=0)$ "0.2587" × "0.5367" or 0.1388 M1
  - If $n = 15$ for both distr's, see next page
  - NB eg 0.0362 implies B(15, 0.05) see below
- If $n = 15$ for both distr's:
  - $B(15, 0.05)$ B0
  - $P(X = 2) = ^{15}C_x\times 0.05^2\times 0.95^{13}$ or 0.1348 or $P(X \geq 1) = 1 - 0.95^{15}$ or 0.5367 M1
  - "0.1348"×"0.5367" or 0.0723 correct method M1
  - their (i)(a) + "0.0732" Dep 1st M1 M1
  - $= 0.1085$ A0
  - NB Also mark subtraction methods if seen.

### (i) (b) (alternative)
**Answer:** Alternative scheme for the case where $n = 15$ is used for both distr's

**Marks:** B0, M1, M1, M1, A0

**Guidance:**
- If $n = 15$ for both distr's:
  - $B(15, 0.05)$ B0
  - $P(X = 2) = ^{15}C_x\times 0.05^2\times 0.95^{13}$ or 0.1348 or $P(X \geq 1) = 1 - 0.95^{15}$ or 0.5367 M1
  - "0.1348"×"0.5367" or 0.0723 correct method M1
  - their (i)(a) + "0.0732" Dep 1st M1 M1
  - $= 0.1085$ A0
  - NB Also mark subtraction methods if seen.

### (ii)
**Answer:**
- Any use of 0.327 or their (i)(b) for 1st M1
- $(1 - 0.327)^3\times 0.327 + (1 - 0.327)^4\times 0.327$
- Allow "correct" use of their (i)(a) or (i)(b) for 2nd M1
- $= 0.167$ (3 sf)

**Marks:** M1, M1, A1[3]

**Guidance:**
- Allow any use of their (i)(b) for 1st M1 then if "correct" use, also 2nd M1
- Allow use of their (i)(a) in "correct" method for M0M1A0
- No marks for use of 0.95 & 0.05

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In a factory, an inspector checks a random sample of 30 mugs from a large batch and notes the number, $X$, which are defective. He then deals with the batch as follows.

• If $X < 2$, the batch is accepted.
• If $X > 2$, the batch is rejected.
• If $X = 2$, the inspector selects another random sample of only 15 mugs from the batch. If this second sample contains 1 or more defective mugs, the batch is rejected. Otherwise the batch is accepted.

It is given that 5\% of mugs are defective.

\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Find the probability that the batch is rejected after just the first sample is checked. [3]

\item Show that the probability that the batch is rejected is 0.327, correct to 3 significant figures. [5]
\end{enumerate}

\item Batches are checked one after another. Find the probability that the first batch to be rejected is either the 4th or the 5th batch that is checked. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR S1 2013 Q7 [11]}}
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