CAIE S1 2015 November — Question 7 12 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2015
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating Binomial to Normal Distribution
TypeFind minimum/maximum n for probability condition
DifficultyStandard +0.3 This is a standard S1 question covering routine binomial distribution calculations and normal approximation. Part (i) is direct binomial probability, part (ii) requires solving P(X≥1)>0.9 using complement rule, part (iii) applies the normal approximation with continuity correction, and part (iv) checks np>5 and nq>5. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial

7 A factory makes water pistols, \(8 \%\) of which do not work properly.
  1. A random sample of 19 water pistols is taken. Find the probability that at most 2 do not work properly.
  2. In a random sample of \(n\) water pistols, the probability that at least one does not work properly is greater than 0.9 . Find the smallest possible value of \(n\).
  3. A random sample of 1800 water pistols is taken. Use an approximation to find the probability that there are at least 152 that do not work properly.
  4. Justify the use of your approximation in part (iii).

Question 7:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(0,1,2) = (0.92)^{19} + {}^{19}C_1(0.08)(0.92)^{18} + {}^{19}C_2(0.08)^2(0.92)^{17}\)M1, M1 Binomial term \({}^{19}C_x p^x(1-p)^{19-x}\) seen, \(0
\(= 0.809\)A1 (3) Correct answer (no working SC B2)
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(\text{at least }1) = 1 - P(0) = 1 - P(0.92)^n > 0.90\)M1 Eqn with their \(0.92^n\), 0.9 or 0.1, 1 not nec
\(0.1>(0.92)^n\)M1 Solving attempt by logs or trial and error, power eqn with one unknown power
\(n > 27.6\)
Ans 28A1 (3) Correct answer, not approx., \(\approx, \geqslant, >, \leqslant, <\)
Part (iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(np = 1800\times 0.08 = 144\); \(npq = 132.48\)B1 Correct unsimplified \(np\) and \(npq\) seen; accept 132.5, 132, 11.5, awrt 11.51
\(P(\text{at least }152) = P\!\left(z>\left(\frac{151.5-144}{\sqrt{132.48}}\right)\right)\)M1 Standardising, with \(\sqrt{}\)
M1Continuity correction 151.5 or 152.5 seen
\(= P(z>0.6516) = 1-0.7429\)M1 Correct area \(1-\Phi\) (probability)
\(= 0.257\)A1 (5) Correct answer
Part (iv):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Use because \(1800\times 0.08\) (and \(1800\times 0.92\) are both) \(> 5\)B1 (1) \(1800\times 0.08>5\) is sufficient; \(np>5\) is sufficient if clearly evaluated in (iii). If \(npq>5\) stated then award B0
## Question 7:

### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(0,1,2) = (0.92)^{19} + {}^{19}C_1(0.08)(0.92)^{18} + {}^{19}C_2(0.08)^2(0.92)^{17}$ | M1, M1 | Binomial term ${}^{19}C_x p^x(1-p)^{19-x}$ seen, $0<p<1$; correct unsimplified expression |
| $= 0.809$ | A1 (3) | Correct answer (no working SC B2) |

### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{at least }1) = 1 - P(0) = 1 - P(0.92)^n > 0.90$ | M1 | Eqn with their $0.92^n$, 0.9 or 0.1, 1 not nec |
| $0.1>(0.92)^n$ | M1 | Solving attempt by logs or trial and error, power eqn with one unknown power |
| $n > 27.6$ | | |
| Ans 28 | A1 (3) | Correct answer, not approx., $\approx, \geqslant, >, \leqslant, <$ |

### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $np = 1800\times 0.08 = 144$; $npq = 132.48$ | B1 | Correct unsimplified $np$ and $npq$ seen; accept 132.5, 132, 11.5, awrt 11.51 |
| $P(\text{at least }152) = P\!\left(z>\left(\frac{151.5-144}{\sqrt{132.48}}\right)\right)$ | M1 | Standardising, with $\sqrt{}$ |
| | M1 | Continuity correction 151.5 or 152.5 seen |
| $= P(z>0.6516) = 1-0.7429$ | M1 | Correct area $1-\Phi$ (probability) |
| $= 0.257$ | A1 (5) | Correct answer |

### Part (iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use because $1800\times 0.08$ (and $1800\times 0.92$ are both) $> 5$ | B1 (1) | $1800\times 0.08>5$ is sufficient; $np>5$ is sufficient if clearly evaluated in (iii). If $npq>5$ stated then award B0 |
7 A factory makes water pistols, $8 \%$ of which do not work properly.\\
(i) A random sample of 19 water pistols is taken. Find the probability that at most 2 do not work properly.\\
(ii) In a random sample of $n$ water pistols, the probability that at least one does not work properly is greater than 0.9 . Find the smallest possible value of $n$.\\
(iii) A random sample of 1800 water pistols is taken. Use an approximation to find the probability that there are at least 152 that do not work properly.\\
(iv) Justify the use of your approximation in part (iii).

\hfill \mbox{\textit{CAIE S1 2015 Q7 [12]}}