CAIE S1 2010 June — Question 2 5 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2010
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeDirect binomial from normal probability
DifficultyModerate -0.8 This is a straightforward application of normal distribution and binomial probability. Part (i) requires a single z-score calculation and table lookup. Part (ii) combines this with basic binomial probability (finding P(X ≥ 2) using complement). Both parts are routine procedures with no conceptual challenges, making this easier than average but not trivial due to the two-part structure.
Spec2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

The lengths of new pencils are normally distributed with mean 11 cm and standard deviation 0.095 cm.
  1. Find the probability that a pencil chosen at random has a length greater than 10.9 cm. [2]
  2. Find the probability that, in a random sample of 6 pencils, at least two have lengths less than 10.9 cm. [3]

AnswerMarks Guidance
(i) \(P(x > 10.9) = P\left(z > \frac{10.9 - 11}{0.095}\right) = P(z > -1.0526) = 0.8538 \text{ (0.854)}\)M1, A1 [2] Standardising, no cc, no sq rt. Rounding to correct answer
(ii) \(P(\text{at least } 2 < 10.9) = 1 - P(0, 1) = 1 - (0.8538)^6 - {^6}C_1(0.1462)(0.8538)^5 = 0.215\)M1, A1ft, A1 [3] Bin expression with \(\sum\) powers = 6, "\(C_r\)", \(p + q = 1\). Reasonably correct unsimplified expression fit their (i). Rounding to correct answer
(i) $P(x > 10.9) = P\left(z > \frac{10.9 - 11}{0.095}\right) = P(z > -1.0526) = 0.8538 \text{ (0.854)}$ | M1, A1 [2] | Standardising, no cc, no sq rt. Rounding to correct answer

(ii) $P(\text{at least } 2 < 10.9) = 1 - P(0, 1) = 1 - (0.8538)^6 - {^6}C_1(0.1462)(0.8538)^5 = 0.215$ | M1, A1ft, A1 [3] | Bin expression with $\sum$ powers = 6, "$C_r$", $p + q = 1$. Reasonably correct unsimplified expression fit their (i). Rounding to correct answer

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The lengths of new pencils are normally distributed with mean 11 cm and standard deviation 0.095 cm.

\begin{enumerate}[label=(\roman*)]
\item Find the probability that a pencil chosen at random has a length greater than 10.9 cm. [2]

\item Find the probability that, in a random sample of 6 pencils, at least two have lengths less than 10.9 cm. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE S1 2010 Q2 [5]}}