CAIE S1 2021 June — Question 2 4 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2021
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeFind standard deviation from probability
DifficultyStandard +0.3 This is a straightforward inverse normal problem requiring students to find the standard deviation given a probability. It involves converting a frequency to a probability (72/2000), finding the corresponding z-score from tables, then solving σ = (x-μ)/z. While it requires multiple steps, each is routine and the problem type is standard in S1 curricula, making it slightly easier than average.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation

2 The weights of bags of sugar are normally distributed with mean 1.04 kg and standard deviation \(\sigma \mathrm { kg }\). In a random sample of 2000 bags of sugar, 72 weighed more than 1.10 kg . Find the value of \(\sigma\).

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(P(X > 1.1) = \frac{72}{2000} (= 0.036)\), \(z = \pm 1.798\)B1 \(1.79 < z \leqslant 1.80\), \(-1.80 \leqslant z < -1.79\) seen
\(\frac{1.1 - 1.04}{\sigma} = 1.798\)B1 1.1 and 1.04 substituted in \(\pm\)standardisation formula, allow continuity correction, not \(\sigma^2\) or \(\sqrt{\sigma}\)
\(\left[\frac{0.06}{\sigma} = 1.798\right]\)M1 Equate *their* \(\pm\)standardisation formula to a \(z\)-value and solve for appropriate area leading to final answer (expect \(\sigma < 0.5\)). Accept \(\pm\frac{0.06}{\sigma} = z\text{-value}\)
\(\sigma = 0.0334\)A1 \(0.03335 \leq \sigma \leq 0.0334\). At least 3 s.f.
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X > 1.1) = \frac{72}{2000} (= 0.036)$, $z = \pm 1.798$ | B1 | $1.79 < z \leqslant 1.80$, $-1.80 \leqslant z < -1.79$ seen |
| $\frac{1.1 - 1.04}{\sigma} = 1.798$ | B1 | 1.1 and 1.04 substituted in $\pm$standardisation formula, allow continuity correction, not $\sigma^2$ or $\sqrt{\sigma}$ |
| $\left[\frac{0.06}{\sigma} = 1.798\right]$ | M1 | Equate *their* $\pm$standardisation formula to a $z$-value and solve for appropriate area leading to final answer (expect $\sigma < 0.5$). Accept $\pm\frac{0.06}{\sigma} = z\text{-value}$ |
| $\sigma = 0.0334$ | A1 | $0.03335 \leq \sigma \leq 0.0334$. At least 3 s.f. |

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2 The weights of bags of sugar are normally distributed with mean 1.04 kg and standard deviation $\sigma \mathrm { kg }$. In a random sample of 2000 bags of sugar, 72 weighed more than 1.10 kg .

Find the value of $\sigma$.\\

\hfill \mbox{\textit{CAIE S1 2021 Q2 [4]}}