Standard +0.3 This is a straightforward application of the Central Limit Theorem requiring students to work backwards from a probability to find sample size. It involves standard normal distribution lookup and basic algebraic manipulation (z = 1.96, then solve for n from the standard error formula), but requires no conceptual insight beyond recognizing the CLT setup. Slightly easier than average due to being a direct single-method problem with clear steps.
3 A population has mean 12 and standard deviation 2.5. A large random sample of size \(n\) is chosen from this population and the sample mean is denoted by \(\bar { X }\). Given that \(\mathrm { P } ( \bar { X } < 12.2 ) = 0.975\), correct to 3 significant figures, find the value of \(n\).
Standardisation. Allow cc. need correct form incl sqrt
\((=)\ 1.96\)
B1
Correct \(z\)
\(\sqrt{n} = 1.96 \times 2.5 \div 0.2\)
M1
Rearrange equation in \(n\) or \(\sqrt{n}\) with numerical \(z\) to the stage \(n=\) or \(\sqrt{n}=\) allow arithmetical slips only
\(n = 600\)
A1
accept 601; SR whole number ans from 595 to 605 can score full marks if fully justified
Total: 4
## Question 3:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{12.2 - 12}{2.5/\sqrt{n}}$ | M1 | Standardisation. Allow cc. need correct form incl sqrt |
| $(=)\ 1.96$ | B1 | Correct $z$ |
| $\sqrt{n} = 1.96 \times 2.5 \div 0.2$ | M1 | Rearrange equation in $n$ or $\sqrt{n}$ with numerical $z$ to the stage $n=$ or $\sqrt{n}=$ allow arithmetical slips only |
| $n = 600$ | A1 | accept 601; SR whole number ans from 595 to 605 can score full marks if fully justified |
| **Total: 4** | | |
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3 A population has mean 12 and standard deviation 2.5. A large random sample of size $n$ is chosen from this population and the sample mean is denoted by $\bar { X }$. Given that $\mathrm { P } ( \bar { X } < 12.2 ) = 0.975$, correct to 3 significant figures, find the value of $n$.\\
\hfill \mbox{\textit{CAIE S2 2018 Q3 [4]}}