| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2007 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Mixed calculations with boundaries |
| Difficulty | Moderate -0.3 This is a straightforward S2 question testing standard normal distribution calculations and hypothesis testing. Part (a) involves routine z-score calculations, probability finding, and weighted averages. Part (b) is a textbook one-sample z-test with clearly stated parameters. All techniques are standard applications with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05a Hypothesis testing language: null, alternative, p-value, significance2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(24 < X < 33) = P\left(\frac{24-28}{4} < Z < \frac{33-28}{4}\right) = P(-1 < Z < 1.25) = \Phi(1.25) - (1 - \Phi(1)) = 0.8944 - (1 - 0.8413) = 0.8944 - 0.1587 = 0.7357\) (4 s.f.) or 0.736 (to 3 s.f.) | M1 for standardizing; A1 for 1.25 and -1; M1 for prob. with tables and correct structure; A1 CAO (min 3 s.f., to include use of difference column) | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Total = £1839 + £198 = £2037 | M1 for either product (with or without price); M1 for sum of both products with price; A1 CAO awrt £2040 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = 26.42\) (4 s.f.) or 26.4 (to 3 s.f.) | B1 for ±1.645 seen; M1 for correct equation in k with positive z-value; A1 CAO | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Where \(\mu\) denotes the mean weight in kilograms of the population of onions of the new variety | B1 for both correct & ito \(\mu\); B1 for definition of \(\mu\) | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| It is reasonable to conclude that the new variety has a higher mean weight. | B1; M1 must include \(\sqrt{25}\); A1FT; B1 for 2.326; M1 For sensible comparison leading to a conclusion; A1 for correct, consistent conclusion in words and in context | 6 |
## (a)(i)
$X \sim N(28,16)$
$P(24 < X < 33) = P\left(\frac{24-28}{4} < Z < \frac{33-28}{4}\right) = P(-1 < Z < 1.25) = \Phi(1.25) - (1 - \Phi(1)) = 0.8944 - (1 - 0.8413) = 0.8944 - 0.1587 = 0.7357$ (4 s.f.) or 0.736 (to 3 s.f.) | M1 for standardizing; A1 for 1.25 and -1; M1 for prob. with tables and correct structure; A1 CAO (min 3 s.f., to include use of difference column) | 4
## (a)(ii)
$25000 \times 0.7357 \times 0.1 = £1839$
$25000 \times 0.1587 \times 0.05 = £198$
Total = £1839 + £198 = £2037 | M1 for either product (with or without price); M1 for sum of both products with price; A1 CAO awrt £2040 | 3
## (a)(iii)
$X \sim N(k, 16)$
From tables $\Phi^{-1}(0.95) = 1.645$
$\frac{33-k}{4} = 1.645$
$33 - k = 1.645 \times 4$
$k = 33 - 6.58$
$k = 26.42$ (4 s.f.) or 26.4 (to 3 s.f.) | B1 for ±1.645 seen; M1 for correct equation in k with positive z-value; A1 CAO | 3
## (b)(i)
$H_0: \mu = 0.155$; $H_1: \mu > 0.155$
Where $\mu$ denotes the mean weight in kilograms of the population of onions of the new variety | B1 for both correct & ito $\mu$; B1 for definition of $\mu$ | 2
## (b)(ii)
Mean weight = $\frac{4.77}{25} = 0.1908$
Test statistic = $\frac{0.1908 - 0.155}{\sqrt{0.005}/\sqrt{25}} = \frac{0.0358}{0.01414} = 2.531$
1% level 1-tailed critical value of $z = 2.326$
$2.531 > 2.236$ so significant.
There is sufficient evidence to reject $H_0$
It is reasonable to conclude that the new variety has a higher mean weight. | B1; M1 must include $\sqrt{25}$; A1FT; B1 for 2.326; M1 For sensible comparison leading to a conclusion; A1 for correct, consistent conclusion in words and in context | 6
---\begin{enumerate}[label=(\alph*)]
\item A farmer grows Brussels sprouts. The diameter of sprouts in a particular batch, measured in mm, is Normally distributed with mean 28 and variance 16. Sprouts that are between 24 mm and 33 mm in diameter are sold to a supermarket.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the diameter of a randomly selected sprout will be within this range. [4]
\item The farmer sells the sprouts in this range to the supermarket for 10 pence per kilogram. The farmer sells sprouts under 24 mm in diameter to a frozen food factory for 5 pence per kilogram. Sprouts over 33 mm in diameter are thrown away. Estimate the total income received by the farmer for the batch, which weighs 25 500 kg. [3]
\item By harvesting sprouts earlier, the mean diameter for another batch can be reduced to $k$ mm. Find the value of $k$ for which only 5\% of the sprouts will be above 33 mm in diameter. You may assume that the variance is still 16. [3]
\end{enumerate}
\item The farmer also grows onions. The weight in kilograms of the onions is Normally distributed with mean 0.155 and variance 0.005. He is trying out a new variety, which he hopes will yield a higher mean weight. In order to test this, he takes a random sample of 25 onions of the new variety and finds that their total weight is 4.77 kg. You should assume that the weight in kilograms of the new variety is Normally distributed with variance 0.005.
\begin{enumerate}[label=(\roman*)]
\item Write down suitable null and alternative hypotheses for the test in terms of $\mu$. State the meaning of $\mu$ in this case. [2]
\item Carry out the test at the 1\% level. [6]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2007 Q2 [18]}}