| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2006 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Standard +0.3 This is a straightforward multi-part S2 question covering standard normal distribution techniques: basic probability calculations using tables/calculator, a simple binomial application, inverse normal to find parameters, and a routine one-sample z-test. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks |
|---|---|
| \(P(X > 51.5) = P\left(Z > \frac{51.5 - 49.7}{1.6}\right) = P(Z > 1.125)\) | M1 |
| \(= 1 - 0.8697 = 0.1303\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(48.0 < X < 51.5)\) | M1 | |
| \(= P(-1.0625 < Z < 1.125)\) | A1 | |
| \(= 0.8697 - (1-0.8564) = 0.7261\) | A1 | awrt 0.726 |
| Answer | Marks | Guidance |
|---|---|---|
| \(B(4, 0.7261)\), \(P(Y=1) = \binom{4}{1}(0.7261)^1(0.2739)^3\) | M1 A1 | |
| \(= 0.0595\) | A1 | awrt 0.0595 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 49.0) = 0.6 \Rightarrow \frac{49.0 - \mu}{\sigma} = 0.2533\) | M1 A1 | |
| \(P(X < 47.5) = 0.3 \Rightarrow \frac{47.5 - \mu}{\sigma} = -0.5244\) | M1 A1 | |
| Solving simultaneously: \(\sigma = \frac{1.5}{0.7777} = 1.929\) | M1 | |
| \(\mu = 49.0 - 0.2533 \times 1.929 = 48.51\) | A1 | awrt \(\mu = 48.5\), \(\sigma = 1.93\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu = 49.7\); \(H_1: \mu > 49.7\) | B1 | \(\mu\) is the population mean head circumference of organically fed boys |
| Test statistic: \(z = \frac{50.45 - 49.7}{1.6/\sqrt{10}} = \frac{0.75}{0.5060} = 1.483\) | M1 A1 | |
| Critical value \(z = 1.282\) at 10% | B1 | |
| \(1.483 > 1.282\), reject \(H_0\) | M1 | |
| Significant evidence to support nutritionist's claim | A1 |
# Question 2:
## Part (i)(A)
$P(X > 51.5) = P\left(Z > \frac{51.5 - 49.7}{1.6}\right) = P(Z > 1.125)$ | M1 |
$= 1 - 0.8697 = 0.1303$ | A1 |
## Part (i)(B)
$P(48.0 < X < 51.5)$ | M1 |
$= P(-1.0625 < Z < 1.125)$ | A1 |
$= 0.8697 - (1-0.8564) = 0.7261$ | A1 | awrt 0.726
## Part (ii)
$B(4, 0.7261)$, $P(Y=1) = \binom{4}{1}(0.7261)^1(0.2739)^3$ | M1 A1 |
$= 0.0595$ | A1 | awrt 0.0595
## Part (iii)
$P(X < 49.0) = 0.6 \Rightarrow \frac{49.0 - \mu}{\sigma} = 0.2533$ | M1 A1 |
$P(X < 47.5) = 0.3 \Rightarrow \frac{47.5 - \mu}{\sigma} = -0.5244$ | M1 A1 |
Solving simultaneously: $\sigma = \frac{1.5}{0.7777} = 1.929$ | M1 |
$\mu = 49.0 - 0.2533 \times 1.929 = 48.51$ | A1 | awrt $\mu = 48.5$, $\sigma = 1.93$
## Part (iv)
$H_0: \mu = 49.7$; $H_1: \mu > 49.7$ | B1 | $\mu$ is the population mean head circumference of organically fed boys
Test statistic: $z = \frac{50.45 - 49.7}{1.6/\sqrt{10}} = \frac{0.75}{0.5060} = 1.483$ | M1 A1 |
Critical value $z = 1.282$ at 10% | B1 |
$1.483 > 1.282$, reject $H_0$ | M1 |
Significant evidence to support nutritionist's claim | A1 |
---2 The head circumference of 3-year-old boys is known to be Normally distributed with mean 49.7 cm and standard deviation 1.6 cm .
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the head circumference of a randomly selected 3 -year-old boy will be\\
(A) over 51.5 cm ,\\
(B) between 48.0 and 51.5 cm .
\item Four 3-year-old boys are selected at random. Find the probability that exactly one of them has head circumference between 48.0 and 51.5 cm .
\item The head circumference of 3-year-old girls is known to be Normally distributed with mean $\mu$ and standard deviation $\sigma$. Given that $60 \%$ of 3-year-old girls have head circumference below 49.0 cm and $30 \%$ have head circumference below 47.5 cm , find the values of $\mu$ and $\sigma$.
A nutritionist claims that boys who have been fed on a special organic diet will have a larger mean head circumference than other boys. A random sample of ten 3 -year-old boys who have been fed on this organic diet is selected. It is found that their mean head circumference is 50.45 cm .
\item Using the null and alternative hypotheses $\mathrm { H } _ { 0 } : \mu = 49.7 \mathrm {~cm} , \mathrm { H } _ { 1 } : \mu > 49.7 \mathrm {~cm}$, carry out a test at the $10 \%$ significance level to examine the nutritionist's claim. Explain the meaning of $\mu$ in these hypotheses. You may assume that the standard deviation of the head circumference of organically fed 3 -year-old boys is 1.6 cm .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2006 Q2 [18]}}