| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Moderate -0.3 This is a straightforward multi-part normal distribution question requiring standard techniques: z-score calculations, binomial probability from normal probabilities, and inverse normal to find standard deviation. All parts are routine applications 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 transformation5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X < 30) = P\!\left(Z < \dfrac{30-30.7}{3.5}\right)\) | M1 | For standardising; penalise erroneous continuity corrections and wrong sd; condone numerator reversed |
| \(= P(Z < -0.20) = \Phi(-0.20) = 1 - \Phi(0.20) = 1 - 0.5793\) | M1 | For correct structure; \(1 - \Phi(\text{positive }z)\) |
| \(= 0.4207\) | A1 | CAO; allow 0.421 www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(25 < X < 35) = P\!\left(\dfrac{25-30.7}{3.5} < Z < \dfrac{35-30.7}{3.5}\right)\) | M1 | Correctly standardising both; penalise erroneous continuity corrections and wrong sd; condone both numerators reversed |
| \(= P(-1.629 < X < 1.229) = \Phi(1.229) - \Phi(-1.629)\) \(= 0.8904 - (1-0.9483) = 0.8904 - 0.0517\) | M1 | For correct structure; \(\Phi(1.23)-\Phi(-1.63)\) leads to \(0.8907 - 0.0516 = 0.8391\) |
| \(= 0.8387\) | A1 | Use of differences column required; only allow 0.839 if 0.8387 is seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{all 5 weigh at least 30kg}) = 0.5793^5\) | M1 | Allow FT \((1 - \text{their (i)(A)})^5\) or \([\text{their } P(X \geq 30)]^5\); FT only \((1 - \text{their (i)(A)})^5\) |
| \(= 0.0652\) | A1 | Allow 0.06524, allow 0.065 www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{weight} > 30) = 0.05\) | ||
| \(P\left(Z > \frac{30 - 26.8}{\sigma}\right) = 0.05\) | ||
| \(\Phi^{-1}(0.95) = 1.645\) | B1 | For 1.645; B0 for \(1 - 1.645\) or \(0.1645\); NOTE use of \(-1.645\) allowed only if numerator reversed |
| \(\frac{30 - 26.8}{\sigma} = 1.645\) | M1* | For equation as seen or equivalent, with their \(z > 1\); Condone use of spurious c.c. if already penalised in parts (i)(A) or (i)(B) |
| \(\sigma = \frac{30 - 26.8}{1.645} = 1.945 \text{ kg}\) | M1dep*, A1 | Rearranging for \(\sigma\); CAO; Allow \(\sigma = 1.95\) www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Two Normal shapes including asymptotic behaviour with horizontal axis at each of the four ends | G1 | Penalise clear asymmetry |
| Means shown explicitly or by scale on a single diagram | G1 | If shown explicitly, positions must be consistent with horizontal scale if present; If not labelled, assume larger mean represents Male |
| Lower max height for Male | G1 | If not labelled, assume larger mean represents Male |
| Visibly greater width for Male | G1 | If not labelled, assume larger mean represents Male |
# Question 3:
## Part (i)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X < 30) = P\!\left(Z < \dfrac{30-30.7}{3.5}\right)$ | M1 | For standardising; penalise erroneous continuity corrections and wrong sd; condone numerator reversed |
| $= P(Z < -0.20) = \Phi(-0.20) = 1 - \Phi(0.20) = 1 - 0.5793$ | M1 | For correct structure; $1 - \Phi(\text{positive }z)$ |
| $= 0.4207$ | A1 | CAO; allow 0.421 www |
## Part (i)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(25 < X < 35) = P\!\left(\dfrac{25-30.7}{3.5} < Z < \dfrac{35-30.7}{3.5}\right)$ | M1 | Correctly standardising both; penalise erroneous continuity corrections and wrong sd; condone both numerators reversed |
| $= P(-1.629 < X < 1.229) = \Phi(1.229) - \Phi(-1.629)$ $= 0.8904 - (1-0.9483) = 0.8904 - 0.0517$ | M1 | For correct structure; $\Phi(1.23)-\Phi(-1.63)$ leads to $0.8907 - 0.0516 = 0.8391$ |
| $= 0.8387$ | A1 | Use of differences column required; only allow 0.839 if 0.8387 is seen |
## Question 3:
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{all 5 weigh at least 30kg}) = 0.5793^5$ | M1 | Allow FT $(1 - \text{their (i)(A)})^5$ or $[\text{their } P(X \geq 30)]^5$; FT only $(1 - \text{their (i)(A)})^5$ |
| $= 0.0652$ | A1 | Allow 0.06524, allow 0.065 www |
**[2 marks]**
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{weight} > 30) = 0.05$ | | |
| $P\left(Z > \frac{30 - 26.8}{\sigma}\right) = 0.05$ | | |
| $\Phi^{-1}(0.95) = 1.645$ | B1 | For 1.645; B0 for $1 - 1.645$ or $0.1645$; NOTE use of $-1.645$ allowed only if numerator reversed |
| $\frac{30 - 26.8}{\sigma} = 1.645$ | M1* | For equation as seen or equivalent, with their $z > 1$; Condone use of spurious c.c. if already penalised in parts (i)(A) or (i)(B) |
| $\sigma = \frac{30 - 26.8}{1.645} = 1.945 \text{ kg}$ | M1dep*, A1 | Rearranging for $\sigma$; CAO; Allow $\sigma = 1.95$ www |
**[4 marks]**
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Two Normal shapes including asymptotic behaviour with horizontal axis at each of the four ends | G1 | Penalise clear asymmetry |
| Means shown explicitly or by scale on a single diagram | G1 | If shown explicitly, positions must be consistent with horizontal scale if present; If not labelled, assume larger mean represents Male |
| Lower max height for Male | G1 | If not labelled, assume larger mean represents Male |
| Visibly greater width for Male | G1 | If not labelled, assume larger mean represents Male |
**[4 marks]**
---3 The random variable $X$ represents the weight in kg of a randomly selected male dog of a particular breed. $X$ is Normally distributed with mean 30.7 and standard deviation 3.5.
\begin{enumerate}[label=(\roman*)]
\item Find\\
(A) $\mathrm { P } ( X < 30 )$,\\
(B) $P ( 25 < X < 35 )$.
\item Five of these dogs are chosen at random. Find the probability that each of them weighs at least 30 kg .
\item The weights of females of the same breed of dog are Normally distributed with mean 26.8 kg . Given that $5 \%$ of female dogs of this breed weigh more than 30 kg , find the standard deviation of their weights.
\item Sketch the distributions of the weights of male and female dogs of this breed on a single diagram.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2015 Q3 [16]}}