| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Standard +0.3 This is a straightforward multi-part normal distribution question requiring standard techniques: z-score calculations, independence for multiple trials, and binomial probability. Part (iii) combines normal and binomial distributions but in a routine way. Parts (iv-v) are inverse normal problems, and (vi) requires minimal interpretation. All techniques are standard S2 material with no novel insight 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 | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X \geq 750) = P\!\left(Z \geq \frac{750 - 751.4}{2.5}\right)\) | M1 | For standardizing |
| M1 | For correct structure (M0 if continuity correction used) | |
| \(= P(Z > -0.56) = \Phi(0.56) = 0.7123\) | A1 | CAO Allow 0.712 www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{all 6 at least 750ml}) = 0.7123^6 = 0.1306\) | M1 | For (their answer to part (i))\(^6\) |
| A1 | FT 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(Y=0) = \binom{25}{0} \times 0.8694^{25}\ (= 0.0302)\) | M1 | For using \(\text{Binomial}(25, p)\) with their \(p\) from part (ii) |
| \(P(Y=1) = \binom{25}{1} \times 0.8694^{24} \times 0.1306\ (= 0.1135)\) | M1 | For correct structure of either \(P(Y=0)\) or \(P(Y=1)\) with their \(p\) from part (ii). M0 if \(p\) and \(q\) reversed |
| \(P(Y=0) + P(Y=1) = 0.144\); \(P(Y \geq 2) = 1 - 0.144 = 0.856\) | M1dep | For \(1 -\) sum of both probabilities |
| A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P\!\left(Z < \frac{750 - \mu}{2.5}\right) = 0.02\); \(\Phi^{-1}(0.02) = -2.054\) | B1 | For \(\pm 2.054\) seen. Allow \(\pm 2.055\) |
| \(\frac{750 - \mu}{2.5} = -2.054\) | M1 | For correct equation as seen or equivalent. FT \(\sigma = \sqrt{2.5}\). M0 if c.c. used. |
| \(\mu = 750 + 2.054 \times 2.5 = 755.1\) | M1 | For correctly rearranging their equation (if 750 used in numerator) for \(\mu\), FT their \(z\) |
| A1 | cao Condone 755 or 5 s.f. rounding to 755.1 www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P\!\left(Z < \frac{750 - 751.4}{\sigma}\right) = 0.02\); \(\frac{750 - 751.4}{\sigma} = -2.054\) | M1 | For correct equation as seen or equivalent |
| \(\sigma = \frac{-1.4}{-2.054}\) | M1 | For correctly rearranging their equation (if 750 used in numerator) for \(\sigma\) unless this leads to \(\sigma < 0\) |
| \(= 0.682\) | A1 | cao Allow answers rounding to 0.68 www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Probably easier to change the mean (as reducing the standard deviation would require a much more accurate filling process). However increasing the mean would result in fewer bottles being filled overall and so less profit for the owners, so reducing the standard deviation would be preferable to the vineyard owners. | E1 | |
| E1 | For "preferable to reduce the standard deviation" with valid reason. |
# Question 3:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X \geq 750) = P\!\left(Z \geq \frac{750 - 751.4}{2.5}\right)$ | M1 | For standardizing |
| | M1 | For correct structure (M0 if continuity correction used) |
| $= P(Z > -0.56) = \Phi(0.56) = 0.7123$ | A1 | CAO Allow 0.712 www |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{all 6 at least 750ml}) = 0.7123^6 = 0.1306$ | M1 | For (their answer to part (i))$^6$ |
| | A1 | FT 3 s.f. |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(Y=0) = \binom{25}{0} \times 0.8694^{25}\ (= 0.0302)$ | M1 | For using $\text{Binomial}(25, p)$ with their $p$ from part (ii) |
| $P(Y=1) = \binom{25}{1} \times 0.8694^{24} \times 0.1306\ (= 0.1135)$ | M1 | For correct structure of either $P(Y=0)$ or $P(Y=1)$ with their $p$ from part (ii). M0 if $p$ and $q$ reversed |
| $P(Y=0) + P(Y=1) = 0.144$; $P(Y \geq 2) = 1 - 0.144 = 0.856$ | M1dep | For $1 -$ sum of both probabilities |
| | A1 | CAO |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P\!\left(Z < \frac{750 - \mu}{2.5}\right) = 0.02$; $\Phi^{-1}(0.02) = -2.054$ | B1 | For $\pm 2.054$ seen. Allow $\pm 2.055$ |
| $\frac{750 - \mu}{2.5} = -2.054$ | M1 | For correct equation as seen or equivalent. FT $\sigma = \sqrt{2.5}$. M0 if c.c. used. |
| $\mu = 750 + 2.054 \times 2.5 = 755.1$ | M1 | For correctly rearranging their equation (if 750 used in numerator) for $\mu$, FT their $z$ |
| | A1 | cao Condone 755 or 5 s.f. rounding to 755.1 www |
## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P\!\left(Z < \frac{750 - 751.4}{\sigma}\right) = 0.02$; $\frac{750 - 751.4}{\sigma} = -2.054$ | M1 | For correct equation as seen or equivalent |
| $\sigma = \frac{-1.4}{-2.054}$ | M1 | For correctly rearranging their equation (if 750 used in numerator) for $\sigma$ unless this leads to $\sigma < 0$ |
| $= 0.682$ | A1 | cao Allow answers rounding to 0.68 www |
## Part (vi)
| Answer | Mark | Guidance |
|--------|------|----------|
| Probably easier to change the mean (as reducing the standard deviation would require a much more accurate filling process). However increasing the mean would result in fewer bottles being filled overall and so less profit for the owners, so **reducing** the **standard deviation** would be **preferable** to the vineyard owners. | E1 | |
| | E1 | For "preferable to reduce the standard deviation" with valid reason. |
---3 At a vineyard, the process used to fill bottles with wine is subject to variation. The contents of bottles are independently Normally distributed with mean $\mu = 751.4 \mathrm { ml }$ and standard deviation $\sigma = 2.5 \mathrm { ml }$.\\
(i) Find the probability that a randomly selected bottle contains at least 750 ml .\\
(ii) A case of wine consists of 6 bottles. Find the probability that all 6 bottles in a case contain at least 750 ml .\\
(iii) Find the probability that, in a random sample of 25 cases, there are at least 2 cases in which all 6 bottles contain at least 750 ml .
It is decided to increase the proportion of bottles which contain at least 750 ml to $98 \%$.\\
(iv) This can be done by changing the value of $\mu$, but retaining the original value of $\sigma$. Find the required value of $\mu$.\\
(v) An alternative is to change the value of $\sigma$, but retain the original value of $\mu$. Find the required value of $\sigma$.\\
(vi) Comment briefly on which method might be easier to implement and which might be preferable to the vineyard owners.
\hfill \mbox{\textit{OCR MEI S2 2012 Q3 [18]}}