| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Fixed container with random contents |
| Difficulty | Standard +0.3 This is a standard S3 question testing routine applications of normal distribution properties and linear combinations. Part (i) is a direct normal probability calculation, (ii) involves straightforward reverse lookup with sum of normals, (iii) applies the standard result for difference of independent normals, and (iv) is a textbook confidence interval calculation. All parts follow well-rehearsed procedures 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 transformation5.04b Linear combinations: of normal distributions5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(A < 90) = P\!\left(Z < \frac{90-80}{11} = 0.9091\right)\) | M1, A1 | For standardising. Award once, here or elsewhere |
| \(= 0.8182\) | A1 | c.a.o. [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(W_B = B_1 + B_2 + \cdots + B_6 + 15 \sim N(435,\ \sigma^2 = 6v^2)\) | B1, B1 | Mean; Expression for variance |
| \(P(\text{this} < 450) = P\!\left(Z < \frac{450 - 435}{v\sqrt{6}}\right) = 0.8463\) | M1 | Formulation of the problem |
| \(\therefore \frac{450 - 435}{v\sqrt{6}} = \Phi^{-1}(0.8463) = 1.021\) | B1 | Inverse Normal |
| \(\therefore v = \frac{15}{1.021 \times \sqrt{6}} = 5.9977 = 6\ \text{grams (nearest gram)}\) | A1 | Convincingly shown, beware A.G. [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(W_A = A_1 + \cdots + A_5 + 25 \sim N(425,\ \sigma^2 = 11^2 \times 5 = 605)\) | ||
| \(D = W_A - W_B \sim N(-10,\ 605 + 216 = 821)\) | B1 | Mean. Accept "\(B - A\)" |
| Variance \(= 821\) | M1, A1 | Accept sd \((= 28.65)\) |
| Want \(P(W_A > W_B) = P(W_A - W_B > 0)\) | M1 | |
| \(= P\!\left(Z > \frac{0-(-10)}{\sqrt{821}} = 0.3490\right) = 1 - 0.6365 = 0.3635\) | A1 | c.a.o. [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \frac{3126.0}{60} = 52.1\), \(s = \sqrt{\frac{164223.96 - 60 \times 52.1^2}{59}} = 4.8\) | B1 | Both correct |
| CI: \(52.1 \pm 1.96 \times \frac{4.8}{\sqrt{60}}\) | M1, B1, M1 | |
| \(= 52.1 \pm 1.2146 = (50.885(4),\ 53.314(6))\) | A1 | c.a.o. Must be expressed as an interval [5] |
# Question 4:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(A < 90) = P\!\left(Z < \frac{90-80}{11} = 0.9091\right)$ | M1, A1 | For standardising. Award once, here or elsewhere |
| $= 0.8182$ | A1 | c.a.o. **[3]** |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $W_B = B_1 + B_2 + \cdots + B_6 + 15 \sim N(435,\ \sigma^2 = 6v^2)$ | B1, B1 | Mean; Expression for variance |
| $P(\text{this} < 450) = P\!\left(Z < \frac{450 - 435}{v\sqrt{6}}\right) = 0.8463$ | M1 | Formulation of the problem |
| $\therefore \frac{450 - 435}{v\sqrt{6}} = \Phi^{-1}(0.8463) = 1.021$ | B1 | Inverse Normal |
| $\therefore v = \frac{15}{1.021 \times \sqrt{6}} = 5.9977 = 6\ \text{grams (nearest gram)}$ | A1 | Convincingly shown, beware A.G. **[5]** |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $W_A = A_1 + \cdots + A_5 + 25 \sim N(425,\ \sigma^2 = 11^2 \times 5 = 605)$ | | |
| $D = W_A - W_B \sim N(-10,\ 605 + 216 = 821)$ | B1 | Mean. Accept "$B - A$" |
| Variance $= 821$ | M1, A1 | Accept sd $(= 28.65)$ |
| Want $P(W_A > W_B) = P(W_A - W_B > 0)$ | M1 | |
| $= P\!\left(Z > \frac{0-(-10)}{\sqrt{821}} = 0.3490\right) = 1 - 0.6365 = 0.3635$ | A1 | c.a.o. **[5]** |
## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \frac{3126.0}{60} = 52.1$, $s = \sqrt{\frac{164223.96 - 60 \times 52.1^2}{59}} = 4.8$ | B1 | Both correct |
| CI: $52.1 \pm 1.96 \times \frac{4.8}{\sqrt{60}}$ | M1, B1, M1 | |
| $= 52.1 \pm 1.2146 = (50.885(4),\ 53.314(6))$ | A1 | c.a.o. Must be expressed as an interval **[5]** |4 The weights of a particular variety (A) of tomato are known to be Normally distributed with mean 80 grams and standard deviation 11 grams.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that a randomly chosen tomato of variety A weighs less than 90 grams.
The weights of another variety (B) of tomato are known to be Normally distributed with mean 70 grams. These tomatoes are packed in sixes using packaging that weighs 15 grams.
\item The probability that a randomly chosen pack of 6 tomatoes of variety B , including packaging, weighs less than 450 grams is 0.8463 . Show that the standard deviation of the weight of single tomatoes of variety B is 6 grams, to the nearest gram.
\item Tomatoes of variety A are packed in fives using packaging that weighs 25 grams. Find the probability that the total weight of a randomly chosen pack of variety A is greater than the total weight of a randomly chosen pack of variety B .
\item A new variety (C) of tomato is introduced. The weights, $c$ grams, of a random sample of 60 of these tomatoes are measured giving the following results.
$$\Sigma c = 3126.0 \quad \Sigma c ^ { 2 } = 164223.96$$
Find a $95 \%$ confidence interval for the true mean weight of these tomatoes.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2010 Q4 [18]}}