| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Comparison involving sums or multiples |
| Difficulty | Standard +0.3 This is a standard S3 question testing routine applications of normal distribution properties (linear combinations, sums of normals) and confidence intervals. Parts (i)-(iii) are straightforward z-score calculations, part (iv) requires recognizing that |2B - A| < 3 needs P(-3 < 2B - A < 3) which is slightly more involved but still mechanical, and part (v) is a textbook confidence interval application. All techniques are standard with no novel insight required, making it slightly easier than average overall. |
| Spec | 2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A < 103) = P\left(Z < \frac{103-100}{1.9} = -1.5789\right) = 0.9429\) | M1, A1, A1 | For standardising. Award once, here or elsewhere. c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{this} > 306) = P\left(Z > \frac{306-300}{3.291} = -1.823\right) = 1 - 0.9658 = 0.0342\) | B1, B1, A1, A1 | Mean. Variance. Accept sd (= 3.291). c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{this} > 147) = P\left(Z > \frac{147-150}{2.302} = -1.303\right) = 0.9037\) | B1, B1, A1 | Mean. Variance. Accept sd (= 2.302). c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(-3 < \text{this} < 3) = P\left(\frac{-3-0}{2.644} < Z < \frac{3-0}{2.644}\right) = P(-1.135 < Z < -1.135) = 2 \times 0.8718 - 1 = 0.7436\) | B1, B1, M1, A1 | Mean. Or \(A - (B_1 + B_2)\). Variance. Accept sd (= 2.644). Formulation of requirement ... two sided. c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| The batch appears not to be as specified since 300 is outside the confidence interval. | M1, B1, A1, E1 | Correct use of 302.3 and \(3.7/\sqrt{100}\). For 1.96. c.a.o. Must be expressed as an interval. |
## (i)
$P(A < 103) = P\left(Z < \frac{103-100}{1.9} = -1.5789\right) = 0.9429$ | M1, A1, A1 | For standardising. Award once, here or elsewhere. c.a.o.
## (ii)
$A_1 + A_2 + A_3 \sim N(300, \sigma^2 = 1.9^2 + 1.9^2 + 1.9^2 = 10.83)$
$P(\text{this} > 306) = P\left(Z > \frac{306-300}{3.291} = -1.823\right) = 1 - 0.9658 = 0.0342$ | B1, B1, A1, A1 | Mean. Variance. Accept sd (= 3.291). c.a.o.
## (iii)
$A + B \sim N(150, \sigma^2 = 1.9^2 + 1.3^2 = 5.3)$
$P(\text{this} > 147) = P\left(Z > \frac{147-150}{2.302} = -1.303\right) = 0.9037$ | B1, B1, A1 | Mean. Variance. Accept sd (= 2.302). c.a.o.
## (iv)
$B_1 + B_2 - A \sim N(0, 1.3^2 + 1.3^2 + 1.9^2 = 6.99)$
$P(-3 < \text{this} < 3) = P\left(\frac{-3-0}{2.644} < Z < \frac{3-0}{2.644}\right) = P(-1.135 < Z < -1.135) = 2 \times 0.8718 - 1 = 0.7436$ | B1, B1, M1, A1 | Mean. Or $A - (B_1 + B_2)$. Variance. Accept sd (= 2.644). Formulation of requirement ... two sided. c.a.o.
## (v)
Given $\bar{x} = 302.3$, $s_{n-1} = 3.7$
CI is given by $302.3 \pm 1.96 \times \frac{3.7}{\sqrt{100}} = 302.3 \pm 0.7252 = (301.57(48), 303.02(52))$
The batch appears not to be as specified since 300 is outside the confidence interval. | M1, B1, A1, E1 | Correct use of 302.3 and $3.7/\sqrt{100}$. For 1.96. c.a.o. Must be expressed as an interval.
---An electronics company purchases two types of resistor from a manufacturer. The resistances of the resistors (in ohms) are known to be Normally distributed. Type A have a mean of 100 ohms and standard deviation of 1.9 ohms. Type B have a mean of 50 ohms and standard deviation of 1.3 ohms.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the resistance of a randomly chosen resistor of type A is less than 103 ohms. [3]
\item Three resistors of type A are chosen at random. Find the probability that their total resistance is more than 306 ohms. [3]
\item One resistor of type A and one resistor of type B are chosen at random. Find the probability that their total resistance is more than 147 ohms. [3]
\item Find the probability that the total resistance of two randomly chosen type B resistors is within 3 ohms of one randomly chosen type A resistor. [5]
\item The manufacturer now offers type C resistors which are specified as having a mean resistance of 300 ohms. The resistances of a random sample of 100 resistors from the first batch supplied have sample mean 302.3 ohms and sample standard deviation 3.7 ohms. Find a 95\% confidence interval for the true mean resistance of the resistors in the batch. Hence explain whether the batch appears to be as specified. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2008 Q2 [18]}}