| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | March |
| Marks | 10 |
| 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 straightforward application of linear combinations of normal distributions. Part (i) requires finding P(8S + 3F > 900) using standard formulas for mean and variance of linear combinations, then a normal probability calculation. Part (ii) requires finding P(S < F) which is equivalent to P(S - F < 0), another routine application. The assumption of independence is standard. While it involves multiple steps, all techniques are direct applications of core S2 content with no novel insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Assume cartons are random sample(s) | B1 | or masses of cartons are independent of each other oe |
| \(E(T) = 816.4\), \(\text{Var}(T) = 1570.08\) | B1 | Both |
| \(z = \frac{900 - \text{"816.4"}}{\sqrt{\text{"1570.08"}}}\) \((= 2.110)\) | M1 | |
| \(1 - \Phi(\text{"2.110"})\) | M1 | |
| \(= 0.0174 = 1.74\%\) (3 sfs) | A1 | % only (accept 1.7% if 0.0174 seen) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(F - S > 0)\) stated or implied | M1 | \(P(S - F < 0)\) |
| \(62.0 - 78.8\ (= -16.8)\) & \(10.0^2 + 12.6^2\ (= 258.76)\) | B1 | \(78.8 - 62.0\ (= 16.8)\) & \(12.6^2 + 10.0^2\ (= 258.76)\) |
| \(z = \frac{0 - (\text{"-16.8"})}{\sqrt{\text{"258.76"}}}\ (= 1.044)\) | M1 | \(z = \frac{0 - \text{"16.8"}}{\sqrt{\text{"258.76"}}}\ (= -1.044)\) |
| \(1 - \Phi(\text{"1.044"})\) | M1 | \(\Phi(\text{"-1.044"}) = 1 - \Phi(\text{"1.044"})\) |
| \((= 1 - 0.8517) = 0.148\) (3 sfs) | A1 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Assume cartons are random sample(s) | B1 | or masses of cartons are independent of each other oe |
| $E(T) = 816.4$, $\text{Var}(T) = 1570.08$ | B1 | Both |
| $z = \frac{900 - \text{"816.4"}}{\sqrt{\text{"1570.08"}}}$ $(= 2.110)$ | M1 | |
| $1 - \Phi(\text{"2.110"})$ | M1 | |
| $= 0.0174 = 1.74\%$ (3 sfs) | A1 | % only (accept 1.7% if 0.0174 seen) |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(F - S > 0)$ stated or implied | M1 | $P(S - F < 0)$ |
| $62.0 - 78.8\ (= -16.8)$ & $10.0^2 + 12.6^2\ (= 258.76)$ | B1 | $78.8 - 62.0\ (= 16.8)$ & $12.6^2 + 10.0^2\ (= 258.76)$ |
| $z = \frac{0 - (\text{"-16.8"})}{\sqrt{\text{"258.76"}}}\ (= 1.044)$ | M1 | $z = \frac{0 - \text{"16.8"}}{\sqrt{\text{"258.76"}}}\ (= -1.044)$ |
| $1 - \Phi(\text{"1.044"})$ | M1 | $\Phi(\text{"-1.044"}) = 1 - \Phi(\text{"1.044"})$ |
| $(= 1 - 0.8517) = 0.148$ (3 sfs) | A1 | |6 The masses, in kilograms, of cartons of sugar and cartons of flour have the distributions $\mathrm { N } \left( 78.8,12.6 ^ { 2 } \right)$ and $\mathrm { N } \left( 62.0,10.0 ^ { 2 } \right)$ respectively.\\
(i) The standard load for a certain crane is 8 cartons of sugar and 3 cartons of flour. The maximum load that can be carried safely by the crane is 900 kg . Stating a necessary assumption, find the percentage of standard loads that will exceed the maximum safe load.\\
(ii) Find the probability that a randomly chosen carton of sugar has a smaller mass than a randomly chosen carton of flour.\\
\hfill \mbox{\textit{CAIE S2 2017 Q6 [10]}}