| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Multiple stage process probability |
| Difficulty | Standard +0.3 This question tests standard application of linear combinations of normal variables and the Central Limit Theorem. Part (i) requires straightforward addition of means and variances (independent variables). Part (ii) applies CLT to find the distribution of the sample mean and then a routine normal probability calculation. While it involves multiple steps, each is a direct application of well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| 34 | B1 | |
| \(2.2^2 + 1.3^2 + 2.6^2\ (= 13.29)\) | B1 [2] | Accept 13.3 or \(3.65^2\). Allow at early stage |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{33 - 34}{\sqrt{\frac{13.29}{70}}}\ (= -2.295)\) | M1 | correct standardisation method for either |
| \(\frac{35 - 34}{\sqrt{\frac{13.29}{70}}}\ (= 2.295)\) | M1 | For attempt to use tables to find the probability between two \(z\) values; may be implied by next line |
| \(\Phi(2.295) - \Phi(-2.295)\) | M1 | For a correct method to find the area between their two \(z\) values |
| \(= \Phi(2.295) - (1 - \Phi(2.295))\) oe | A1 [4] | |
| \(= 0.978\) (3 sf) |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 34 | B1 | |
| $2.2^2 + 1.3^2 + 2.6^2\ (= 13.29)$ | B1 [2] | Accept 13.3 or $3.65^2$. Allow at early stage |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{33 - 34}{\sqrt{\frac{13.29}{70}}}\ (= -2.295)$ | M1 | correct standardisation method for either |
| $\frac{35 - 34}{\sqrt{\frac{13.29}{70}}}\ (= 2.295)$ | M1 | For attempt to use tables to find the probability between two $z$ values; may be implied by next line |
| $\Phi(2.295) - \Phi(-2.295)$ | M1 | For a correct method to find the area between their two $z$ values |
| $= \Phi(2.295) - (1 - \Phi(2.295))$ oe | A1 [4] | |
| $= 0.978$ (3 sf) | | |
---3 The daily times, in minutes, that Yu Ming takes showering, getting dressed and having breakfast are independent and have the distributions $\mathrm { N } \left( 9,2.2 ^ { 2 } \right) , \mathrm { N } \left( 8,1.3 ^ { 2 } \right)$ and $\mathrm { N } \left( 17,2.6 ^ { 2 } \right)$ respectively. The total daily time that Yu Ming takes for all three activities is denoted by $T$ minutes.\\
(i) Find the mean and variance of $T$.\\
(ii) Yu Ming notes the value of $T$ on each day in a random sample of 70 days and calculates the sample mean. Find the probability that the sample mean is between 33 and 35 .
\hfill \mbox{\textit{CAIE S2 2015 Q3 [6]}}