| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Single sum threshold probability |
| Difficulty | Challenging +1.2 This question requires understanding of linear combinations of normal distributions and calculating probabilities for sums of independent normal variables. While it involves multiple steps (finding mean and variance of combined loads, standardizing, using tables), the techniques are standard for S3. The second part adds mild complexity with conditional probability (unknown gender), but the approach is methodical rather than requiring novel insight. Slightly above average difficulty due to the multi-variable setup and careful arithmetic required. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Use \(\sum F + \sum M \sim N(\mu, \sigma^2)\) | M1 | Sum of indep normal variables is normal |
| \(\mu = 1104.9\) | A1 | |
| \(\sigma^2 = 6\times9.3^2 + 9\times8.5^2\) | M1 | |
| \(= 1169.2\) | A1 | |
| \(P(>1150) = 1 - \Phi([1150 - 1104.9]/\sqrt{1169.2})\) | M1 | Standardise, correct tail. M0 \(\sigma/\sqrt{15}\) |
| \(= 0.0937\) | A1 6 | Accept \(.094\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| If unknown M, prob \(\frac{1}{2}\), 6F and 9M as before. If unknown W, prob \(\frac{1}{2}\), 7W and 8M | M1 | Considering two cases |
| Having \(N(1093.3, 1183.4)\) | B1 B1 | Mean and variance |
| A1 | ||
| \(P(>1150) = 1 - \Phi(1.648) = 0.0497\) | M1 | Use of \(\frac{1}{2}\) |
| \(P = \frac{1}{2}\times0.0936 + \frac{1}{2}\times0.0497\) | A1 | ART \(0.072\) |
| \(= 0.07165\) | 6 |
# Question 7:
## Part (i):
| Working | Marks | Guidance |
|---------|-------|----------|
| Use $\sum F + \sum M \sim N(\mu, \sigma^2)$ | M1 | Sum of indep normal variables is normal |
| $\mu = 1104.9$ | A1 | |
| $\sigma^2 = 6\times9.3^2 + 9\times8.5^2$ | M1 | |
| $= 1169.2$ | A1 | |
| $P(>1150) = 1 - \Phi([1150 - 1104.9]/\sqrt{1169.2})$ | M1 | Standardise, correct tail. M0 $\sigma/\sqrt{15}$ |
| $= 0.0937$ | A1 **6** | Accept $.094$ |
## Part (ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| If unknown M, prob $\frac{1}{2}$, 6F and 9M as before. If unknown W, prob $\frac{1}{2}$, 7W and 8M | M1 | Considering two cases |
| Having $N(1093.3, 1183.4)$ | B1 B1 | Mean and variance |
| | A1 | |
| $P(>1150) = 1 - \Phi(1.648) = 0.0497$ | M1 | Use of $\frac{1}{2}$ |
| $P = \frac{1}{2}\times0.0936 + \frac{1}{2}\times0.0497$ | A1 | ART $0.072$ |
| $= 0.07165$ | **6** | |
---7 The employees of a certain company have masses which are normally distributed. Female employees have a mean of 66.7 kg and standard deviation 9.3 kg , and male employees have a mean of 78.3 kg and standard deviation 8.5 kg . It may be assumed that all employees' masses are independent. On the ground floor 6 women and 9 men enter the empty staff lift for which it is stated that the maximum load is 1150 kg .\\
(i) Calculate the probability that the maximum load is exceeded.
At the first floor all 15 passengers leave and 6 women, 8 men and an unknown employee enter.\\
(ii) Assuming that the unknown employee is equally likely to be a woman or a man, calculate the probability that the maximum load is now exceeded.
\hfill \mbox{\textit{OCR S3 2010 Q7 [12]}}