| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Normal distribution probability then binomial/normal approximation on sample |
| Difficulty | Standard +0.3 This is a straightforward normal distribution question with standard techniques: (i) uses inverse normal with given percentage, (ii) applies direct normal probability, (iii) requires normal approximation to binomial with continuity correction. All are routine S1 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| For \(\pm 1.282\) seen, or 1.28, 1.281, not 1.29 or 1.30 | B1 | |
| For standardising, with or without sq rt, squared, no cc | M1 | |
| For correct answer | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| For standardising, criteria as above, can include cc | M1 | |
| For correct area found using tables, ie < 0.5 ft on wrong (i) | M1 | |
| For correct answer, accept 0.0274 | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| For 60 and 54 seen (could be sd or variance) | B1 | |
| For using 64.5 or 65.5 in a standardising process | M1 | |
| For standardising, must have \(\sqrt{\text{their 54}}\) in denom | M1 | |
| For correct answer | A1 | 4 marks |
**(i)** $1.282 = (5130-\mu) / 40.6$; $\mu = 5080$ (5078) rounding to 5080
| For $\pm 1.282$ seen, or 1.28, 1.281, not 1.29 or 1.30 | B1
| For standardising, with or without sq rt, squared, no cc | M1
| For correct answer | A1 | 3 marks
**(ii)** $P(<5000) = \Phi[(5000-5078) / 40.6] = \Phi(-1.921) = 1 - 0.9727 = 0.0273$ or 2.73%
| For standardising, criteria as above, can include cc | M1
| For correct area found using tables, ie < 0.5 ft on wrong (i) | M1
| For correct answer, accept 0.0274 | A1 | 3 marks
**(iii)** $\mu = 60$, var $= 54$; $P(\text{fewer than 65}) = \Phi[(64.5 - 60) / \sqrt{54}] = \Phi(0.6123) = 0.730$ accept 0.73
| For 60 and 54 seen (could be sd or variance) | B1
| For using 64.5 or 65.5 in a standardising process | M1
| For standardising, must have $\sqrt{\text{their 54}}$ in denom | M1
| For correct answer | A1 | 4 marks7 In tests on a new type of light bulb it was found that the time they lasted followed a normal distribution with standard deviation 40.6 hours. 10\% lasted longer than 5130 hours.\\
(i) Find the mean lifetime, giving your answer to the nearest hour.\\
(ii) Find the probability that a light bulb fails to last for 5000 hours.\\
(iii) A hospital buys 600 of these light bulbs. Using a suitable approximation, find the probability that fewer than 65 light bulbs will last longer than 5130 hours.
\hfill \mbox{\textit{CAIE S1 2005 Q7 [10]}}