Question 5 (b)
Normal distribution of water-heater lifetimes
9 marks
- Time to write
- About 18 minutes
- Topic
- Normal distribution
Question
Question 5 (b) · 9 marksA certain kind of water heater has a mean lifetime of 500 hours with a standard deviation of 40 hours. Assuming that the distribution of lifetime is normal, find:
the percentage of water heaters having a lifetime of more than 570 hours,
the percentage of water heaters having a lifetime between 480 and 520,
the minimum life of the best 10% of the water heaters.
[Given: Table Values of area for Z = 1.75 is 0.4599, Z= 0.50 is 0.1915, Z= 1.29 is 0.4015]
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
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Read off the question and its marks
Read from the question
- Assuming that the distribution of lifetime…The paper fixes this: use it as given, and say so in your answer.
- findFormula first, then the working, then the answer with its unit (₹, %, units).
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- Read the data; note exactly what is asked 1 min
- 1 Standardisation and use of the table 1½ marks 3 min
- 2 (i) percentage above 570 hours 2½ marks 5 min
- 3 (ii) percentage between 480 and 520 hours 2½ marks 4 min
- 4 (iii) minimum life of the best 10% 2½ marks 4 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
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Solution
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Standardising. Lifetime X is normal with μ = 500 hours and σ = 40 hours. Each value is converted to the standard normal variate z = (X − μ) / σ, and probabilities are areas under the standard normal curve; the curve is symmetrical about the mean, so the area on each side of the mean is 0.5. The given table values are areas between z = 0 and z.
(i) Lifetime of more than 570 hours
z = (570 − 500) / 40 = 1.75; area from 0 to 1.75 = 0.4599. P(X > 570) = 0.5 − 0.4599 = 0.0401, that is 4.01% of the heaters.
(ii) Lifetime between 480 and 520 hours
z₁ = (480 − 500) / 40 = −0.5 and z₂ = (520 − 500) / 40 = 0.5, one on each side of the mean:
| Part of the range | Area |
|---|---|
| z = −0.5 to 0 | 0.1915 |
| z = 0 to 0.5 | 0.1915 |
| z = −0.5 to 0.5 | 0.383 |
So 38.30% of the heaters last between 480 and 520 hours.
(iii) Minimum life of the best 10%
The best 10% are the longest-lasting, the top tail with area 0.10. The area between the mean and the cut-off is then 0.5 − 0.10 = 0.40, which the table gives (0.4015) at z = 1.29.
Minimum life = μ + zσ = 500 + 1.29 × 40 = 551.6 hours. A heater must last at least about 552 hours to be in the best 10%.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Standardisation and use of the table | 1.5 |
| 2 | (i) percentage above 570 hours | 2.5 |
| 3 | (ii) percentage between 480 and 520 hours | 2.5 |
| 4 | (iii) minimum life of the best 10% | 2.5 |
| Total | 9 |
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