Question 5 (b)
Normal distribution of hospital bills: numbers and probabilities
9 marks
- Time to write
- About 18 minutes
- Topic
- Normal distribution
Question
Question 5 (b) · 9 marksSanjeevani care, a hospital specialized in heart surgery, admitted 5,000 patients in the year 2023-24. The average payment made by a patient was ₹3,00,000, with a variance of ₹40,00,00,000. Assuming that the distribution of payments is normal, find and show under the normal curve:
The number of patients who paid between ₹2,70,000 and ₹3,50,000.
The probability that the bill of a patient exceeds ₹3,10,000.
The number of patients who paid more than ₹230,000 but less than ₹2,70,000
(Given: Table values of area: Z=1.5 gives 0.4332, Z=2.5 gives 0.4938, Z=0.5 gives 0.1915, Z=3.5 gives 0.4998)
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.
Before you write
Read off the question and its marks
Read from the question
- Assuming that the distribution of payments…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).
What the paper leaves open
As printed, the lower limit in item (iii) is written without lakh grouping, unlike the other amounts.
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Σ and the z values 2 marks 4 min
- 2 (i) area and number of patients, with sketch 2½ marks 4 min
- 3 (ii) probability, with sketch 2 marks 4 min
- 4 (iii) area and number of patients, with sketch 2½ marks 4 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Try it first. About 18 minutes for 9 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Standardising. Mean μ = ₹3,00,000; variance ₹40,00,00,000, so σ = √40,00,00,000 = ₹20,000. A normal variable X is converted to the standard normal variate z = (X − μ) / σ, and probabilities are read as areas under the standard normal curve, the curve being symmetrical about the mean. The table values given are areas between z = 0 and the stated z.
| Bill (₹) | z = (X − 3,00,000) / 20,000 | Area from 0 to z |
|---|---|---|
| 2,30,000 | −3.5 | 0.4998 |
| 2,70,000 | −1.5 | 0.4332 |
| 3,10,000 | 0.5 | 0.1915 |
| 3,50,000 | 2.5 | 0.4938 |
(i) Patients paying between ₹2,70,000 and ₹3,50,000
The two values lie on opposite sides of the mean, so the areas are added:
| Part of the range | Area |
|---|---|
| z = −1.5 to 0 | 0.4332 |
| z = 0 to 2.5 | 0.4938 |
| z = −1.5 to 2.5 | 0.927 |
Number of patients = 5,000 × 0.9270 = 4,635.
Under the normal curve: the shaded area runs from z = −1.5 on the left of the mean to z = 2.5 on the right.
(ii) Probability that a bill exceeds ₹3,10,000
P(X > 3,10,000) = P(z > 0.5) = 0.5 − 0.1915 = 0.3085 (about 1,543 of the 5,000 patients).
Under the normal curve: the shaded area is the right-hand tail beyond z = 0.5.
(iii) Patients paying between ₹2,30,000 and ₹2,70,000
Both values lie below the mean, so the smaller area is subtracted from the larger: P(−3.5 < z < −1.5) = 0.4998 − 0.4332 = 0.0666.
Number of patients = 5,000 × 0.0666 = 333.
Under the normal curve: the shaded strip lies in the left tail, between z = −3.5 and z = −1.5.
Where the 9 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | σ and the z values | 2 |
| 2 | (i) area and number of patients, with sketch | 2.5 |
| 3 | (ii) probability, with sketch | 2 |
| 4 | (iii) area and number of patients, with sketch | 2.5 |
| Total | 9 |
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