JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year STATISTICAL METHODS FOR DATA SCIENCE Previous Question Paper (March - 2021 - R18)
Answer any five questions. All questions carry equal marks.
Main
In a factory employing 3,000 persons, 5 percent earn less than Rs. 3 per hour, 580 earn from Rs. 3.01 to Rs. 4.50 per hour, 30 percent earn from Rs. 4.51 to Rs. 6.00 per hour, 500 earn from Rs. 6.01 to Rs. 7.50 per hour, 20 percent earn from Rs. 7.51 to Rs. 9.00 per hour, and the rest earn Rs. 9.01 or more per hour. What is the median wage?
Answer
To calculate the median wage, we first organize the data into a frequency distribution table. The total number of employees is 3,000.
| Wage Interval (Rs.) | Frequency (f) | Cumulative Frequency (cf) |
|---|---|---|
| 0.00 - 3.00 | 150 (5% of 3000) | 150 |
| 3.01 - 4.50 | 580 | 730 |
| 4.51 - 6.00 | 900 (30% of 3000) | 1630 |
| 6.01 - 7.50 | 500 | 2130 |
| 7.51 - 9.00 | 600 (20% of 3000) | 2730 |
| 9.01 and above | 270 (Remainder) | 3000 |
The median position is N/2 = 3000/2 = 1,500. Looking at the cumulative frequency column, the value 1,500 falls within the 4.51 - 6.00 interval. Therefore, the median class is 4.51 - 6.00.
Using the median formula:
Where:
- L (lower boundary of median class) = 4.505
- N/2 = 1500
- cf (cumulative frequency of class preceding median class) = 730
- f (frequency of median class) = 900
- h (width of median class) = 1.50
Substituting the values:
The median wage is approximately Rs. 5.79 per hour.
State and prove Baye’s theorem.
Answer
Baye's theorem states that for a set of mutually exclusive and exhaustive events A1, A2, ..., An associated with a sample space S, and an arbitrary event B occurring with these events such that P(B) > 0, the posterior probability of an event Ai given B is given by:
Proof: By the definition of conditional probability, P(Ai|B) = P(Ai ∩ B) / P(B). Similarly, P(B|Ai) = P(Ai ∩ B) / P(Ai), which implies P(Ai ∩ B) = P(Ai) * P(B|Ai). Substituting this into the first equation, we get the numerator. Since the events Ai are mutually exclusive and exhaustive, the total probability of B is given by the Law of Total Probability: P(B) = Σj=1n P(Aj ∩ B) = Σj=1n P(Aj) * P(B|Aj). Substituting this into the denominator completes the proof.
Find the mode of the following frequency distribution:
| Size(X) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| Frequency(f) | 3 | 8 | 15 | 23 | 35 | 40 | 32 | 28 | 20 | 45 | 14 | 6 |
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The marks obtained by 10 students in Mathematics and Statistics are given below. Find the Coefficient of correlation between the two subjects.
| Marks in Maths | 75 | 30 | 60 | 80 | 53 | 35 | 15 | 40 | 38 | 48 |
| Marks in Statistics | 85 | 45 | 54 | 91 | 38 | 63 | 35 | 43 | 45 | 44 |
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The joint p.d.f of the random variables X and Y is given by f(x,y) = K e-(x+y), x>0, y>0. Find:
- K
- P(X > 2)
- P(X + Y < 3)
- prove that X and Y are independent.
Answer
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Discuss various commonly used sampling distributions.
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Define exponential distribution. Obtain its moment generating function. Hence obtain its mean and variance.
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The nicotine content in milligrams of the samples of tobacco were found as follows:
| Sample 1 | 24 | 27 | 26 | 21 | 25 |
| Sample 2 | 30 | 28 | 31 | 22 | 36 |
Can it be said that the two samples come from normal population with the same mean?
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1000 babies were born during a certain week in a city of which 600 were boys and 400 girls. Use χ2 test to examine the correctness of the hypothesis that the sex-ratio is 1:1 in newly born babies.
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The percentage of smokers in a certain city was 90. A random sample of 100 persons was selected in which 85 persons were found to be smokers. Is the sample proportion significantly different from the proportion of smokers in the city? Answer on the basis of Chi-square test.
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Calculate the coefficient of rank correlation:
| x | 68 | 64 | 75 | 50 | 64 | 80 | 75 | 40 | 55 | 64 |
| y | 62 | 58 | 68 | 45 | 81 | 60 | 68 | 48 | 50 | 70 |
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