JNTU
JNTU B.Tech II COMPUTATIONAL STATISTICS Previous Question Paper (August/September -2022 - R18)
Answer any five questions. All questions carry equal marks.
Main
Use Fermat’s Factorization method to factorize 119143.
Answer
Fermat's factorization method relies on representing an odd integer n as a difference of two squares: n = x2 - y2 = (x - y)(x + y). We start by calculating the ceiling of the square root of 119143.
We test values of x starting from 346 to find a value such that x2 - n = y2 is a perfect square:
- x = 346: 3462 - 119143 = 119716 - 119143 = 573 (not a square)
- x = 347: 3472 - 119143 = 120409 - 119143 = 1266 (not a square)
- x = 348: 3482 - 119143 = 121104 - 119143 = 1961 (not a square)
- x = 349: 3492 - 119143 = 121801 - 119143 = 2658 (not a square)
- x = 350: 3502 - 119143 = 122500 - 119143 = 3357 (not a square)
- x = 351: 3512 - 119143 = 123201 - 119143 = 4058 (not a square)
- x = 352: 3522 - 119143 = 123904 - 119143 = 4761 = 692
Since y2 = 692, we have x = 352 and y = 69. The factors are (x - y) and (x + y):
Thus, 119143 = 283 × 421.
Solve the System of congruences
Answer
The system is X ≡ 3 (mod 10), X ≡ 8 (mod 15), X ≡ 5 (mod 84). First, check for consistency. The moduli 10, 15, 84 are not pairwise coprime (gcd(10, 15)=5, gcd(10, 84)=2, gcd(15, 84)=3). We decompose the system into prime power moduli:
- X ≡ 3 (mod 10) → X ≡ 1 (mod 2), X ≡ 3 (mod 5)
- X ≡ 8 (mod 15) → X ≡ 2 (mod 3), X ≡ 3 (mod 5)
- X ≡ 5 (mod 84) → X ≡ 1 (mod 4), X ≡ 2 (mod 3), X ≡ 5 (mod 7)
Collecting the conditions for each prime power:
- Mod 4: X ≡ 1 (mod 4) (implies X ≡ 1 mod 2)
- Mod 3: X ≡ 2 (mod 3)
- Mod 5: X ≡ 3 (mod 5)
- Mod 7: X ≡ 5 (mod 7)
Using the Chinese Remainder Theorem on the coprime moduli 4, 3, 5, 7 (product M = 420):
X = (1×105×1 + 2×140×2 + 3×84×4 + 5×60×2) mod 420
X = (105 + 560 + 1008 + 600) mod 420 = 2273 mod 420 = 173.
State and prove the fundamental theorem of arithmetic.
Answer
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Find if there is any significant correlation between the heights and weights given below.
| Height in inches | 57 | 59 | 62 | 63 | 64 | 65 | 55 | 58 | 57 |
|---|---|---|---|---|---|---|---|---|---|
| Weight in lbs | 113 | 117 | 126 | 126 | 130 | 129 | 111 | 116 | 112 |
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A random variable X has the following probability function.
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| P(x) | K | 3k | 5k | 7k | 9k | 11k | 13k |
(i) Determine k (ii) Evaluate P(X<4), P(X≥5), P(3 0.3? (iv) Mean (v) Variance.
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Out of 800 families with 5 children each, how many would you expect to have (i) 3 boys (ii) 5 girls (iii) either 2 or 3 boys (iv) at least one boy. Assume equal probabilities for boys and girls.
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Fit a parabola of the form
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Y | 23 | 5.2 | 9.7 | 16.5 | 29.4 | 35.5 | 54.4 |
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In a normal distribution 31% of the items are under 45 and 8% are over 64. Find the Mean and Variance of the distribution.
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A random sample size 100 is taken from population with standard deviation 5.1 and the sample mean is 21.6. Construct a 95% confidence interval for the population mean.
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A population consisting of five numbers 5, 10, 14, 18, 13, 24. Consider all the possible samples of size two which can be drawn with replacement from this population, find:
- a) The mean of population
- b) The standard deviation of the population
- c) The mean of sampling distribution means
- d) The standard deviation of the sampling distribution of the means.
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The mean height of 50 male students who participated in sports is 68.2 inches with a standard deviation of 2.5 inches. The mean height of 50 male students who have not participated in sports is 67.2 inches with a standard deviation of 2.8 inches. Test the hypothesis that the height of students who participated in sports is more than the students who have not participated in sports.
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Define Type-I and Type-II Errors, Critical region, Level of Significance, Null Hypothesis, Alternative Hypothesis.
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A fair die is tossed repeatedly. If
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The Transition probability matrix of Markov Chain {
| 0.1 | 0.5 | 0.4 |
| 0.6 | 0.2 | 0.2 |
| 0.3 | 0.4 | 0.3 |
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