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JNTU B.Tech II COMPUTATIONAL STATISTICS Previous Question Paper (August/September -2022 - R18)

B.Tech | II | II-I | R18 | August/September -2022

Answer PaperPaper Code: 153CEMaximum Marks: 75Time: 3 Hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

Use Fermat’s Factorization method to factorize 119143.

7 marks

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.

√119143 ≈ 345.17, so x = 346.

We test values of x starting from 346 to find a value such that x2 - n = y2 is a perfect square:

  1. x = 346: 3462 - 119143 = 119716 - 119143 = 573 (not a square)
  2. x = 347: 3472 - 119143 = 120409 - 119143 = 1266 (not a square)
  3. x = 348: 3482 - 119143 = 121104 - 119143 = 1961 (not a square)
  4. x = 349: 3492 - 119143 = 121801 - 119143 = 2658 (not a square)
  5. x = 350: 3502 - 119143 = 122500 - 119143 = 3357 (not a square)
  6. x = 351: 3512 - 119143 = 123201 - 119143 = 4058 (not a square)
  7. 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):

352 - 69 = 283
352 + 69 = 421

Thus, 119143 = 283 × 421.

b

Solve the System of congruences

X ≡ 3(mod 10), X ≡ 8(mod 15), X ≡ 5(mod 84)
.

8 marks

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:

  1. X ≡ 3 (mod 10) → X ≡ 1 (mod 2), X ≡ 3 (mod 5)
  2. X ≡ 8 (mod 15) → X ≡ 2 (mod 3), X ≡ 3 (mod 5)
  3. 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):

M1 = 105, 105y1 ≡ 1 (mod 4) → 1y1 ≡ 1 → y1 = 1
M2 = 140, 140y2 ≡ 1 (mod 3) → 2y2 ≡ 1 → y2 = 2
M3 = 84, 84y3 ≡ 1 (mod 5) → 4y3 ≡ 1 → y3 = 4
M4 = 60, 60y4 ≡ 1 (mod 7) → 4y4 ≡ 1 → y4 = 2

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.

2

State and prove the fundamental theorem of arithmetic.

15 marks

Answer

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315 marks
a

Find if there is any significant correlation between the heights and weights given below.

Height in inches575962636465555857
Weight in lbs113117126126130129111116112
7 marks

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b

A random variable X has the following probability function.

X0123456
P(x)K3k5k7k9k11k13k

(i) Determine k (ii) Evaluate P(X<4), P(X≥5), P(3 0.3? (iv) Mean (v) Variance.

8 marks

Answer

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415 marks
a

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.

7 marks

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b

Fit a parabola of the form

y = a + bx + cx2
to the following data.

X1234567
Y235.29.716.529.435.554.4
8 marks

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515 marks
a

In a normal distribution 31% of the items are under 45 and 8% are over 64. Find the Mean and Variance of the distribution.

8 marks

Answer

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b

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.

7 marks

Answer

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6

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.
15 marks

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715 marks
a

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.

8 marks

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b

Define Type-I and Type-II Errors, Critical region, Level of Significance, Null Hypothesis, Alternative Hypothesis.

7 marks

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815 marks
a

A fair die is tossed repeatedly. If

Xn
denote the maximum of the numbers occurring in the first n tosses, find the transition probability matrix P of the Markov chain {
Xn
}. Find also
P2
and
P(X2 = 6)
.

8 marks

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b

The Transition probability matrix of Markov Chain {

Xn
}; n = 1, 2, 3, . . . . having three states 1, 2 and 3 is

0.10.50.4
0.60.20.2
0.30.40.3
7 marks

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