JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year PROBABILITY AND STATISTICS Previous Question Paper (September - 2021 - R13)

B.Tech | II Year | I Semester | R13 | September - 2021

Answer PaperPaper Code: 113ANMaximum Marks: 75Time: 3 hours

Answer any five questions. All questions carry equal marks.

Main

1

A random variable X has the following probability distribution.

X: 0 1 2 3 4

P(X): 3k 3k K 2k 6k

Find: a) k, b) var(X), c) var(2X-4), d) P(X < 2) and e) P(X > 1).

15 marks

Answer

a) Since the sum of probabilities must equal 1:

3k + 3k + k + 2k + 6k = 1
15k = 1 ⇒ k = 1/15

b) First, calculate the mean E(X):

E(X) = Σ xP(x) = (0)(3/15) + (1)(3/15) + (2)(1/15) + (3)(2/15) + (4)(6/15) = (0 + 3 + 2 + 6 + 24)/15 = 35/15 = 7/3

Next, calculate E(X2):

E(X2) = Σ x2P(x) = (02)(3/15) + (12)(3/15) + (22)(1/15) + (32)(2/15) + (42)(6/15) = (0 + 3 + 4 + 18 + 96)/15 = 121/15

Variance Var(X) = E(X2) - [E(X)]2:

Var(X) = 121/15 - (7/3)2 = 121/15 - 49/9 = (363 - 245)/45 = 118/45 ≈ 2.622

c) Using the property Var(aX + b) = a2Var(X):

Var(2X - 4) = 22Var(X) = 4(118/45) = 472/45 ≈ 10.489

d) P(X < 2) = P(X=0) + P(X=1):

P(X < 2) = 3k + 3k = 6k = 6/15 = 2/5 = 0.4

e) P(X > 1) = P(X=2) + P(X=3) + P(X=4):

P(X > 1) = k + 2k + 6k = 9k = 9/15 = 3/5 = 0.6
215 marks
a

If 2% of the items of a factory are defective and the items are packed in boxes, find the probability that there will be atleast 3 defective items in a box of 100 items.

7 marks

Answer

Given that the probability of a defective item is p = 0.02 and the number of items in a box is n = 100, we use the Poisson distribution as an approximation since n is large and p is small. The parameter λ is calculated as:

λ = n × p = 100 × 0.02 = 2

We need to find the probability of having at least 3 defective items, P(X ≥ 3). Using the complement rule:

P(X ≥ 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]

Using the Poisson formula P(X=k) = (e⁻λ × λᵏ) / k!:

P(X=0) = e⁻² × 2⁰ / 0! = e⁻² ≈ 0.1353
P(X=1) = e⁻² × 2¹ / 1! = 2e⁻² ≈ 0.2707
P(X=2) = e⁻² × 2² / 2! = 2e⁻² ≈ 0.2707

Summing these probabilities: 0.1353 + 0.2707 + 0.2707 = 0.6767. Therefore:

P(X ≥ 3) = 1 - 0.6767 = 0.3233
b

In a Normal distribution, 7% of the items are under 35 and 89% are under 63. Find the mean and variance of the distribution.

8 marks

Answer

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3

For the following bivariate probability distribution of X and Y:

X\Y0123
01/322/322/323/32
12/328/321/41/32
22/323/3216/320

Find: a) P(X ≤ 1, Y = 3), b) P(Y ≤ 2), c) P(X ≤ 1, Y ≤ 2), d) FX(2) and e) FY(1).

15 marks

Answer

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4

Find the coefficient of correlation and the two lines of regression.

X: 10 12 18 24 23 27

Y: 13 18 12 25 30 10

15 marks

Answer

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5

If samples of size 2 are taken from the population 1, 2, 3, 4, 5, 6 without replacement, then find:

15 marks
a

the mean and standard deviation of population.

7 marks

Answer

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b

the mean and standard deviation of the sampling distribution of means.

8 marks

Answer

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6

Fit a Poisson distribution to the following data and test for goodness of fit at 5% level of significance.

x: 0 1 2 3 4

f: 214 92 20 3 1

15 marks

Answer

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7

Cars arrive at a petrol pump with exponential interval times having mean 1/2 minute. The attendant takes an average of 1/5 minute per car to supply petrol and the service time being exponentially distributed. Find:

15 marks
a

the average number of cars waiting to be served

5 marks

Answer

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b

the average number of cars in the system and

5 marks

Answer

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c

the proportion of time for which the pump attendant is idle.

5 marks

Answer

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8

A gambler has Rs. 2. He bets 1 at a time and wins Rs. 1 with probability 1/2. He stops playing if loses Rs. 2 or wins Rs. 4. Find:

15 marks
a

the tpm of the related Markov chain

5 marks

Answer

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b

the probability that he has lost his money at the end of 5 plays.

5 marks

Answer

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c

the probability that the game lasts more than 7 plays.

5 marks

Answer

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About this paper

This page contains the question paper of PROBABILITY AND STATISTICS from JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD for B.Tech AME II Year I Semester, September - 2021, R13.

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