Jawaharlal Nehru Technological University Hyderabad

Jawaharlal Nehru Technological University Hyderabad B.Tech I MATHEMATICS - III Previous Question Paper (July/August - 2021 - R16)

B.Tech | I | I-II | R16 | July/August - 2021

Answer PaperPaper Code: 132ACMaximum Marks: 75Time: 3 Hours

Answer any five questions. All questions carry equal marks.

Main

115 marks
a

If the pdf of a continuous random variable is given by f(x) = e-x, 0 ≤ x < ∞, find the mean and variance of the random variable.

7 marks

Answer

For a continuous random variable with probability density function f(x) = e-x for x ≥ 0, the mean E(X) is calculated as:

E(X) = ∫0∞ x e-x dx

Using integration by parts, E(X) = [-x e-x]0∞ + ∫0∞ e-x dx = 0 + [-e-x]0∞ = 1. Thus, the mean is 1.

To find the variance, we first calculate E(X2):

E(X2) = ∫0∞ x2 e-x dx

Using the gamma function property ∫0∞ xn e-ax dx = n! / an+1, we get E(X2) = 2! / 13 = 2. The variance V(X) is given by:

V(X) = E(X2) - [E(X)]2 = 2 - (1)2 = 1
b

Compute the first four moments about the mean from the following data:

x12345
f23541
8 marks

Answer

First, calculate the mean (μ):

μ = Σ fixi / Σ fi = (2(1) + 3(2) + 5(3) + 4(4) + 1(5)) / (2+3+5+4+1) = 45 / 15 = 3

The moments about the mean μr are defined as Σ fi(xi - μ)r / N, where N = 15:

  • μ1 = Σ fi(xi - 3) / 15 = (2(-2) + 3(-1) + 5(0) + 4(1) + 1(2)) / 15 = 0
  • μ2 = Σ fi(xi - 3)2 / 15 = (2(4) + 3(1) + 5(0) + 4(1) + 1(4)) / 15 = 20 / 15 = 1.333
  • μ3 = Σ fi(xi - 3)3 / 15 = (2(-8) + 3(-1) + 5(0) + 4(1) + 1(8)) / 15 = -5 / 15 = -0.333
  • μ4 = Σ fi(xi - 3)4 / 15 = (2(16) + 3(1) + 5(0) + 4(1) + 1(16)) / 15 = 55 / 15 = 3.667
215 marks
a

A continuous random variable X is normally distributed with mean 25 and standard deviation 8. Find the probability that i) 20 ≤ X ≤ 40 and ii) X ≥ 35.

7 marks

Answer

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b

A random sample of size 100 is taken from an infinite population having the mean 76 and variance 256. Find the probability that mean of sample will be between 75 and 78.

8 marks

Answer

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315 marks
a

Determine a 95% confidence interval for the mean of a normal distribution with variance 0.25 using a sample of 100 values with mean 212.3.

7 marks

Answer

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b

A sample of 64 students have a mean weight 70 kgs. Can this be regarded as a sample from a population with mean weight 56 kgs and standard deviation 25 kgs?

8 marks

Answer

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4

The following are the values of skills of 2 samples with individuals 5 and 6. Test the difference between the means.

Sample 174.177.774.474.073.8
Sample 270.874.974.270.469.272.2
15 marks

Answer

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5

Random samples of 500 men and 500 women were asked whether they would like to have a fly over near their residence. 150 men and 200 women are in favor of it. Test the difference between the proportions.

15 marks

Answer

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615 marks
a

Perform three iterations of the Gauss-Seidel method to solve the system of equations:

4x + 2y + z = 1; x + 5y + z = 5; 2x + y + 4z = 3
7 marks

Answer

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b

The equation x ln(x) - 3 = 0 has a root in (4,5). Find the root correct to three decimal places by Newton-Raphson method.

8 marks

Answer

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715 marks
a

Evaluate the integral of x dx from 1.0 to 1.6 using Simpson’s 1/3rd rule with h = 0.1.

7 marks

Answer

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b

Find the approximate value of y(x) at x = 1.2, 1.4, 1.6 for the initial value problem y' = x + y, y(1) = 1, h = 0.2 by Euler’s method.

8 marks

Answer

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8

Apply Runge-Kutta method of order 4 to find the approximate value of y(1.3) for y' = x2 + y2, y(1) = 2 with h = 0.1.

15 marks

Answer

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About this paper

This page contains the question paper of MATHEMATICS - III from Jawaharlal Nehru Technological University Hyderabad for B.Tech PTM I I-II, July/August - 2021, R16.

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