IGNOU MTE-11 Solved Question Paper PDF Download

IGNOU MTE-11 Previous Year Solved Question Paper in Hindi

IGNOU MTE-11 Previous Year Solved Question Paper in English

Q1. (a) The marks (out of 100) of 25 students in an exam are given below: 6 40, 65, 57, 2, 26, 42, 75, 49, 50, 35, 52, 42, 5, 70, 63, 40, 38, 56, 60, 70, 51, 5, 44, 42, 22 (i) Prepare frequency distribution of the marks. Take the first class interval as 0—20. (ii) Construct ‘less than’ and ‘more than’ cumulative frequency distributions. (iii) Estimate the proportion of students who got: (A) less than 50 marks. (B) more than 70 marks. (b) An incomplete distribution is given below: 4 Class Frequency 0—10 10 10–20 20 20—30 ? 30—40 40 40—50 ? 50—60 25 60—70 5 Find out missing frequencies if median value is 35.

Ans. (a) Given data: 40, 65, 57, 2, 26, 42, 75, 49, 50, 35, 52, 42, 5, 70, 63, 40, 38, 56, 60, 70, 51, 5, 44, 42, 22. Total students = 25.

(i) Frequency Distribution: We create class intervals with a class width of 20, as instructed.

• Class 0-20: 2, 5, 5 (Frequency = 3)
• Class 20-40: 26, 35, 38, 22 (Frequency = 4)
• Class 40-60: 40, 42, 49, 50, 52, 42, 57, 40, 56, 51, 44, 42 (Frequency = 12)
• Class 60-80: 65, 75, 70, 63, 60, 70 (Frequency = 6)

Frequency Table:

Class Interval	Frequency (f)
0-20	3
20-40	4
40-60	12
60-80	6
Total	25

(ii) Cumulative Frequency Distributions:

• ‘Less than’ Cumulative Frequency:
  • Less than 20: 3
  • Less than 40: 3 + 4 = 7
  • Less than 60: 7 + 12 = 19
  • Less than 80: 19 + 6 = 25
• ‘More than’ Cumulative Frequency:
  • More than 0: 25
  • More than 20: 25 – 3 = 22
  • More than 40: 22 – 4 = 18
  • More than 60: 18 – 12 = 6

(iii) Proportion of students: (A) Less than 50 marks: We count the students with marks less than 50 directly from the raw data. They are: 40, 2, 26, 42, 49, 35, 42, 5, 40, 38, 5, 44, 42, 22. There are 14 such students. Proportion = 14 / 25 = 0.56.

(B) More than 70 marks: We count the students with marks strictly greater than 70. That is: 75. There is only 1 such student. Proportion = 1 / 25 = 0.04.

(b) Finding the missing frequencies. Let the missing frequency for the class 20-30 be f₁ and for 40-50 be f₂. Given: Median = 35. Since the median is 35, it lies in the median class 30-40 .

The formula for the median is: Median = L + [ (N/2 – C) / f ] × h Where: L = Lower limit of the median class = 30 f = Frequency of the median class = 40 h = Class width = 10 C = Cumulative frequency of the class preceding the median class = 10 + 20 + f₁ = 30 + f₁ N = Total frequency = 10 + 20 + f₁ + 40 + f₂ + 25 + 5 = 100 + f₁ + f₂

Plugging the values into the formula: 35 = 30 + [ ( (100 + f₁ + f₂) / 2 – (30 + f₁) ) / 40 ] × 10 5 = [ (50 + 0.5f₁ + 0.5f₂ – 30 – f₁) / 40 ] × 10 5 = [ (20 – 0.5f₁ + 0.5f₂) / 4 ] 20 = 20 – 0.5f₁ + 0.5f₂ 0 = -0.5f₁ + 0.5f₂ 0.5f₁ = 0.5f₂ f₁ = f₂

This relation shows that the two missing frequencies are equal. However, the question does not provide the total frequency (N), without which a unique value for f₁ and f₂ cannot be found. The problem is ill-posed. Let’s assume a total frequency (N) was given, for instance, N = 170 (a common setup for such problems). Then, N = 100 + f₁ + f₂ 170 = 100 + f₁ + f₂ f₁ + f₂ = 70

Since f₁ = f₂, we have: 2f₁ = 70 f₁ = 35 Therefore, f₁ = 35 and f₂ = 35 . (Note: This solution is based on the assumption that the total frequency was 170.)

Q2. (a) In a certain distribution, the first four central moments are 0, 16, –64 and 162 respectively. Calculate β₁ and β₂ coefficients. Check whether the distribution is skewed. 3 (b) A and B appear in an interview for two vacancies for the same post. The probability of selection of A is 1/5 and that of B is 1/2. Find the probability that: 4 (i) both A and B will be selected. (ii) at least one of them will be selected. (c) Let p.m.f. of a random variable X is: 3 f(x) = (3-x)/10, x = -1, 0, 1, 2 Calculate E(X) and E(X²). Also, find E[ (4X +5) ].

Ans. (a) The given central moments are: μ₁ = 0 μ₂ = 16 (Variance) μ₃ = -64 μ₄ = 162

Coefficient of Skewness (β₁): The formula for β₁ is: β₁ = μ₃² / μ₂³ β₁ = (-64)² / (16)³ = (64 × 64) / (16 × 16 × 16) β₁ = (4 × 16) × (4 × 16) / (16 × 16 × 16) = (4 × 4) / 16 = 16 / 16 = 1 β₁ = 1

Coefficient of Kurtosis (β₂): The formula for β₂ is: β₂ = μ₄ / μ₂² β₂ = 162 / (16)² = 162 / 256 Simplifying β₂ by dividing by 2: β₂ = 81 / 128

Checking for Skewness: Skewness is checked by the value of μ₃ or β₁. Since μ₃ = -64 , which is not zero, the distribution is skewed . Because μ₃ is negative, the distribution is negatively skewed . Additionally, the coefficient of skewness γ₁ = √β₁ = √1 = 1. Since γ₁ ≠ 0, the distribution is skewed.

(b) Given: Probability of A’s selection, P(A) = 1/5 Probability of B’s selection, P(B) = 1/2 We assume the selections of A and B are independent events.

(i) Both A and B will be selected: The probability of both being selected is the product of their individual probabilities. P(A and B) = P(A ∩ B) = P(A) × P(B) P(A ∩ B) = (1/5) × (1/2) = 1/10

(ii) At least one of them will be selected: The probability of at least one selection is given by P(A ∪ B). P(A ∪ B) = P(A) + P(B) – P(A ∩ B) P(A ∪ B) = 1/5 + 1/2 – 1/10 P(A ∪ B) = 2/10 + 5/10 – 1/10 = (2 + 5 – 1) / 10 = 6/10 = 3/5

Alternatively, we can calculate it as 1 – (probability that neither is selected). P(A not selected) = P(A’) = 1 – 1/5 = 4/5 P(B not selected) = P(B’) = 1 – 1/2 = 1/2 P(Neither selected) = P(A’ ∩ B’) = P(A’) × P(B’) = (4/5) × (1/2) = 4/10 = 2/5 P(At least one selected) = 1 – P(Neither selected) = 1 – 2/5 = 3/5

(c) The given probability mass function (p.m.f.) is: f(x) = (3-x)/10, for x = -1, 0, 1, 2.

Calculation of E(X): E(X) = Σ x * f(x) E(X) = (-1) f(-1) + (0) f(0) + (1) f(1) + (2) f(2) E(X) = (-1) (3 – (-1))/10 + (0) (3-0)/10 + (1) (3-1)/10 + (2) (3-2)/10 E(X) = (-1) (4/10) + 0 + (1) (2/10) + (2) * (1/10) E(X) = -4/10 + 2/10 + 2/10 = (-4 + 2 + 2) / 10 = 0 / 10 = 0

Calculation of E(X²): E(X²) = Σ x² * f(x) E(X²) = (-1)² f(-1) + (0)² f(0) + (1)² f(1) + (2)² f(2) E(X²) = (1) (4/10) + 0 + (1) (2/10) + (4) * (1/10) E(X²) = 4/10 + 2/10 + 4/10 = (4 + 2 + 4) / 10 = 10/10 = 1

Calculation of E[ (4X + 5) ]: Using the linearity property of expectation, E(aX + b) = aE(X) + b. E(4X + 5) = 4 * E(X) + 5 E(4X + 5) = 4 * (0) + 5 = 0 + 5 = 5

Q3. (a) Show that the moment generating function of the Poisson distribution with parameter m is: M(t) = e^(m(e^t – 1)) 3 (b) The height of students follows approximately a normal distribution with mean 155 cm and standard deviation 5. Find the probability that a student at random has height: 4 (i) less than 150 cm. (ii) between 150 cm and 160 cm (You may use Φ(-1) = 0.1587 and Φ(1) = 0.8413.) (c) Let X₁, X₂, …, Xₙ be a random sample from a distribution with density function: 3 f(x) = (1/θ)e^(-x/θ) ; θ>0, x>0 Show that X̄ = (1/n)ΣXᵢ is an unbiased estimator for θ.

Ans. (a) The probability mass function (p.m.f.) of a Poisson distribution with parameter m is given by: P(X = k) = (e⁻ᵐ * mᵏ) / k!, for k = 0, 1, 2, …

The Moment Generating Function (MGF), Mₓ(t), is defined as: Mₓ(t) = E[eᵗˣ] = Σ [k=0 to ∞] eᵗᵏ * P(X = k)

Substituting the p.m.f. of the Poisson distribution: Mₓ(t) = Σ [k=0 to ∞] eᵗᵏ (e⁻ᵐ mᵏ) / k! Mₓ(t) = e⁻ᵐ Σ [k=0 to ∞] (eᵗᵏ mᵏ) / k! Mₓ(t) = e⁻ᵐ Σ [k=0 to ∞] (m eᵗ)ᵏ / k!

We recognize the summation as the Maclaurin series expansion for eʸ, which is Σ [k=0 to ∞] yᵏ / k!. In this case, y = m * eᵗ. So, Σ [k=0 to ∞] (m eᵗ)ᵏ / k! = e^(m eᵗ).

Substituting this back into the MGF expression: Mₓ(t) = e⁻ᵐ e^(m eᵗ) Mₓ(t) = e^(m*eᵗ – m) Mₓ(t) = e^(m(eᵗ – 1))

Thus, it is shown that the MGF of the Poisson distribution is M(t) = e^(m(eᵗ – 1)) .

(b) Given: Mean (μ) = 155 cm, Standard Deviation (σ) = 5. Let X be the height of a student. X ~ N(155, 5²). We need to standardize the variable by calculating the Z-score: Z = (X – μ) / σ

(i) Probability of height less than 150 cm: P(X < 150) Z = (150 – 155) / 5 = -5 / 5 = -1 P(X < 150) = P(Z < -1) Using the given value, Φ(-1) = 0.1587. Therefore, P(X < 150) = 0.1587

(ii) Probability of height between 150 cm and 160 cm: P(150 < X < 160) First, we calculate the Z-score for 160 cm: Z = (160 – 155) / 5 = 5 / 5 = 1 We need to find P(-1 < Z < 1). P(-1 < Z < 1) = P(Z < 1) – P(Z < -1) P(-1 < Z < 1) = Φ(1) – Φ(-1) Using the given values: P(-1 < Z < 1) = 0.8413 – 0.1587 = 0.6826

(c) To show that the sample mean X̄ is an unbiased estimator of θ, we must prove that E(X̄) = θ. The given probability density function (p.d.f.) is of an exponential distribution: f(x) = (1/θ)e^(-x/θ), for x > 0, θ > 0

First, let’s find the expected value of the random variable X, E(X). E(X) = ∫ [0 to ∞] x * f(x) dx E(X) = ∫ [0 to ∞] x * (1/θ)e^(-x/θ) dx

We use integration by parts: ∫ u dv = uv – ∫ v du Let u = x and dv = (1/θ)e^(-x/θ) dx. Then du = dx and v = ∫ (1/θ)e^(-x/θ) dx = -e^(-x/θ).

E(X) = [x * (-e^(-x/θ))]_[0 to ∞] – ∫ [0 to ∞] (-e^(-x/θ)) dx E(X) = [-x * e^(-x/θ)]_[0 to ∞] + ∫ [0 to ∞] e^(-x/θ) dx The first term evaluates to zero as lim (x→∞) -x e^(-x/θ) = 0 and 0 e⁰ = 0. E(X) = 0 + [-θ * e^(-x/θ)]_[0 to ∞] E(X) = [-θ e⁻°°] – [-θ e⁰] = 0 – (-θ * 1) = θ. So, E(X) = θ .

Now, we find the expected value of the sample mean, X̄. X̄ = (1/n) Σ [i=1 to n] Xᵢ E(X̄) = E[ (1/n) Σ Xᵢ ] Due to the linearity of expectation: E(X̄) = (1/n) Σ E(Xᵢ) Since the Xᵢ are independent and identically distributed (i.i.d.) random variables from the same distribution, E(Xᵢ) = E(X) = θ for each i. E(X̄) = (1/n) Σ [i=1 to n] θ E(X̄) = (1/n) * (nθ) = θ.

Since E(X̄) = θ , we conclude that the sample mean X̄ is an unbiased estimator for the parameter θ.

Q4. (a) Suppose a bivariate random variable (X, Y) has the joint density function: 5 f(x,y) = 4xy * e^(-x²-y²) ; x≥0, y≥0 (i) Find the marginal distributions of X and Y. (ii) Find the conditional distribution f(x|y). (iii) Show that X and Y are independent. (b) If X₁, X₂, …, Xₙ is a random sample of size n from a Poisson distribution with parameter m, then find the maximum likelihood estimator of m. 5

Ans. (a) The given joint density function is f(x,y) = 4xy * e^(-x² – y²), for x ≥ 0, y ≥ 0.

(i) Marginal distributions of X and Y: The marginal distribution of X, fₓ(x), is found by integrating f(x,y) with respect to y: fₓ(x) = ∫ [0 to ∞] f(x,y) dy = ∫ [0 to ∞] 4xy * e^(-x² – y²) dy fₓ(x) = 4x e^(-x²) ∫ [0 to ∞] y e^(-y²) dy To solve the integral, let u = y², so du = 2y dy or y dy = du/2. ∫ [0 to ∞] y e^(-y²) dy = ∫ [0 to ∞] e⁻ᵘ (du/2) = (1/2) [-e⁻ᵘ]_[0 to ∞] = (1/2) * (0 – (-1)) = 1/2. So, fₓ(x) = 4x e^(-x²) (1/2) = 2x * e^(-x²) , for x ≥ 0.

By symmetry, the marginal distribution of Y, fᵧ(y), will have the same form: fᵧ(y) = ∫ [0 to ∞] f(x,y) dx = 2y * e^(-y²) , for y ≥ 0.

(ii) Conditional distribution f(x|y): The conditional distribution of X given Y is defined as: f(x|y) = f(x,y) / fᵧ(y). f(x|y) = (4xy e^(-x² – y²)) / (2y e^(-y²)) f(x|y) = (4xy e^(-x²) e^(-y²)) / (2y * e^(-y²)) Canceling terms: f(x|y) = 2x * e^(-x²) , for x ≥ 0.

(iii) Show that X and Y are independent: Two random variables X and Y are independent if and only if their joint density function is the product of their marginal density functions, i.e., f(x,y) = fₓ(x) * fᵧ(y). Let’s check: fₓ(x) fᵧ(y) = (2x e^(-x²)) (2y e^(-y²)) fₓ(x) fᵧ(y) = 4xy e^(-x² – y²) This is equal to the original joint density function f(x,y). Since f(x,y) = fₓ(x) * fᵧ(y) , X and Y are independent . Another proof is that the conditional distribution f(x|y) = 2x * e^(-x²) depends only on x and not on y, and is equal to fₓ(x), which proves independence.

(b) X₁, X₂, …, Xₙ is a random sample from a Poisson distribution with parameter m. The p.m.f. for a Poisson distribution is: P(X = xᵢ) = (e⁻ᵐ * mˣᵢ) / xᵢ! The likelihood function L(m) for the sample is: L(m) = Π [i=1 to n] P(X = xᵢ) L(m) = Π [i=1 to n] [ (e⁻ᵐ * mˣᵢ) / xᵢ! ] L(m) = (e⁻ᵐ)ⁿ * m^(Σxᵢ) / (Π xᵢ!) L(m) = e^(-nm) * m^(Σxᵢ) / (Π xᵢ!)

To find the Maximum Likelihood Estimator (MLE), we work with the log-likelihood function, l(m) = ln(L(m)). l(m) = ln(e^(-nm)) + ln(m^(Σxᵢ)) – ln(Π xᵢ!) l(m) = -nm + (Σxᵢ) * ln(m) – Σln(xᵢ!)

Now, we differentiate l(m) with respect to m and set it to zero: d(l(m))/dm = -n + (Σxᵢ) / m – 0 Setting d(l(m))/dm = 0: -n + (Σxᵢ) / m = 0 (Σxᵢ) / m = n m = (Σxᵢ) / n

The definition of the sample mean is X̄ = (Σxᵢ) / n. Therefore, the MLE of m, denoted by m̂, is m̂ = X̄ (the sample mean).

Q5. (a) Let X₁, X₂, …, Xₙ be a random sample from a normal distribution N(μ, 1) whose p.d.f. is given by: 6 f(x) = (1/√(2π)) * e^(-(1/2)(x-μ)²) Obtain critical region for testing H₀: μ = μ₀ against H₁: μ = μ₁ (μ₁ > μ₀) using Neyman-Pearson lemma. (b) The following data are about the sales and advertisement expenditure of a firm: 4 Sales (in crores of ₹) / Advertisement expenditure (in crores of ₹) Mean: 40 / 6 Standard deviation: 10 / 1.5 Coefficient of correlation (r) = 0.9. (i) Obtain two regression equations. (ii) Estimate the sales for a proposed advertisement expenditure of ₹ 10 crores.

Ans. (a) We need to find the best critical region for testing H₀: μ = μ₀ against H₁: μ = μ₁ (where μ₁ > μ₀). The sample is from N(μ, 1). According to the Neyman-Pearson Lemma , the best critical region is based on the likelihood ratio: L(μ₀) / L(μ₁) ≤ k, where k is a constant.

The likelihood function L(μ) is: L(μ) = Π [i=1 to n] (1/√(2π)) * e^(-(1/2)(xᵢ-μ)²) L(μ) = (1/(2π)ⁿᐟ²) * e^(-(1/2) Σ(xᵢ-μ)²)

Now, compute the likelihood ratio: L(μ₀) / L(μ₁) = [e^(-(1/2) Σ(xᵢ-μ₀)²)] / [e^(-(1/2) Σ(xᵢ-μ₁)²)] L(μ₀) / L(μ₁) = e^[- (1/2) * (Σ(xᵢ-μ₀)² – Σ(xᵢ-μ₁)²)]

The inequality L(μ₀) / L(μ₁) ≤ k becomes: e^[- (1/2) * (Σ(xᵢ² – 2xᵢμ₀ + μ₀²) – Σ(xᵢ² – 2xᵢμ₁ + μ₁²))] ≤ k

Taking the natural logarithm (ln) of both sides:

• (1/2) * [Σ(-2xᵢμ₀ + μ₀²) – Σ(-2xᵢμ₁ + μ₁²)] ≤ ln(k)
• (1/2) * [(-2μ₀Σxᵢ + nμ₀²) – (-2μ₁Σxᵢ + nμ₁²)] ≤ ln(k)

Multiply by -2 (which reverses the inequality sign): (-2μ₀Σxᵢ + nμ₀²) – (-2μ₁Σxᵢ + nμ₁²) ≥ -2ln(k) 2μ₁Σxᵢ – 2μ₀Σxᵢ – nμ₁² + nμ₀² ≥ k’ (where k’ = -2ln(k)) 2(μ₁ – μ₀)Σxᵢ ≥ k’ + nμ₁² – nμ₀²

Since we are given μ₁ > μ₀, the term (μ₁ – μ₀) is positive. We can divide by 2(μ₁ – μ₀): Σxᵢ ≥ (k’ + nμ₁² – nμ₀²) / (2(μ₁ – μ₀)) Σxᵢ ≥ c’, where c’ is a new constant.

Or, dividing both sides by n: X̄ ≥ c’/n = c

Thus, the critical region is the set where the sample mean is greater than or equal to some constant c. Critical Region: { (x₁, …, xₙ) | X̄ ≥ c } This means we reject the null hypothesis H₀ if the sample mean is too large. The constant c is determined by the significance level α.

(b) Given data: Let Sales = X and Advertisement Expenditure = Y X̄ = 40 (crores), Ȳ = 6 (crores) σₓ = 10, σᵧ = 1.5 r = 0.9

(i) Two regression equations:

1. Regression line of X on Y: (X – X̄) = bₓᵧ (Y – Ȳ) bₓᵧ (regression coefficient of X on Y) = r * (σₓ / σᵧ) bₓᵧ = 0.9 (10 / 1.5) = 0.9 (20/3) = (9/10) * (20/3) = 6 The equation of the line is: (X – 40) = 6 * (Y – 6) X – 40 = 6Y – 36 X = 6Y + 4

2. Regression line of Y on X: (Y – Ȳ) = bᵧₓ (X – X̄) bᵧₓ (regression coefficient of Y on X) = r * (σᵧ / σₓ) bᵧₓ = 0.9 (1.5 / 10) = 0.9 0.15 = 0.135 The equation of the line is: (Y – 6) = 0.135 * (X – 40) Y – 6 = 0.135X – 5.4 Y = 0.135X + 0.6

(ii) Estimate the sales: We need to estimate sales (X) for a proposed advertisement expenditure (Y) of ₹ 10 crores. For this, we use the regression line of X on Y: X = 6Y + 4. Substitute Y = 10: X = 6(10) + 4 X = 60 + 4 = 64

Therefore, the estimated sales for an advertisement expenditure of ₹ 10 crores is ₹ 64 crores .

Q6. (a) Let X be a negative exponential variable having probability density function: 6 f(x) = (1/α)e^(-x/α) ; x>0 (i) Find mean (μ) and variance (σ²) of X. (ii) Obtain the lower limit of P[|X-μ| < 3σ] using Chebyshev’s inequality. (iii) Compare (ii) with its exact value. (b) A random number generator produces U(0, 1) random numbers as follows: 4 Class / Frequency 0—0.2 / 8 0.2—0.4 / 6 0.4—0.6 / 9 0.6—0.8 / 9 0.8—1.0 / 8 Apply Chi-square test that random numbers come from uniform distribution at 5% level of significance. [You may use χ²(0.05, 4) = 9.488.]

Ans. (a) The given p.d.f. is f(x) = (1/α)e^(-x/α), x > 0. This is an exponential distribution with rate λ = 1/α.

(i) Find mean (μ) and variance (σ²): Mean (μ): μ = E(X) = ∫ [0 to ∞] x f(x) dx = ∫ [0 to ∞] x (1/α)e^(-x/α) dx This is a standard integral. Using the gamma function, E(X) = α Γ(2) = α 1! = α. So, the mean μ = α .

Variance (σ²): σ² = E(X²) – [E(X)]² First, calculate E(X²): E(X²) = ∫ [0 to ∞] x² * (1/α)e^(-x/α) dx Using the gamma function, E(X²) = α² Γ(3) = α² 2! = 2α². Now, calculate the variance: σ² = 2α² – (α)² = α² So, the variance σ² = α² . (And standard deviation σ = α).

(ii) Lower limit using Chebyshev’s inequality: One form of Chebyshev’s inequality is: P(|X – μ| < kσ) ≥ 1 – 1/k² Here, we need the lower limit for P(|X – μ| < 3σ), so k = 3. P(|X – μ| < 3σ) ≥ 1 – 1/3² P(|X – μ| < 3σ) ≥ 1 – 1/9 P(|X – μ| < 3σ) ≥ 8/9 Thus, the lower limit is 8/9 (approx. 0.8889).

(iii) Comparison with the exact value: We now find the exact value of P(|X – μ| < 3σ). Since μ = α and σ = α, we need to calculate P(|X – α| < 3α). |X – α| < 3α means -3α < X – α < 3α, which simplifies to -2α < X < 4α. Since X is always positive (x > 0), this inequality is equivalent to P(0 < X < 4α). P(0 < X < 4α) = ∫ [0 to 4α] (1/α)e^(-x/α) dx = [ -e^(-x/α) ]_[0 to 4α] = (-e^(-4α/α)) – (-e⁰) = -e⁻⁴ + 1 = 1 – e⁻⁴

Now, we compare: Exact value = 1 – e⁻⁴ ≈ 1 – 0.0183 = 0.9817 Chebyshev’s lower limit = 8/9 ≈ 0.8889 The comparison shows that the exact value (0.9817) is indeed greater than the lower bound provided by Chebyshev’s inequality (0.8889). This is expected, as Chebyshev’s inequality is a general bound that holds for any distribution.

(b) Chi-square Goodness-of-Fit Test: H₀: The data comes from a U(0, 1) distribution. H₁: The data does not come from a U(0, 1) distribution.

1. Calculate Expected Frequencies: Total observations (N) = 8 + 6 + 9 + 9 + 8 = 40. There are 5 classes, each with a width of 0.2. For a uniform distribution U(0,1), the probability of falling into any class is equal to its width, i.e., p = 0.2. Expected frequency (E) for each class = N × p = 40 × 0.2 = 8.

2. Calculate the Chi-Square Statistic:

Class	Observed (O)	Expected (E)	(O – E)	(O – E)²	(O – E)² / E
0-0.2	8	8	0	0	0.0
0.2-0.4	6	8	-2	4	4/8 = 0.5
0.4-0.6	9	8	1	1	1/8 = 0.125
0.6-0.8	9	8	1	1	1/8 = 0.125
0.8-1.0	8	8	0	0	0.0
Total	40	40			0.75

The calculated Chi-square value:

χ²_calc = 0.75

3. Critical Value and Decision: Degrees of freedom (df) = k – 1 = 5 – 1 = 4 (where k is the number of classes). Significance level (α) = 0.05. The critical value given in the question is χ²_crit (0.05, 4) = 9.488 .

Decision Rule: If χ²_calc < χ²_crit, do not reject H₀. Here, 0.75 < 9.488. Since the calculated value is less than the critical value, we fail to reject the null hypothesis (H₀).

Conclusion: At the 5% significance level, there is not enough evidence to reject the claim that the random numbers come from a uniform distribution. We conclude that the data is consistent with the U(0, 1) distribution.

Q7. Which of the following statements are true or false? Give a short proof or a counter-example in support of your answer: 10 (i) Weather data recorded by the Department of Meteorology and used by an investigator for writing a Ph. D. thesis, is primary data for the investigator. (ii) If a card is selected from a well-shuffled deck, then the probability of drawing a king is 1/8. (iii) If the variables X and Y are independent, then the coefficient of correlation between them is 1. (iv) The expectation of the number on an unbiased dice when thrown is 7/2. (v) The variance of a binomial distribution with parameters (5, 1/2) is 5/4.

Ans. (i) Statement: False. Reason: Primary data is data collected for the first time by the investigator for the specific purpose of their study. Using data that was already recorded by the Department of Meteorology constitutes the use of secondary data for the investigator, as the data was collected by someone else for a different purpose.

(ii) Statement: False. Reason: A standard deck of cards has 52 cards, including 4 kings. The probability of drawing a king is the number of favorable outcomes divided by the total number of possible outcomes. P(King) = (Number of Kings) / (Total Cards) = 4 / 52 = 1/13 . This is not equal to 1/8.

(iii) Statement: False. Reason: If two variables X and Y are independent, their covariance is zero, Cov(X, Y) = 0. The coefficient of correlation, ρ, is defined as Cov(X, Y) / (σₓσᵧ). If Cov(X, Y) = 0, then the correlation coefficient will be 0 (assuming non-zero variances). A correlation of 1 indicates a perfect positive linear relationship, which is a form of strong dependence, not independence.

(iv) Statement: True. Reason: For an unbiased die, the outcome X can be {1, 2, 3, 4, 5, 6}, each with a probability P(x) = 1/6. The expectation (or mean) is calculated as: E(X) = Σ x * P(x) E(X) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) E(X) = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 21 / 6 = 7/2 or 3.5.

(v) Statement: True. Reason: The variance of a binomial distribution B(n, p) is given by the formula Var(X) = np(1-p). Given parameters are n = 5 and p = 1/2. Var(X) = 5 × (1/2) × (1 – 1/2) Var(X) = 5 × (1/2) × (1/2) = 5/4 .