TGPSC MPSO / ASO · Paper II · Topic 4
TGPSC MPSO / ASO Random Variables, Distributions & Inference questions with answers
29 questions from TGPSC ASO Paper II Mock 1 with answers and TGPSC ASO Paper II Challenge 1 with answers. Try each one, then tap Show answer.
4.1Random variables & mathematical expectationNotesPractise this chapter · Set 1 free
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For a random variable X, E(X) = 2 and E(X²) = 8. Find Var(2X + 1).
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Answer C. 16
Var(X) = 8 − 2² = 4, so Var(2X + 1) = 2² × 4 = 16.
Source: Standard result (BRAOU Sem2 Mathematical Expectation, units on random variables and expectation)
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Consider the following statements about a continuous random variable X with pdf f(x):
1. f(x) can be greater than 1 for some values of x.
2. P(a < X < b) = P(a ≤ X ≤ b).
Which of the above is/are correct?
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Answer B. Both 1 and 2
A density only needs to be non-negative with total area 1, so it can exceed 1. Endpoints have zero probability, so including them changes nothing.
Source: Standard definition (BRAOU Sem2 Mathematical Expectation, units on random variables, pmf, pdf and distribution function)
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f(x) = kx(1 − x) for 0 ≤ x ≤ 1 (and 0 elsewhere) is a pdf. Find k.
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Answer D. 6
The integral of x − x² from 0 to 1 is 1/2 − 1/3 = 1/6, so k/6 = 1 and k = 6.
Source: Standard result (BRAOU Sem2 Mathematical Expectation, units on random variables and expectation)
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Which of the following pairs of a random variable and its type is NOT correctly matched?
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Answer A. Number of goals scored in a football match : Continuous
The number of goals is a count (0, 1, 2, …), so it is a discrete random variable, not continuous.
Source: Standard definition (BRAOU Sem2 Mathematical Expectation, units on random variables, pmf, pdf and distribution function)
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Match List I with List II:
List I:
A. Probability mass function
B. Probability density function
C. Distribution function
D. Mathematical expectation of X
List II:
1. Σ x p(x)
2. P(X ≤ x)
3. Point probabilities of a discrete variable that add to 1
4. Area under its curve over an interval gives a probability
Choose the correct answer:
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Answer A. A-3, B-4, C-2, D-1
A pmf gives point probabilities that add to 1; a pdf gives probabilities as areas; F(x) = P(X ≤ x); E(X) = Σ x p(x).
Source: Standard definition (BRAOU Sem2 Mathematical Expectation, units on random variables, pmf, pdf and distribution function)
4.1Bernoulli & binomial distributionsNotesPractise this chapter · Set 1 free
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X ~ B(6, p) and 9 P(X = 4) = P(X = 2). Find p.
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Answer B. 1/4
9 × 15 p⁴q² = 15 p²q⁴ gives 9p² = q², so q = 3p, 4p = 1 and p = 1/4.
Source: Standard result (BRAOU Sem3 Discrete Distributions, unit on Bernoulli and binomial distributions)
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X ~ B(6, 1/3). Using the recurrence relation, P(X = 3) = k × P(X = 2). Find k.
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Answer B. 2/3
P(x + 1)/P(x) = [(n − x)/(x + 1)] × (p/q) = (4/3) × (1/2) = 2/3.
Source: Standard result (BRAOU Sem3 Discrete Distributions, unit on Bernoulli and binomial distributions)
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X ~ B(5, 1/2). Find P(X ≤ 1).
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Answer C. 3/16
P(X ≤ 1) = (1 + 5)/32 = 6/32 = 3/16.
Source: Standard result (BRAOU Sem3 Discrete Distributions, unit on Bernoulli and binomial distributions)
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Match List I with List II:
List I:
A. Mean of a Bernoulli distribution
B. Variance of a Bernoulli distribution
C. Mean of a binomial distribution
D. Variance of a binomial distribution
List II:
1. npq
2. p
3. np
4. pq
Choose the correct answer:
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Answer D. A-2, B-4, C-3, D-1
Bernoulli: mean p, variance pq. Binomial (n trials): mean np, variance npq.
Source: Standard definition (BRAOU Sem3 Discrete Distributions, unit on Bernoulli and binomial distributions)
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A test has 8 questions, each with 4 options of which one is correct. A student guesses every answer. What is the expected number of correct answers?
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Answer B. 2
X ~ B(8, 1/4), so E(X) = np = 8 × 1/4 = 2.
Source: Standard result (BRAOU Sem3 Discrete Distributions, unit on Bernoulli and binomial distributions)
4.1Poisson distributionPractise this chapter · Set 1 free
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For a Poisson distribution with λ = 4, consider:
1. The mean is 4.
2. The standard deviation is 2.
3. P(X = 3) = P(X = 4).
4. The only mode is 4.
Which of the statements given above are correct?
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Answer D. 1, 2 and 3 only
Statements 1, 2 and 3 are correct. The distribution has two modes, 3 and 4.
Source: Standard definition (BRAOU Sem3 Discrete Distributions, unit on the Poisson distribution)
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Assertion (A): For a Poisson distribution, the mean and the variance are equal.
Reason (R): The Poisson is the limit of a binomial with p → 0, so q → 1 and the binomial variance npq approaches the mean np.
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Answer A. Both A and R are true, and R is the correct explanation of A
Mean = variance = λ for the Poisson. As p → 0, q → 1 and npq → np = λ, which is why the two become equal.
Source: Standard definition (BRAOU Sem3 Discrete Distributions, unit on the Poisson distribution)
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Which of the following pairs about the Poisson distribution is NOT correctly matched?
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Answer A. Mode when λ is a whole number : a single mode at λ
For a whole-number λ, P(λ − 1) = P(λ), so the distribution is bimodal at λ − 1 and λ.
Source: Standard definition (BRAOU Sem3 Discrete Distributions, unit on the Poisson distribution)
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Which of the following are typical Poisson situations?
1. The number of accidents per day at a road crossing
2. The number of defects per metre of cloth
3. The marks obtained by students in an examination
4. The number of telephone calls received per minute
Which of the statements given above are correct?
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Answer D. 1, 2 and 4 only
Statements 1, 2 and 4 are correct. Marks are measurements with an upper limit, not counts of rare events.
Source: Standard definition (BRAOU Sem3 Discrete Distributions, unit on the Poisson distribution)
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The number of defects per sheet of cloth follows a Poisson distribution with mean 1. Three sheets are inspected, independently of one another. What is the probability that exactly one of the three sheets is free of defects?
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Answer A. 3e^−1(1 − e^−1)²
The probability is 3e^−1(1 − e^−1)². A sheet is free of defects with probability P(X = 0) = e^−1. The number of defect-free sheets among three is then binomial with n = 3 and p = e^−1. So P(exactly one) = C(3, 1) e^−1 (1 − e^−1)², about 0.44.
Source: Standard properties of the Poisson and binomial distributions (BRAOU Sem3 Discrete Distributions)
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Calls reach desk A as a Poisson variable with mean 1 per hour and desk B as an independent Poisson variable with mean 2 per hour. What is the probability that the two desks together receive exactly 2 calls in an hour? (Take e^−3 = 0.0498.)
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Answer C. 0.2241
The probability is about 0.2241. The sum of independent Poisson variables is Poisson with mean 1 + 2 = 3. So P(total = 2) = e^−3 × 3²/2! = 0.0498 × 4.5 ≈ 0.2241. The same answer comes from adding the cases (0, 2), (1, 1) and (2, 0): e^−3(2 + 2 + 0.5).
Source: Standard method (BRAOU Sem3 Discrete Distributions)
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Misprints per page in book A follow a Poisson distribution with mean 1, and in book B a Poisson distribution with mean 2. One of the two books is chosen at random and a page is opened at random. The page has no misprint. What is the probability that the book chosen was A?
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Answer C. e/(1 + e)
The probability is e/(1 + e), about 0.73. P(no misprint | A) = e^−1 and P(no misprint | B) = e^−2. By Bayes' theorem with equal priors, P(A | no misprint) = e^−1/(e^−1 + e^−2). Multiplying top and bottom by e², this is e/(e + 1).
Source: Poisson distribution with Bayes' theorem (BRAOU Sem3 Discrete Distributions; Sem1 Probability)
4.1Normal distribution & its usesNotesPractise this chapter · Set 1 free
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A population has mean 50 and SD 10. For a random sample of 25, find P(x̄ > 52). (Area from 0 to z: z = 0.2 → 0.0793, z = 1 → 0.3413)
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Answer A. 0.1587
x̄ ~ N(50, 10²/25), SE = 2. z = (52 − 50)/2 = 1. P = 0.5 − 0.3413 = 0.1587.
Source: Standard result (BRAOU Sem4 Continuous Distributions, unit on the normal distribution)
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Kiran scores 70 in a test with mean 60 and SD 5. Ravi scores 80 in another test with mean 65 and SD 10. Both tests have normal scores. Who did relatively better?
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Answer B. Kiran, with z = 2 against Ravi's 1.5
z = (x − μ)/σ: Kiran (70 − 60)/5 = 2; Ravi (80 − 65)/10 = 1.5. Kiran stands higher relative to his group.
Source: Standard result (BRAOU Sem4 Continuous Distributions, unit on the normal distribution)
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Assertion (A): The normal approximation to the binomial works well when n is large and p is close to 0.5.
Reason (R): For a normal distribution, the mean, median and mode coincide.
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Answer B. Both A and R are true, but R is not the correct explanation of A
With p near 0.5 the binomial is nearly symmetrical, so the normal fits it well even for moderate n. R is a true property of the normal, but it does not say why the binomial approaches it.
Source: Standard definition (BRAOU Sem4 Continuous Distributions, unit on the normal distribution)
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For a normal distribution, consider:
1. The area between μ and μ + σ equals the area between μ + σ and μ + 2σ.
2. The total area under the normal curve is 1.
Which of the statements given above is/are correct?
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Answer B. 2 only
The first area is about 34.13%, the second only about 13.59%. The density integrates to 1 over −∞ to ∞.
Source: Standard definition (BRAOU Sem4 Continuous Distributions, unit on the normal distribution)
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Consider the following statements:
I. If X is normal with mean μ and variance σ², then for constants a ≠ 0 and b, aX + b is normal with mean aμ + b and variance a²σ².
II. If X is normal with mean 50 and variance 16, then 2X has the same distribution as X1 + X2, where X1 and X2 are independent, each normal with mean 50 and variance 16.
Which of the statements given above is/are correct?
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Answer A. Only I is correct
Only statement I is correct. A linear function of a normal variable is normal, with mean aμ + b and variance a²σ². 2X has mean 100 and variance 2² × 16 = 64. X1 + X2 also has mean 100, but its variance is 16 + 16 = 32, because independent variances add. So the two distributions differ, and statement II is false.
Source: Standard method (BRAOU Sem4 Continuous Distributions, normal distribution)
4.2Estimators: unbiasedness, consistency, efficiency, sufficiencyNotesPractise this chapter · Set 1 free
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Consider the following statements about the Cramér–Rao lower bound:
1. An unbiased estimator that attains the bound is a minimum variance unbiased estimator.
2. Every unbiased estimator attains the bound.
3. The bound is derived under regularity conditions, such as the range of the variable not depending on θ.
Which of the statements given above are correct?
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Answer A. 1 and 3 only
Statements 1 and 3 are correct. Attaining the bound makes an estimator the MVUE, but not every unbiased estimator attains it, and the bound needs regularity conditions.
Source: Standard result (BRAOU Sem6 Estimation & Testing of Hypothesis, Block I unit 2: sufficiency, Fisher–Neyman factorisation, MVUE)
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For a sample of size 9 from a population with mean μ = 50, the estimator T = Σx/(n + 1) is used. What is the bias of T?
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Answer B. −5
E(T) = 9 × 50/10 = 45, so bias = 45 − 50 = −5.
Source: Standard result (BRAOU Sem6 Estimation & Testing of Hypothesis, Block I unit 1: consistency, unbiasedness, efficiency)
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Consider the following statements about S² = Σ(x − x̄)²/n as an estimator of σ²:
1. It is a negatively biased estimator of σ².
2. Its bias tends to zero as n tends to infinity.
Which of the statements given above is/are correct?
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Answer C. Both 1 and 2
Both are correct. E(S²) = (n − 1)σ²/n, so the bias −σ²/n is negative and tends to zero as n → ∞.
Source: Standard result (BRAOU Sem6 Estimation & Testing of Hypothesis, Block I unit 1: consistency, unbiasedness, efficiency)
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Consider the following statements:
1. A statistic is a random variable because its value changes from sample to sample.
2. A parameter is a random variable because it is unknown.
Which of the statements given above is/are correct?
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Answer A. Only 1
Only statement 1 is correct. A statistic varies from sample to sample and has a sampling distribution; a parameter is a fixed population constant.
Source: Standard definition (BRAOU Sem6 Estimation & Testing of Hypothesis, Block I unit 1: basic terminology)
4.2Hypotheses, large & small sample testsNotesPractise this chapter · Set 1 free
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Which one of the following pairs is NOT correctly matched?
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Answer B. F-test : named after 'Student'
The F-distribution is named after Fisher (by Snedecor); 'Student' is Gosset's pen-name for t.
Source: Standard definition (BRAOU Sem 6 Estimation & Testing of Hypothesis, Unit 9 F-test and chi-square test)
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Consider the following statements about large sample tests.
1. Under H0, Z = (t − E(t))/SE(t) is approximately a standard normal variate when n is large.
2. At a given level of significance, the critical value of Z does not depend on the sample size n.
Which of the statements given above is/are correct?
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Answer C. Both 1 and 2
For large n, Z = (t − E(t))/SE(t) is approximately N(0, 1) under H0, and its critical values (such as 1.96 at 5%, two-tailed) depend only on α and the tails, not on n.
Source: Standard definition (BRAOU Sem 6 Estimation & Testing of Hypothesis, Unit 6 tests for single mean and proportion)
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A 95% confidence interval for a population mean μ, based on a random sample of 100 and the known population SD σ, is (48.08, 55.92), using z = 1.96. What is σ, and what is the decision on H0: μ = 50 against H1: μ ≠ 50 at the 5% level?
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Answer D. σ = 20; do not reject H0
σ = 20 and H0 is not rejected. The interval is x̄ ± 1.96σ/√n, so x̄ = 52 and the half-width 3.92 = 1.96 × σ/10 gives σ = 20. The two-tailed 5% test of μ = 50 rejects H0 exactly when 50 lies outside the 95% interval, and 50 lies inside (48.08, 55.92). Directly, z = (52 − 50)/(20/10) = 1 < 1.96.
Source: Standard concept (BRAOU Sem6 Estimation & Testing of Hypothesis)
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