CA Foundation P3 · Chapter 16

Theoretical Distributions MCQs with Answers

9 multiple-choice questions on Theoretical Distributions for CA Foundation P3 Quantitative Aptitude. Try each one before revealing the answer and explanation.

1. Question 1

   A fair coin is tossed 4 times. The probability of getting exactly 2 heads is:

   • A) 0.25
   • B) 0.375
   • C) 0.5
   • D) 0.0625
   Show answer & explanation

   Answer: B) 0.375

   P(X = 2) = ⁴C₂ (1/2)² (1/2)² = 6 x 1/16 = 6/16 = 0.375. Omitting the ⁴C₂ factor gives 1/16 = 0.0625.

2. Question 2

   A Poisson variable X has mean 2. Using e⁻² = 0.1353, P(X ≥ 1) is:

   • A) 0.1353
   • B) 0.2707
   • C) 0.7293
   • D) 0.8647
   Show answer & explanation

   Answer: D) 0.8647

   P(X = 0) = e⁻ᵐ m⁰/0! = e⁻² = 0.1353. P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.1353 = 0.8647. 0.2707 is P(X = 1) = 2e⁻².

3. Question 3

   The standard deviation of a Poisson distribution is 3. Its mean is:

   • A) 3
   • B) 9
   • C) √3
   • D) 6
   Show answer & explanation

   Answer: B) 9

   In a Poisson distribution, mean = variance = m. Variance = SD² = 9, so the mean is 9. Equating the mean to the SD (3) is the typical error.

4. Question 4

   For a binomial distribution with n = 10 and p = 0.4, the variance is:

   • A) 2.4
   • B) 4
   • C) 6
   • D) 1.55
   Show answer & explanation

   Answer: A) 2.4

   Variance = npq = 10 x 0.4 x 0.6 = 2.4. The mean is np = 4, and √2.4 ≈ 1.55 is the standard deviation.

5. Question 5

   The mean and variance of a binomial distribution are 6 and 4 respectively. The number of trials n is:

   • A) 9
   • B) 12
   • C) 18
   • D) 24
   Show answer & explanation

   Answer: C) 18

   np = 6 and npq = 4, so q = 4/6 = 2/3 and p = 1/3. Then n = 6/(1/3) = 18. Using p = 2/3 by mistake gives n = 9.

6. Question 6

   X is normally distributed with mean 60 and standard deviation 10. Given that the area under the standard normal curve from z = 0 to z = 1.5 is 0.4332, P(X > 75) is:

   • A) 0.9332
   • B) 0.4332
   • C) 0.0668
   • D) 0.1336
   Show answer & explanation

   Answer: C) 0.0668

   z = (75 − 60)/10 = 1.5. P(X > 75) = P(Z > 1.5) = 0.5 − 0.4332 = 0.0668. 0.9332 is P(Z < 1.5), and 0.1336 is the two-tailed area P(|Z| > 1.5).

7. Question 7

   For a normal distribution with standard deviation 15, the mean deviation about the mean is approximately (use the conventions MD ≈ 0.8σ and QD ≈ 0.675σ):

   • A) 12
   • B) 10
   • C) 15
   • D) 18.75
   Show answer & explanation

   Answer: A) 12

   For a normal distribution, mean deviation ≈ 0.8σ (more precisely √(2/π)σ ≈ 0.7979σ). MD ≈ 0.8 x 15 = 12. Quartile deviation ≈ 0.675σ = 0.675 x 15 ≈ 10.1, so 10 is the QD, which is the usual confusion; 18.75 results from dividing by 0.8 instead of multiplying, and 15 is σ itself.

8. Question 8

   For a Poisson variable X, P(X = 1) = P(X = 2). Using e⁻² = 0.1353, P(X = 3) is (to four decimals):

   • A) 0.2707
   • B) 0.1804
   • C) 0.0902
   • D) 0.1353
   Show answer & explanation

   Answer: B) 0.1804

   e⁻ᵐ m = e⁻ᵐ m²/2 gives m = 2. P(X = 3) = e⁻² 2³/3! = 0.1353 x 8/6 = 0.1353 x 1.3333 = 0.1804. 0.2707 is P(X = 1) = P(X = 2), and 0.0902 is P(X = 4).

9. Question 9

   Marks of 2,000 students are normally distributed with mean 50 and SD 5. Given areas from z = 0: 0.3413 for z = 1 and 0.4772 for z = 2, the number of students scoring between 45 and 60 is about:

   • A) 1365
   • B) 954
   • C) 1637
   • D) 1909
   Show answer & explanation

   Answer: C) 1637

   z for 45 = (45 − 50)/5 = −1 and z for 60 = (60 − 50)/5 = 2. Area = 0.3413 + 0.4772 = 0.8185. Number = 0.8185 x 2,000 = 1,637. Using ±1 (0.6826) gives 1,365, and 0.4772 alone gives 954.