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Class 12 · Tamil Nadu State Board

Samacheer Kalvi 12th Statistics Chapter 8 Exercise Solutions

Tamil Nadu Class 12 Statistics Chapter 8 Random Variables and Probability Distributions exercise solutions — discrete continuous distributions, Binomial Poisson Normal, expectation MCQs.

Updated 18 June 20264 min read

Chapter 8 — Random Variables and Probability Distributions


Key Concepts Quick Recap

DistributionParametersMeanVarianceWhen to Use
Binomial B(n, p)n trials, p success probnpnpqFixed trials, two outcomes
Poisson P(λ)λ = mean eventsλλRare events, large n, small p
Normal N(μ, σ²)μ mean, σ² varianceμσ²Continuous data, large samples

I. Choose the Best Answer (MCQs)

1. A random variable X follows Binomial distribution with n = 5, p = 0.3. The mean of X is: (a) 0.15 (b) 1.5 (c) 3 (d) 5

Answer(b) 1.5 Mean of B(n,p) = np = 5 × 0.3 = 1.5. Variance = npq = 5 × 0.3 × 0.7 = 1.05.

2. In a Poisson distribution with parameter λ = 3, the mean and variance are both: (a) 3 and 6 (b) 3 and 3 (c) 6 and 3 (d) 9 and 3

Answer(b) 3 and 3 Key property of Poisson: Mean = Variance = λ. So both = 3.

3. The normal distribution is symmetric about: (a) Zero (b) Standard deviation (c) Mean (μ) (d) Median

Answer(c) Mean (μ) Normal curve is symmetric about the mean. Mean = Median = Mode for normal distribution. Total area under curve = 1.

4. For a standard normal distribution Z ~ N(0,1), the standard normal variate is: (a) Z = (X − σ)/μ (b) Z = (X − μ)/σ (c) Z = μ/σ (d) Z = X × σ

Answer(b) Z = (X − μ)/σ Standardisation converts any normal variable to standard normal (mean=0, SD=1) by subtracting mean and dividing by SD.

5. The area under the normal curve between μ−σ and μ+σ is approximately: (a) 50% (b) 68.27% (c) 95.45% (d) 99.73%

Answer(b) 68.27% Empirical rule: μ±1σ = 68.27%, μ±2σ = 95.45%, μ±3σ = 99.73%. This is the "68-95-99.7 rule."

6. Which distribution is used for modelling the number of defects per unit in quality control? (a) Binomial (b) Normal (c) Poisson (d) Uniform

Answer(c) Poisson Poisson is used for rare events in a fixed interval of time or space: defects per unit, calls per hour, accidents per day. Applicable when n is very large and p is very small (but np = λ is moderate).

7. If X ~ B(n=10, p=0.5), the probability of getting exactly 5 successes, P(X=5), uses: (a) P(X=5) = λ⁵e⁻λ/5! (b) P(X=5) = C(10,5)(0.5)⁵(0.5)⁵ (c) P(X=5) = 1/10 (d) Area under normal curve

Answer(b) P(X=5) = C(10,5)(0.5)⁵(0.5)⁵ Binomial formula: P(X=x) = nCx · pˣ · q^(n-x). Here: C(10,5) × (0.5)⁵ × (0.5)⁵ = 252 × (1/32) × (1/32) = 252/1024 ≈ 0.246.

II. Short Answer Questions

Q1. Explain the conditions for using Poisson distribution as an approximation to Binomial distribution.
Answer

Poisson approximation: when n is large (n > 50) and p is small (p < 0.1), B(n,p) can be approximated by Poisson with λ = np.

Example: Probability of a defective item is 0.005 (p = 0.005). Sample of 200 items (n = 200). λ = np = 200 × 0.005 = 1. Use Poisson with λ = 1.

Q2. A coin is tossed 6 times. What is the probability of getting exactly 4 heads?
Answer

X ~ B(n=6, p=0.5), find P(X=4).

P(X=4) = C(6,4) × (0.5)⁴ × (0.5)²

= 15 × 0.0625 × 0.25

= 15 × 0.015625

= 0.234375 ≈ 0.23


Exam Checklist

  • Binomial: n fixed, 2 outcomes, constant p, independent; Mean=np, Var=npq.
  • Poisson: λ = np, mean = variance = λ; for rare events.
  • Normal: bell-shaped, symmetric; μ±1σ=68.27%, μ±2σ=95.45%, μ±3σ=99.73%.
  • Z = (X−μ)/σ — standardise to use Z-table.
  • Poisson approximates Binomial when n large, p small, λ = np moderate.

Frequently Asked Questions

Binomial distribution B(n, p) requires: (1) Fixed number of trials (n). (2) Each trial has only two outcomes (success or failure). (3) Probability of success (p) is constant for each trial. (4) Trials are independent. Mean = np; Variance = npq (q = 1-p); SD = √(npq).
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