Theoretical DistributionsPYQ Sept 25Question 4489 of 230
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The probability mass function of a distribution is given below in a tabular form: p(x)\displaystyle p(x) is k\displaystyle k at x=0\displaystyle x=0, 2k+k2\displaystyle 2k+k^2 at x=1\displaystyle x=1, 3k\displaystyle 3k at x=2\displaystyle x=2, 2k+k2\displaystyle 2k+k^2 at x=3\displaystyle x=3, and k\displaystyle k at x=4\displaystyle x=4. Where k\displaystyle k is a non-negative constant. The median of the distribution is

Options

A3k
B2
C2k
D3
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Correct Answer

✅ Option b — 2

All Options:

  • A3k
  • B2
  • C2k
  • D3

Detailed Solution & Explanation

Let us analyze the given probability mass function (PMF) of the discrete random variable X\displaystyle X:
- p(0)=k\displaystyle p(0) = k
- p(1)=2k+k2\displaystyle p(1) = 2k + k^2
- p(2)=3k\displaystyle p(2) = 3k
- p(3)=2k+k2\displaystyle p(3) = 2k + k^2
- p(4)=k\displaystyle p(4) = k
1. **Examine the symmetry of the distribution**:
Notice that the probabilities are perfectly symmetric about the point x=2\displaystyle x = 2, since:
- p(0)=p(4)=k\displaystyle p(0) = p(4) = k
- p(1)=p(3)=2k+k2\displaystyle p(1) = p(3) = 2k + k^2
For any symmetric probability distribution, the median (which divides the total probability of 1 into two equal halves of 0.5) lies at the center of symmetry. Therefore, the median is 2\displaystyle 2.
2. **Verification by calculating k\displaystyle k**:
The sum of all probabilities must equal 1\displaystyle 1:
∑p(x)=p(0)+p(1)+p(2)+p(3)+p(4)=1\sum p(x) = p(0) + p(1) + p(2) + p(3) + p(4) = 1
k+(2k+k2)+3k+(2k+k2)+k=1k + (2k + k^2) + 3k + (2k + k^2) + k = 1
2k2+9k−1=02k^2 + 9k - 1 = 0
Solving the quadratic equation for non-negative k\displaystyle k:
k=−9+92−4(2)(−1)2(2)=−9+81+84=−9+894k = \frac{-9 + \sqrt{9^2 - 4(2)(-1)}}{2(2)} = \frac{-9 + \sqrt{81 + 8}}{4} = \frac{-9 + \sqrt{89}}{4}
Since 89≈9.434\displaystyle \sqrt{89} \approx 9.434:
k≈−9+9.4344≈0.1085k \approx \frac{-9 + 9.434}{4} \approx 0.1085
Let us compute the cumulative distribution function (CDF) F(x)=P(X≤x)\displaystyle F(x) = P(X \le x):
- For x=0\displaystyle x = 0: F(0)=p(0)=k≈0.1085\displaystyle F(0) = p(0) = k \approx 0.1085
- For x=1\displaystyle x = 1: F(1)=p(0)+p(1)=3k+k2≈3(0.1085)+0.10852≈0.3255+0.0118=0.3373<0.5\displaystyle F(1) = p(0) + p(1) = 3k + k^2 \approx 3(0.1085) + 0.1085^2 \approx 0.3255 + 0.0118 = 0.3373 < 0.5
- For x=2\displaystyle x = 2: F(2)=F(1)+p(2)=F(1)+3k≈0.3373+3(0.1085)=0.3373+0.3255=0.6628≥0.5\displaystyle F(2) = F(1) + p(2) = F(1) + 3k \approx 0.3373 + 3(0.1085) = 0.3373 + 0.3255 = 0.6628 \ge 0.5
Since the cumulative probability reaches and exceeds 0.5\displaystyle 0.5 at x=2\displaystyle x = 2, the median of the distribution is indeed 2\displaystyle 2.
Hence, **Option B** is the correct answer.

About This Chapter: Theoretical Distributions

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Binomial, Poisson, Normal Distribution

This chapter covers Binomial, Poisson, Normal Distribution and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

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