ISS PREVIOUS YEAR 2016 PAPER-1 SET-A Q.no.- 8
- Shivani Rana
- Mar 29
- 2 min read
Updated: 2 days ago
ISS PREVIOUS YEAR 2016 PAPER-1 SET-A
![For the distribution: f(x) = [1 / B(p, q)] × [x^(p − 1) / (1 + x)^(p + q)],where 0 < x < ∞, p > 0, q > 0, find the harmonic mean. Options: (a) p / (p + q) (b) 1 / p (c) (p − 1) / q (d) (p + 1) / (q − 1)](https://static.wixstatic.com/media/8ffd4d_929597f8730a4e7fb508b389d8141e98~mv2.jpg/v1/fill/w_980,h_551,al_c,q_85,usm_0.66_1.00_0.01,enc_avif,quality_auto/8ffd4d_929597f8730a4e7fb508b389d8141e98~mv2.jpg)
Probability Question Solution
Question 8
Let \( X \) be a Poisson variate with parameter \( \lambda \) such that
\[ P(X = 2) = 2P(X = 4) + 20P(X = 6) \]
What is the coefficient of skewness?
Options:
(a) \( \frac{1}{\sqrt{3}} \)
(b) 1
(c) \( \frac{1}{2} \)
(d) \( -\frac{1}{\sqrt{3}} \)
Solution
For a Poisson distribution with parameter \( \lambda \), the probability mass function is:
\[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]
Step 1: Write the probabilities
\( P(X = 2) = \frac{e^{-\lambda} \cdot \lambda^2}{2!} \)
\( P(X = 4) = \frac{e^{-\lambda} \cdot \lambda^4}{4!} \)
\( P(X = 6) = \frac{e^{-\lambda} \cdot \lambda^6}{6!} \)
Step 2: Substitute into the given equation
\[
\frac{e^{-\lambda} \cdot \lambda^2}{2!} = 2\left[\frac{e^{-\lambda} \cdot \lambda^4}{4!}\right] + 20\left[\frac{e^{-\lambda} \cdot \lambda^6}{6!}\right]
\]
Cancel \( e^{-\lambda} \) from both sides:
\[
\frac{\lambda^2}{2} = 2\left(\frac{\lambda^4}{24}\right) + 20\left(\frac{\lambda^6}{720}\right)
\]
Step 3: Simplify
\[
\frac{\lambda^2}{2} = \frac{\lambda^4}{12} + \frac{\lambda^6}{36}
\]
Multiply both sides by 36:
\[
18\lambda^2 = 3\lambda^4 + \lambda^6
\]
Step 4: Rearranging
\[
\lambda^6 + 3\lambda^4 - 18\lambda^2 = 0
\]
Factor out \( \lambda^2 \):
\[
\lambda^2(\lambda^4 + 3\lambda^2 - 18) = 0
\]
Let \( y = \lambda^2 \):
\[
y^2 + 3y - 18 = 0
\]
Factoring gives:
\[
(y + 6)(y - 3) = 0
\]
Thus,
\[
y = 3 \quad \Rightarrow \quad \lambda^2 = 3 \quad \Rightarrow \quad \lambda = \sqrt{3}
\]
Step 5: Coefficient of Skewness
For a Poisson distribution, the formula for skewness is:
\[
\text{Skewness} = \frac{1}{\sqrt{\lambda}}
\]
Substituting \( \lambda = \sqrt{3} \):
\[
\text{Skewness} = \frac{1}{\sqrt{3}}
\]
Final Answer
Correct option: (a) \( \frac{1}{\sqrt{3}} \)
Additional Insights on Poisson Distribution
Understanding Poisson Distribution
The Poisson distribution is essential in statistics. It models the number of events in a fixed interval. These events occur independently. The parameter \( \lambda \) represents the average number of events.
Applications of Poisson Distribution
You can find the Poisson distribution in various fields. It is used in queuing theory, telecommunications, and even in natural sciences. Understanding this distribution can help in solving real-world problems.
Importance of Skewness
Skewness measures the asymmetry of a distribution. A positive skew indicates a longer tail on the right. A negative skew indicates a longer tail on the left. For the Poisson distribution, skewness is always positive.
Conclusion
Mastering the Poisson distribution is crucial for competitive exams. It helps in understanding complex statistical concepts. Practice problems like these to enhance your skills.


Comments