Question

Let mean and standard deviation of probability distribution

\[ \begin{array}{|c|c|c|c|c|} \hline \mathrm{X}=x & -3 & 0 & 1 & \alpha \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{1}{4} & \mathrm{~K} & \frac{1}{4} & \frac{1}{3} \\ \hline \end{array} \]

be \(\mu\) and \(\sigma\) respectively and if \(\sigma-\mu=2\) then \(\sigma=\)

• A. \(\frac{3}{2}\)
• B. \(\frac{5}{2}\)
• C. \(\frac{7}{2}\)
• D. \(\frac{9}{2}\)

Hint:

First find K from the sum of probabilities, then write the mean and variance in terms of alpha and use sigma minus mu equals 2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The probabilities must add to 1: \(\tfrac14+K+\tfrac14+\tfrac13=1\), so \(K=\tfrac16\).

Step 2: Mean and variance
\(\mu=-\tfrac34+0+\tfrac14+\tfrac\alpha3=\tfrac\alpha3-\tfrac12\).
\(E(X^2)=\tfrac94+\tfrac14+\tfrac{\alpha^2}{3}=\tfrac52+\tfrac{\alpha^2}{3}\).
\(\sigma^2=E(X^2)-\mu^2=\tfrac{\alpha^2}{9}+\tfrac{\alpha}{3}+\tfrac94\).

Step 3: Use the condition
\(\sigma=\mu+2=\tfrac\alpha3+\tfrac32\), so \(\sigma^2=\tfrac{\alpha^2}{9}+\alpha+\tfrac94\).
Equating: \(\tfrac{\alpha^2}{9}=\tfrac{2\alpha}{3}\), so \(\alpha=6\) (\(\alpha=0\) would repeat the value 0).

Step 4: Find sigma
\(\sigma=\tfrac63+\tfrac32=\tfrac72\).

Final Answer:
The standard deviation is \(\tfrac72\). \[ \boxed{\text{(C) }\dfrac{7}{2}} \]