Question:
A random variable \(X\) has the following probability distribution :
\[ \begin{array}{|l|c|c|c|c|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & 0.1 & 0.2 & 0.3 & 0.4 \\ \hline \end{array} \]
The mean and standard deviation of \(X\) are respectively
A random variable \(X\) has the following probability distribution :
\[ \begin{array}{|l|c|c|c|c|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & 0.1 & 0.2 & 0.3 & 0.4 \\ \hline \end{array} \]
The mean and standard deviation of \(X\) are respectively
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Find the mean as the sum of x times P, then the variance as E(X squared) minus mean squared.
Updated On: Nov 13, 2026
- 2 and 3
- 3 and 1
- 3 and \(\sqrt{2}\)
- 2 and 1
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Mean \(\mu=\sum xP(x)\) and variance \(\sigma^2=\sum x^2P(x)-\mu^2\).
Step 2: Mean
\(\mu=1(0.1)+2(0.2)+3(0.3)+4(0.4)=0.1+0.4+0.9+1.6=3\).
Step 3: Variance
\(\sum x^2P=0.1+0.8+2.7+6.4=10\), so \(\sigma^2=10-9=1\).
Step 4: Result
The standard deviation is \(\sigma=1\), so mean 3 and SD 1, option (B).
Final Answer:
The mean is 3 and the standard deviation is 1. \[ \boxed{\text{(B) }3\text{ and }1} \]
Mean \(\mu=\sum xP(x)\) and variance \(\sigma^2=\sum x^2P(x)-\mu^2\).
Step 2: Mean
\(\mu=1(0.1)+2(0.2)+3(0.3)+4(0.4)=0.1+0.4+0.9+1.6=3\).
Step 3: Variance
\(\sum x^2P=0.1+0.8+2.7+6.4=10\), so \(\sigma^2=10-9=1\).
Step 4: Result
The standard deviation is \(\sigma=1\), so mean 3 and SD 1, option (B).
Final Answer:
The mean is 3 and the standard deviation is 1. \[ \boxed{\text{(B) }3\text{ and }1} \]
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