MHT CET · Maths · Probability
A random variable X has the following probability distribution :
\(\begin{array}{|c|c|c|c|c|} \hline \mathrm{X}=x & 1 & 2 & 3 & 4 \\ \hline \mathrm{P}(\mathrm{X}=x) & 0 \cdot 1 & 0 \cdot 2 & 0 \cdot 3 & 0 \cdot 4 \\ \hline \end{array}\)
The mean and standard deviation of X are respectively
- A 2 and 3
- B 3 and 1
- C 3 and \(\sqrt{2}\)
- D 2 and 1
Answer & Solution
Correct Answer
(B) 3 and 1
Step-by-step Solution
Detailed explanation
\( \mu = E[X] = \sum x P(X=x) \) \( \mu = (1)(0.1) + (2)(0.2) + (3)(0.3) + (4)(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3 \)
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