A random variable X has the following probability distribution
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | 0 | k | 2k | 2k | 3k | \(k^2\) | \(2k^2\) | \(7k^2+k\) |
then value of E(X) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3.66
The probabilities of a distribution add up to 1:
\(0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1\)
\(10k^2 + 9k - 1 = 0\)
\((10k - 1)(k + 1) = 0\)
A probability cannot be negative, so k = −1 is rejected and
\( \displaystyle k = \frac{1}{10} \)
The expected value is \(E(X) = \sum x\,P(X)\):
\(E(X) = 1(k) + 2(2k) + 3(2k) + 4(3k) + 5(k^2) + 6(2k^2) + 7(7k^2 + k)\)
\(= 30k + 66k^2\)
Putting k = 0.1:
\(E(X) = 3 + 0.66 = 3.66\)
NCERT Exemplar:
Probability
