Question
Download Solution PDFA random variable y has a known probability distribution given by
y | 2 | 4 | 6 | 8 | 10 |
P(y) | 0.17 | 0.23 | 0.2 | 0.3 | 0.1 |
Then the expected value of y is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation
We know expected value of random variable
f(x) = ∑xipi
where I = 1, 2, 3,-----n
f(x) = P1x1 + P2x2 + P3x3 + P4x4 + P5x5
⇒ f(x) = 0.17 × 2 + 0.23 × 4 + 0.2 × 6 + 0.3 × 8 + 0.1 × 10
∴ f(x) = 5.86
Last updated on Jul 2, 2025
-> ESE Mains 2025 exam date has been released. As per the schedule, UPSC IES Mains exam 2025 will be conducted on August 10.
-> UPSC ESE result 2025 has been released. Candidates can download the ESE prelims result PDF from here.
-> UPSC ESE admit card 2025 for the prelims exam has been released.
-> The UPSC IES Prelims 2025 will be held on 8th June 2025.
-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.
-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.