Sum: 2k+4k+3k+k=10k=1, so k=0.1. Then E(X)=1(0.2)+2(0.4)+3(0.3)+4(0.1)=0.2+0.8+0.9+0.4=2.3.
A random variable X has the following probability distribution: | x | 1 | 2 | 3 | 4 | |---|---|---|---|---| | p(x) | 2k | 4k | 3k | k | The value of…
CUET UG 2022 — Mathematics Algebra
A random variable X has the following probability distribution:
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| p(x) | 2k | 4k | 3k | k |
The value of E(X) is:
Held on 16 Jul 2022 · Verified 13 Jul 2026.
Options
- A
0.23
- B
0.1
- C
1.6
- D
2.3
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