Question
Download Solution PDFA discrete random variable X has the probability functions as:
|
X |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
|
f(x) |
K |
2k |
3k |
5k |
5k |
4k |
3k |
2k |
k |
The value of E(X) is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
|
x |
F(x) |
xf(x) |
|
0 |
k |
0 |
|
1 |
2k |
2k |
|
2 |
3k |
6k |
|
3 |
5k |
15k |
|
4 |
5k |
20k |
|
5 |
4k |
20k |
|
6 |
3k |
18k |
|
7 |
2k |
14k |
|
8 |
k |
8k |
|
Total |
26k |
103k |
We know that for Fpr random variable X sum of Probaboility = 1
⇒ ∑Pi = 1
⇒ k + 2k + 3k + 5k + 5k + 4k + 3k + 2k + k = 1
⇒ 26k = 1
⇒ k = 1/26
∴ xf(x) = 103k = 103 × 1/26
∴ E(x) is 103/26
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