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प्रश्न
Solve the following problem :
The following is the c.d.f of a r.v.X.
| x | – 3 | – 2 | – 1 | 0 | 1 | 2 | 3 | 4 |
| F (x) | 0.1 | 0.3 | 0.5 | 0.65 | 0.75 | 0.85 | 0.9 | 1 |
Find the probability distribution of X and P(–1 ≤ X ≤ 2).
उत्तर
P(X = –3) = F(–3) = 0.1
P(X = –2) = F(–2) – F(–3) = 0.3 – 0.1 = 0.2
P(X = –1) = F(–1) – F(–2) = 0.5 – 0.3 = 0.2
P(X = 0) = F(0) – F(–1) = 0.65 – 0.5 = 0.15
P(X = 1) = F(1) – F(0) = 0.75 – 0.65 = 0.1
P(X = 2) = F(2) – F(1) = 0.85 – 0.75 = 0.1
P(X = 3) = F(3) – F(2) = 0.9 – 0.85 = 0.05
P(X = 4) = F(4) – F(3) = 1 – 0.9 = 0.1
∴ The probability distribution of X is as follows:
| X = x | – 3 | – 2 | – 1 | 0 | 1 | 2 | 3 | 4 |
| P(X = x) | 0.1 | 0.2 | 0.2 | 0.15 | 0.1 | 0.1 | 0.05 | 0.1 |
P(–1 ≤ X ≤ 2)
= P(X = –1 or X = 0 or X = 1 or X = 2)
= P(X = –1) + P(X = 0) + P(X = 1) + P(X = 2)
= 0.2 + 0.15 + 0.1 + 0.1
= 0.55
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