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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). - Mathematics and Statistics

Sum

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).

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Solution

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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APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 8 Probability Distributions
Part I | Q 1.09 | Page 155
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