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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Solve the following problem : The p.d.f. of the r.v. X is given by f(x) = {kxfor 0<x<4.0otherwise.Determine k, the c.d.f. of X, and hence find P(X ≤ 2) and P(X ≥ 1). - Mathematics and Statistics

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प्रश्न

Solve the following problem :

The p.d.f. of the r.v. X is given by

f(x) = `{("k"/sqrt(x), "for"  0 < x < 4.),(0, "otherwise".):}`
Determine k, the c.d.f. of X, and hence find P(X ≤ 2) and P(X ≥ 1).

बेरीज

उत्तर

Given that f(x) represents p.d.f. of r.v. X.

∴ `int_0^4 "k"/sqrt(x)*dx` = 1

∴ `"k"*[2sqrt(x)]_0^4` = 1

∴ `2"k"[sqrt(x)]_0^4` = 1

∴ 2k (2 – 0) = 1

∴ k = `(1)/(4)`
By definition of c.d.f.,
F(x) = P(X ≤ x)

= `int_0^4 "k"/sqrt(x)*dx`

= `"k"[2sqrt(x)]_0^x`

= `(1)/(4)[2sqrt(x)]_0^x`

= `sqrt(x)/(2)`

P(X ≤ 2) = F (2) = `sqrt(2)/(2)  = (1)/sqrt(2)`

P(X ≥ 1) = 1 – P(X < 1)
= 1 – F(1)

= `1 - (sqrt1)/(2)`

= `(1)/(2)`.

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Probability Distribution of a Continuous Random Variable
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Probability Distributions - Part I [पृष्ठ १५६]

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बालभारती Mathematics and Statistics 2 (Commerce) [English] 12 Standard HSC Maharashtra State Board
पाठ 8 Probability Distributions
Part I | Q 1.16 | पृष्ठ १५६

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