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Suppose error involved in making a certain measurement is continuous r.v. X with p.d.f. f(x)kfor –2 ≤ x ≤ 2,otherwisef(x)={k(4-x2) for –2 ≤ x ≤ 2,0 otherwise. P(–1 < x < 1)

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Question

Suppose error involved in making a certain measurement is continuous r.v. X with p.d.f.

`"f(x)" = {("k"(4 - x^2)      "for –2 ≤ x ≤ 2,"),(0                                 "otherwise".):}`

P(–1 < x < 1)

Sum
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Solution

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

∴ `int_(-2)^2 "f(x) dx" = 1`

∴ `int_(-2)^2  "k" (4 − x^2) "dx" = 1`

∴ `"k"[4x − x^3/3]_(-2)^(2) = 1`

∴ `"k"[(8 − 8/3) − (−8 + 8/3)] = 1`

∴ `"k"(16/3 + 16/3) = 1`

∴ `"k"(32/3) = 1`

∴ k = `3/32`

`"F(x)" = int_(-2)^(x) "f(x) dx"`

`= int_(-2)^(x) "k" (4 − x^2) "dx"`

`= 3/32 [4x − x^3/3]_(-2)^(x)`

`= 3/32 [4x − x^3/3 + 8 − 8/3]`

∴ `"F(x)" = 3/32 [4x − x^3/3 + 16/3]`

P(–1 < x < 1)

= F(1) – F(– 1)

= `3/32 (4  –  1/3 + 16/3) – 3/32 (– 4 + 1/3 + 16/3)`

= `3/32 (9  –  5/3)`

= `3/32 (22/3)`

= `11/16`

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Chapter 7: Probability Distributions - Exercise 7.2 [Page 239]

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Balbharati Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 7 Probability Distributions
Miscellaneous Exercise 2 | Q 13.2 | Page 244

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