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प्रश्न
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 and = 0 otherwise.
P (–0·5 < x or x > 0·5)
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उत्तर
Since, f is the p.d.f. of X,
` int_(-∞)^∞ f (x) dx` = 1
∴ ` int_(-∞)^(-2) f (x) dx` +` int_(-2)^(2) f (x) dx` + `int_2^∞f(x) dx = 1`
= 0 + ` int_(-2)^(2) k (4 -x^2) dx + 0 = 1`
∴ k ` int_(-2)^2 (4 -x^2) dx + 0 = 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`
P (–0·5 < x or x > 0·5)
= P (x < –0·5) + P (x > – 0·5)
= `int_(-∞)^-0.5f (x) dx + int_(0.5)^∞f (x) dx`
=` int_(-∞)^-2 f (x) dx + int_(-2)^-0.5 f (x) dx +int_(0.5)^2f (x) dx + int_(2)^∞ f (x) dx`
= 0+` int_(-2)^(-1/2) k (4 -x^2) dx+ int_(1/2)^2 k (4 -x^2) dx` + 0
= ` k int_(-2)^(-1/2) (4 -x^2) dx+ k int_(1/2)^2 (4 -x^2) dx`
= `3/32[4x-(x^3)/3]_-2^(-1/2)+3/32[4x-(x^3)/3]_(1/2)^2` .......[∵ k =`3/32`]
= `3/32[(-2+1/24)-(-8+8/3)] + 3/32[(8-8/3)-(2-1/24)]`
= `3/32((-47)/24+16/3)+ 3/32(16/3-47/24)`
= `3/32((-47)/24+16/3+16/3-47/24)`
= `3/32((-47+128+128 -47)/24)`
= `3/32(162/24) = 81/128`
= 0.6328
Alternative Method :
P (x< – 0·5 or x > 0·5)
= 1 -P( - 0.5 ≤ x ≤ 0.5)
= 1 -` int_(-0.5)^0.5 f (x) dx `
= 1 - ` int_(-1/2)^(1/2) k (4 - x^2) dx`
= 1 - k ` int_(-1/2)^(1/2) (4 - x^2) dx`
= `1-3/32[4x-x^3/3]_(-1/2)^(1/2)` ......[∵ k = `3/32`]
= `1 - 3/32[(2-1/24)-(-2+1/24)]`
= `1 - 3/32(2-1/24+2-1/24)`
= `1 - 3/32(4-1/12)`
= `1 - 3/32 xx 47/12`
= `1 - 47/128`
= `(128-47)/128`
= `81/128`
= 0.6328
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