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Find k, if the following function represents p.d.f. of r.v. X.

f(x) = kx(1 – x), for 0 < x < 1 and = 0, otherwise.

Also, find `P(1/4 < x < 1/2) and P(x < 1/2)`.

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#### Solution

Since, the function f is the p.d.f. of r.v. X,

` int_(-∞)^∞ f(x)dx` = 1

∴ ` int_(-∞)^0 f(x) dx + int_(0)^1 f(x)dx + int_(1)^∞f (x)dx` = 1

∴ `0 + int_(0)^1 kx(1 - x)dx + 0` = 1

∴ `k int_(0)^1 (x - x^2)dx` = 1

∴ `k [x^2/2 - x^3/3]_0^1` = 1

∴ `k (1/2 - 1/3 - 0)` = 1

∴ `k/6` = 1

∴ k = 6.

`P(1 /4 < x < 1 /2)` = ` int_(1/4)^(1/2) f(x)dx`

= ` int_(1/4)^(1/2) kx(1 - x)dx`

= k` int_(1/4)^(1/2) (x - x^2)dx`

= `6[x^2/2 - x^3/3]_(1/4)^(1/2)` ..........[∵ k = 6]

= `6[(1/8 - 1/24) - (1/32 - 1/192)]`

= `6[2/24 - 5/192]`

= `6(11/192)`

∴ `P(1 /4 < x < 1 /2)` = `11/32`

`P(x < 1/2) = int_(-∞)^(1/2) f(x)dx`

= `int_(-∞)^0 f(x)dx + int_0^(1/2) f(x)dx`

= `0 + int_(0)^(1/2) kx(1 - x)dx`

= `kint_(0)^(1/2) (x - x^2)dx`

= `k[x^2/2 - (x^3)/3]_0^(1/2)`

=`k[1/8 - 1/24 - 0]`

= `k(2/24)`

= `6(1/12) ` ............[∵ k = 6]

∴ `P(x < 1/2) = 1/2`

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