Math, asked by saranshsharma222001, 5 months ago

(b) A random variable gives measurements X
between 'O' and '1' with a probability function
f(x) = 12x3 - 21x2 +10x and 0, 0<xsi
(i) Find PX
Find P(x * ) and P(x)
1
(ii) Find a number k such that P(X <k)
2​

Answers

Answered by formless
6

Answer:

i) P(X<=1/2) = 9/16 and P(X>1/2) = 7/16

Step-by-step explanation:

Given data:

f(x) = 12x^3 - 21x^2 + 10x ; 0<=x<=1

i)  

Now,

P(x<=1/2) = lim 0 to 1/2 [∫(12x^3 - 21x^2 + 10x)dx]    

              = lim 0 to 1/2 [3x^4-7x^3+5x^2]

              = 3{1/16-0} - 7{1/8-0} + 5{1/4-0}

              = (3 - 14 + 20)/16

              = 9/16.

And,

P(x>1/2) = lim 1/2 to 1 [∫(12x^3 - 21x^2 + 10x)dx]

             = lim 1/2 to 1 [3x^4-7x^3+5x^2]

             = 3{1-1/16} - 7{1-1/8} + 5{1-1/4}

             = 45/16 - 49/8 + 15/4

             = (45 - 98 + 60)/16

             = 7/16.

ii)

P(x<=k) = 1/2     (given)

=> lim 0 to k [∫f(x)dx] = 1/2

=> lim 0 to k [∫(12x^3 - 21x^2 + 10x)dx] = 1/2

=> lim 0 to k [3x^4-7x^3+5x^2] = 1/2

=> 3k^4 - 7k^3 + 5k^2 = 1/2

=> k^2 (3k^2 - 7k + 5) = 1/2

=> k = 0.45175, k = -2.6587.

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