Question

Suppose that f(x) - or 0<X<8 256 Determine the following probabilities. Round your answers to 3 decimal places (e.g. 98.765) (a) P(X < 2)=7456 (b) P(X< 9) = (d) P(X > 5)- 316 (e) Determine such that P(x x)-0.90 X6.302

Answer 1

e]

Determine X such that P( X < x) = 0.90

Consder,

P( X < x) = 0.90

$\int_{0}^{x}f(x)dx$ = 0.90

$\int_{0}^{x} \frac{3(8x-x^2)}{256} dx$ = 0.90

$\frac{3}{256}{[\frac{8x^2}{2}-\frac{x^3}{3}]}_{0}^{x}$ = 0.90

${[\frac{24x^2}{6}-\frac{2x^3}{6}]} = 0.9*\frac{3}{256}$

$24x^2-2x^3 = 0.9*\frac{3}{256}*6 = 0.06328$

$12x^2-x^3 = 0.03164$

$x^3-12x^2&plus;0.03164 = 0$

After solving this equation, we get X = -0.05124, 0.05146 and 11.9998

But if we consider the range of x for this perticulater density 0 < x < 8 then there is only one value of X is possible that is = 0.05146.

Hence X = 0.05146, is the final answer