A petrol pump is supplied the petrol once a day.if its daily volume of sells (x) in thousands of lires is distributed by f(x)=5(1-x)^4,0<=x<=1, what must be the capacity of its tank in order that its supply will be exhausted in a given day shall be 0.01.
kvnmurty:
are u sure the probability 0.01 ?
Answers
Answered by
43
Supply is exhausted in a day , means sales >= capacity.
Area under the curve f(x) gives the total probability P( x > Capacity) = 0.01
f(x) = 5 at x =0 and f(x) = 0 at x = 1. As x is > 0 (volume of sale) and x < 1, the integration limits are from C < x < 1.
Let 1 - x = t then - dx = dt
![\\ \int\limits^1_C {f(x)} \, dx = 0.01 \\ \\ \int\limits^1_C {5(1-x)^4} \, dx \\ \\ 5 \int\limits^0_{1-C} {t^4} \, (-dt) \\ \\ 5 * [ \frac{1}{5}* t^5 ] \\ \\ (1-C)^5 = 0.01 \\ \\ 1- C = 0.398 \\ \\ C = 0.602 * thousand liters = 602 liters \\](https://tex.z-dn.net/?f=+%5C%5C+%5Cint%5Climits%5E1_C+%7Bf%28x%29%7D+%5C%2C+dx+%3D+0.01+%5C%5C+%5C%5C++%5Cint%5Climits%5E1_C+%7B5%281-x%29%5E4%7D+%5C%2C+dx+%5C%5C+%5C%5C+5++%5Cint%5Climits%5E0_%7B1-C%7D+%7Bt%5E4%7D+%5C%2C+%28-dt%29+%5C%5C+%5C%5C+5+%2A+%5B+%5Cfrac%7B1%7D%7B5%7D%2A+t%5E5+%5D+%5C%5C+%5C%5C+%281-C%29%5E5+%3D+0.01+%5C%5C+%5C%5C+1-+C+%3D+0.398+%5C%5C+%5C%5C+C+%3D+0.602+%2A+thousand+liters+%3D+602+liters+%5C%5C+ "\\ \int\limits^1_C {f(x)} \, dx = 0.01 \\ \\ \int\limits^1_C {5(1-x)^4} \, dx \\ \\ 5 \int\limits^0_{1-C} {t^4} \, (-dt) \\ \\ 5 * [ \frac{1}{5}* t^5 ] \\ \\ (1-C)^5 = 0.01 \\ \\ 1- C = 0.398 \\ \\ C = 0.602 * thousand liters = 602 liters \\")
Area under the curve f(x) gives the total probability P( x > Capacity) = 0.01
f(x) = 5 at x =0 and f(x) = 0 at x = 1. As x is > 0 (volume of sale) and x < 1, the integration limits are from C < x < 1.
Let 1 - x = t then - dx = dt
Similar questions