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.

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are u sure the probability 0.01 ?

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2014-08-27T06:43:53+05:30

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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 \\

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