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[Solved]Continuous random variable={-b-(b^2-4ac)^.5}/(2a)=x?
A factory is supplied with grain at the beginning of ea week.The weekly demand,X thousand tonnes for grain from this factory is a continuous random variable having the probability density function given by
f(x)=2(1-x),0<x<1
0 ,otherwise
Find the quantity of grain in tonnes the factory should've in stock in the beginning of a week in order to be 98% certain that the demand in that week will be met.
*I've found 2 answers but the answer sheet only chose 1 of 'em,found by using the (-b-(b^2-4ac)^.5)/(2a)
(-b+(b^2-4ac)^.5)/(2a)
I've integrated f(x) with the lower x value and higher x value 0 and 1 respectively and got:
[2x-x^2]=0.98
0=x^2-2x+0.98
(-b+(b^2-4ac)^.5)/(2a)
x=1.14 and 0.859
*I don't know why the answer sheet chose only the 0.859,why?
Homework Statement
A factory is supplied with grain at the beginning of ea week.The weekly demand,X thousand tonnes for grain from this factory is a continuous random variable having the probability density function given by
f(x)=2(1-x),0<x<1
0 ,otherwise
Find the quantity of grain in tonnes the factory should've in stock in the beginning of a week in order to be 98% certain that the demand in that week will be met.
*I've found 2 answers but the answer sheet only chose 1 of 'em,found by using the (-b-(b^2-4ac)^.5)/(2a)
Homework Equations
(-b+(b^2-4ac)^.5)/(2a)
The Attempt at a Solution
I've integrated f(x) with the lower x value and higher x value 0 and 1 respectively and got:
[2x-x^2]=0.98
0=x^2-2x+0.98
(-b+(b^2-4ac)^.5)/(2a)
x=1.14 and 0.859
*I don't know why the answer sheet chose only the 0.859,why?
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