Question

et the random variable X have a continuous uniform distribution with a minimum value of 115 and a maximum value of 165.

What isP(X>127.89|X<140.97)?

Round your response to at least 3 decimal places.

   

Answer 1

If X follows a Uniform distribution (a,b)

The probability density of X

P(X = x) =

And Cumulative Distribution function,

F(x) = P(X < x) =*

X follows Uniform Distribution (115,165)

- 11 - -

F(x) = P(XS x)=x-a b-a X-115 X-115 165 - 115 = 50

P(127.89 < X < 140.97) P(X> 127.89|X< 140.97) = P(X < 140.97)

P(127.89 < X <140.97) = P(X<140.97)-P(X<127.89)

25.97 F(140.97) = P(X < 140.97) == 140.97 - 115 50 50 0.5194

127.89 - 115 F(127.89) = P(X 127.89) = 12.89 50 50 = 0.2578

P(127.89 < X <140.97) = P(X<140.97)-P(X<127.89)=0.5194-0.2578=0.2616

P(127.89 < X < 140.97) 0.2616 P(X> 127.891X < 140.97) = P(X<140.97) =5100 = 0.503658067

P(X> 127.89|X < 140.97) = 0.503658067

P(X> 127.89|X<140.97) = 0.503658067 0.504

Ans:

P(X>127.89 | X<140.97) =0.504