I have a pdf: 0 x<=4 kx 4<x<=9 0 x>9 I found k=2/65 So I thought the cdf would be (x^2)/32 but apparently not Can someone please help me with the cdf, thanks
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Hello, For , the cdf is 0. For , the cdf is 1. For , the cdf is Compute that again
ok thanks i've got that now. So for a cdf where 0<x<4 & 4<x<8 have different functions, what limits would i use?
would the limits just be 0 - t and 4 - t?
can someone please help me as I am almost sure that the limits for 0 to 4 and 4 to 8 would be either 0 to t and 4 to t or 0 to t and t to 8 but I dont know which 1, thanks
from 0 to 4, there is no probability I don't know why you keep writing 8, but the probabilities exist between 4 and 9. Hence F(x)=0 for any x less than 4, since f(t)=0 in that region.......... If 4<x<9, then and after 9, we have
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