Question 637397
A. Let f(x)=kx^3 with x e {1,3}
f(1) = k*1^3 = k and f(3) = k*3^3 = 27k
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B. Determine k such that f(x)=kx^3 is a probability density function
Solve k + 27k = 1
28k = 1
k = 1/28
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C. Determine P(X<=2) x<= (Means less than or equal to)
P(x<=2) = P(x = 1) = k = 1/28
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D. Determine the expected value of X
E(x) = 1*(1/28) + 3(27/28) = 82/28 = 2.93
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E. Determine the variance of X::: 0.3712^2 = 0.1378
Cheers,
Stan H.