The frequency function of the continuous stochastic variable X is defined as follows: 0 for x < 0 f(x) = k. (x2 - 6x) fo

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The frequency function of the continuous stochastic variable X is defined as follows: 0 for x < 0 f(x) = k. (x2 - 6x) fo

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The Frequency Function Of The Continuous Stochastic Variable X Is Defined As Follows 0 For X 0 F X K X2 6x Fo 1
The Frequency Function Of The Continuous Stochastic Variable X Is Defined As Follows 0 For X 0 F X K X2 6x Fo 1 (39.42 KiB) Viewed 104 times
The frequency function of the continuous stochastic variable X is defined as follows: 0 for x < 0 f(x) = k. (x2 - 6x) for 0 SXS 3 0 3<X »={«.« for 1) Show that the constant k must have the value requirements of a frequency function. 18 for f(x) to meet the 2) Draw the graph of f (x) for -3<x<6. sk.(x2 - 6x)dx = k·(4- x3 – 3x2) + C. b) Given the indefinite integral 1) Find the distribution function F(x). 2) Draw the graph of F(x) for -3 <x6. c) a) Calculate the me mean value u = E(X) and the variance o? = Var(X) b) Calculate the median Xm. i.e. the 50% fraction in the distribution of X: Xm=X0.50 d) 1) Calculate the probability P(XS 1.00). 2) Calculate the conditional probability P(2.00 SX 1.00 SX).
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