Question

Consider a continuous random variable X with the density function (exponential) ?(?)={?^−? ?? ?≥0 , 0...

Consider a continuous random variable X with the density function (exponential) ?(?)={?^−? ?? ?≥0 , 0 ??ℎ??????}
a) Find and sketch the CDF for X
b) Find the mean and variance of X (I want to see your calculation)
c) Find ?(1≤?≤2)

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Answer #1

(a) The probability density functi on of the exponential distributi on is 0; otherwise (1). e-%20 f (x) , otherwise The cumulative di stribution function is 1-e).xz0 0, otherwise 0, otherwise The mean of the exponential di stribution is E(X) 1 The rate parameter andisA-1 The Step-by-Step Instructions to obtain a Cumulative di stribution function using Excel 2010 * Select the x and F(x) in the columns of the Excel2010 spread sheet * Click Insert and then click Line-> 2D-Line * Click Layout7 * Write the heading The cumulative distributi on function,F(x * Write the heading of the X -axisis *Write the heading of the Y -axis is: CDF,F( * Click ok The cumulative distribution function using Excel 2010

The cumulative distribution,F(X) 0.8 L 0.6 0.4 0.2 F(X) 1 23 4 567 8 9

Interpretation:

The measures of the central tendency of the exponential distribution are mean and variance. The mean of the exponential distribution is E(X)=1, the parameter of the exponential distribution is lambda. The variance of the exponential distribution is Var(X)=1.

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(b) The formula for calcul ating the mean of the continuous random variable(x)is Therefore, the mean is E[X)-1 The formula for calculating the mean of the continuous random variable(x)is The formula for caclulating the variance of X using the expectations is Var(X)=E(X)-| E(X Therefore, the variance of X is Var(X)-

(c) The probability that the random vari able is between 1 and 2 is 1-e-1+e 2-1 >e-1-e-2 0.367879441-0.135335 0.232544158 P(1sX S2)0.23254 Therefore, the probability that the random vaiable is between 1 and 2 is P (1SXS2)s |023254|

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