**Given W is a uniformly distributed random variable with mean 33 and variance 3. determine: (a) probability density function for W (b) cumulative distribution function for W**

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(a) probability density function for W

The probability distribution function of a uniform distribution is defined as below

f(w) = 1/ (b-a) for a≤ w ≤ b

The mean = ( a + b)/ (3) = 33 for our case.

The variance = (b-a)2/12 = 3 for our case.

solving the two equations above to obtain a and b we have

a + b = 99 (first equation)

(b - a)2 = 36 which becomes (b-a) = 6 which becomes b = a + 6 (second equation)

substituting the second equation in the first equation, we have

a + a + 6 = 99

2a = 93

a = 46.5

Thus b = (46.5 + 6) = 52.5

substituting b = 52.5, a = 46.5 in f(x) = 1/ (b-a) for a≤ w ≤ b, we obtain

f(w) = 1/6 for 46.5 ≤ w ≤ 52.5

0, otherwise

which is the required probability distribution of w

(b) cumulative distribution function for W

we integrate the probability distribution function of w to obtain the cumulative distribution function for W .

we obtain as below

$\smallint$ 46.5w (1/6) dw

integrating we obtain (1/6)w46.5 = ( (1/6)w - 7.75 )

Thus , the cumulative distribution of w is F(w) = ( (1/6)w - 7.75 ) 46.5 ≤ w ≤ 52.5

0, otherwise