Question

For the probability density function f defined on the random variable x, find (a) the mean of x, (b) the standard deviation of x, and (c) the probability that the random variable x is within one standard deviation of the mean. f(x) = 1 30 x, [2,8] a) Find the mean. u = (Round to three decimal places as needed.) b) Find the standard deviation. = (Round to three decimal places as needed.) c) Find the probability that the random variable x is within one standard deviation of the mean. The probability is 17

Answer 1

The solution to this problem is given below

The PDF ob pandom variable Х is given by - 22 2s x 58. 30 Now, mean of X is, M E(x) ferica) de ha fade de 2 8 30 23 30 3 8 - 504 90 5.6 U = 5.6.

The standard deviation of X is, var ) E) - CECAST x² f(x) dx 3 2. x dx 30 2 8 3 dx 30 4. 4 30 2 4 4 To 4080/120 = 34 var (x) = f(x?) - ELS = 34-5.62 2.64 i The standard deviation of x is, √ 2.64 1.625 = 1.625 2019

pandom Now, the probability that the variable the within one © X is standard deviation of the mean is a P(4-0 < x <uto) P ( 5.6-10 5.6 -1.625 < x < 5.6 + 1.625 P (3.975 < x < 7.225 V 7.225 1 x dx 30 3.975 7.225 30 30075 7.225 30 - fx x dx HE 7.225°- 39733] 3h 36.4/60 13.975 [ 60 0.607 The probability is 0.607.