Based on extensive testing, it is determined by the manufacturer of a washing machine that the time Y (in years) before a major repair is required is characterized by the probability density function
𝑓(𝑦) = { 0.25 𝑒 ^−𝑦/4, 𝑦 ≥ 0
0, 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
(A) Find the mean and standard deviation of Y.
Alan
Favorite Answer
both standard deviation and mean = 4
mean =∫ integral y*f(y) dx
mean = from 0 to infinity ∫0.25y *e^(-y/4) dy
mean = from 0 to infinity 0.25 ∫y *e^(-y/4) dy
integration by parts
let u = y
du = 1
let v' = e^(-y/4)
v = 4e^(-y/4)
∫udv = uv - ∫vdu
∫y *e^(-y/4) dy =from 0 to inf. | (y*4e^(-y/4) - ∫e^(-y/4) dy
∫y *e^(-y/4) dy = from 0 to inf. | y*4e^(-y/4) - 16e^(-y/4)
∫y *e^(-y/4) dy = infintiy*e^(-infinity/4) - 16e^(-infinity/4)- (-16e^(0))
- 16e^(-infinity/4) =0
infinty*e^(-infinity/4)= infinity * zero
so you must use limit to show , it is really zero
y/ e^(y/4) = l'hopital rule 1/e^(y/4) = e^(-y/4) = 0
∫y *e^(-y/4) dy = 16e^0 = 16
mean = from 0 to infinity 0.25*16 = 4
mean = 4
variance = E(y^2) -(mean)^2 = E(y^2) -16
E(y^2) =0.25∫y ^2*e^(-y/4) dy
if you
let u = y^2
du = 2y
v' = e^(-y/4)
v = -4e^(-y/4)
E(x^2) = 0.25*[ y^2*-4e^(-y/4) - ∫2y*-4e^(-y/4) dy ]
E(x^2) = 0.25*[ y^2*-4e^(-y/4) - 8∫y*e^(-y/4) dy ]
we already did ∫y *e^(-y/4) dy and know it equals -16
and again the first term contributes 0 if you work it out again
E(x^2) = 0.25 *[0 -(-8*16) ] = 32
Variance = 32 -16 = 16
standard deviation = sqrt(variance) = sqrt(16) = 4