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Integration by parts is required. The probability density
Chapter 4, Problem 49E(choose chapter or problem)
Problem 49E
Integration by parts is required. The probability density function for the diameter of a drilled hole in millimeters is 10e −10 (x − 5 ) for x > 5 mm. Although the target diameter is 5 millimeters, vibrations, tool wear, and other nuisances produce diameters greater than 5 millimeters.
(a) Determine the mean and variance of the diameter of the holes.
(b) Determine the probability that a diameter exceeds 5.1 millimeters.
Questions & Answers
QUESTION:
Problem 49E
Integration by parts is required. The probability density function for the diameter of a drilled hole in millimeters is 10e −10 (x − 5 ) for x > 5 mm. Although the target diameter is 5 millimeters, vibrations, tool wear, and other nuisances produce diameters greater than 5 millimeters.
(a) Determine the mean and variance of the diameter of the holes.
(b) Determine the probability that a diameter exceeds 5.1 millimeters.
ANSWER:
Solution:
Step 1 of 2:
We have the probability density function
f(x) = 10 , x > 5
- The claim is to find the mean and variance of the diameter of the holes.
Then, E(X) = x 10
dx
By computing indefinite integral
E(X) = x -
+ c
By applying integration by parts
u
= uv -
= 10 ( x ( ) -
1 . (-
)dx)
= 10 ( -x -
-
dx
Where, u = -10(x - 5), and du = -10 dx
Then,
E(X) = 10 ( -x -
-
du
= 10 ( -x - ( -
( -
-
du)))
Where, du =
Then, E(X) = 10 ( -x - ( -
( -
)))
= 10 ( -x - ( -
( -
)))
= - x -
= - x -
+ C
= 0 - ( - )
= 5.1
Then, V(X) =
U = and du =
V(X) = + 2
From the definition of E(X) the integral above is recognised equal to zero.
Then, V(X) =
= 0.01
Hence the mean is 5.1, and the variance is 0.01.