Question

The error in the length of a part (absolute value of the difference between the actual length and the target length), in mm, is a random variable with probability density function 120x2-x 0 〈 x 〈 1 f(x) = 0 otherwise a. What is the probability that the error is less than 0.2 mm? b. Find the mean error. c. Find the variance of the error. d. Find the cumulative distribution function of the error. e.The specification for the error is 0 to 0.8 mm, What is the probability that the specification is met?

Answer 1

The corrected parts of the question has been attached to this answer.

Answer:

A) Probability that the error is less than 0.2 mm; P(X < 0.2) = 0.0272

B) Mean Error (E(X)) = 0.6

C) Variance Error (V(X)) = 0.04

D) Answer properly written in attachment (Page 2)

E) P(0<X<0.8) = 0.8192

Step-by-step explanation:

The probability density function of X is;

f(x) = { 12(x^(2) −x^(3) ; 0<x<1

So, due to the integral symbol and for clarity sake, i have attached all the explanations for answers A - D.

E) The probability that the specification for the error to be between 0 to 0.8 mm is met will be;

P(0<X<0.8) = F(0.8) − F(0) =12([(0.8)^(3)] /3] −[(0.8)^(4)]/4]

= 0.8192

So, the probability is 0.8192.