Help?? Due tomorrow!!
1 answer:
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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.
