Answer:
0.3830,0.6170
Step-by-step explanation:
Given that a process for manufacturing an electronic component yields items of which 1% are defective.
n =100 and p = 0.01
Here X no of defectives is binomial since independence and two outcomes.
Approximation to normal would be
X is N([tex]100(0.01) , \sqrt{100(0.01)(0.99)} )[/tex]
X is N(1,0.995)
a) the probability that the process continues given the sampling plan described
= [tex]P(X=0)[/tex]
(with continuity correction)
=[tex]P(|x|<0.5)\\=P(|Z|<0.50)\\= 2(0.1915)\\=0.3830[/tex]
b) the probability that the process continues even if the process has gone bad (i.e., if the frequency of defective components has shifted to 5.0% defective)
1-0.3830
=0.6170
