Answer:
Probability that fewer than 2 of these parts are defective is 0.604.
Step-by-step explanation:
We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.
A random sample of 7 parts produced by this machine is chosen.
The above situation can be represented through Binomial distribution;
where, n = number of trials (samples) taken = 7 parts
r = number of success = fewer than 2
p = probability of success which in our question is % of defective
parts produced by one of the machine, i.e; 19%
<em>LET X = Number of parts that are defective</em>
<u>So, it means X ~ Binom(n = 7, p = 0.19)</u>
Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)
P(X < 2) = P(X = 0) + P(X = 1)
=
=
= <u>0.604</u>
<em>Therefore, the probability that fewer than 2 of these parts are defective is 0.604.</em>
