Answer:
(a) P(X 20) = 0.9319
(b) Expected number of defective light bulbs = 15
Step-by-step explanation:
We are given that a manufacturer of Christmas light bulbs knows that 10% of these bulbs are defective. It is known that light bulbs are defective independently. A box of 150 bulbs is selected at random.
Firstly, the above situation can be represented through binomial distribution, i.e.;
where, n = number of samples taken = 150
r = number of success
p = probability of success which in our question is % of bulbs that
are defective, i.e. 10%
<em>Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.</em>
So, Let X = No. of defective bulbs in a box
<u>Mean of X</u>, =
=
= 15
<u>Standard deviation of X</u>, =
=
= 3.7
So, X ~ N(
Now, the z score probability distribution is given by;
Z = ~ N(0,1)
(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X 20) = P(X < 20.5) ---- using continuity correction
P(X < 20.5) = P( <
) = P(Z < 1.49) = 0.9319
(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = =
= 15.
