Answer:
(a) P(X  20) = 0.9319
 20) = 0.9319
(b) Expected number of defective light bulbs = 15
(c) Standard deviation of defective light bulbs = 3.67
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
 = 15
<u>Standard deviation of X</u>,  =
 =  =
 =  = 3.7
 = 3.7
So, X ~ N(
Now, the z score probability distribution is given by;
                 Z =  ~ N(0,1)
 ~ 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
 20) = P(X < 20.5)  ---- using continuity correction
     P(X < 20.5) = P(  <
 <  ) = P(Z < 1.49) = 0.9319
 ) = 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.
 = 15.
 Standard deviation of defective light bulbs is given by = S.D. =  =
 =  = 3.67
 = 3.67