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.

a. Find the probability that this box will contain at most 20 defective light bulbs. Show your work or calculator input. (Round your answer to 4 places after the decimal point).
b. How many defective light bulbs are expected to be found in such boxes, on average

(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%

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.

So, Let X = No. of defective bulbs in a box

Mean of X, = = = 15

Standard deviation of X, = = = 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.

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.