Question

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. Answer the following questions. Question 2 (4 points). 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). Question 3 (2 points): How many defective light bulbs are expected to be found in such boxes, on average? Show your work Question 4 (3 points): Find the standard deviation of defective light bulbs in this case. Show your work. (Round your answer to 2 places after the decimal point).

Answer 1

Answer:

(a) P(X $\leq$ 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.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

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, $\mu$ = $n \times p$ = $150 \times 0.10$ = 15

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.7

So, X ~ N($\mu = 15, \sigma^{2} = 3.7^{2})$

Now, the z score probability distribution is given by;

                Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that this box will contain at most 20 defective light bulbs is given by = P(X $\leq$ 20) = P(X < 20.5)  ---- using continuity correction

    P(X < 20.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20.5-15}{3.7}$ ) = P(Z < 1.49) = 0.9319

(b) Expected number of defective light bulbs found in such boxes, on average is given by = E(X) = $n \times p$ = $150 \times 0.10$ = 15.

Standard deviation of defective light bulbs is given by = S.D. = $\sqrt{np(1-p)}$ = $\sqrt{150 \times 0.10 \times (1-0.10)}$ = 3.67