Question

A quality control inspector has drawn a sample of 16 light bulbs from a recent production lot. Suppose 20% of the bulbs in the lot are defective. What is the probability that between 5 and 7 (both inclusive) bulbs from the sample are defective? Round your answer to four decimal places.

Answer 1

Answer:

Probability that between 5 and 7 (both inclusive) bulbs from the sample are defective is 0.1948.

Step-by-step explanation:

We are given that quality control inspector has drawn a sample of 16 light bulbs from a recent production lot.

Suppose 20% of the bulbs in the lot are defective.

The above situation can be represented through Binomial distribution;

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

where, n = number of trials (samples) taken = 16 light bulbs

            r = number of success = between 5 and 7 (both inclusive)

           p = probability of success which in our question is % of bulbs in

                 the lot that are defective, i.e; 20%

LET X = Number of bulbs that are defective

So, it means X ~ Binom($n=16,p=0.20$)

Now, probability that between 5 and 7 (both inclusive) bulbs from the sample are defective is given by = P(5 $\leq$ X $\leq$ 7)

P(5 $\leq$ X $\leq$ 7) = P(X = 5) + P(X = 6) + P(X = 7)

= $\binom{16}{5}\times 0.20^{5} \times (1-0.20)^{16-5}+ \binom{16}{6}\times 0.20^{6} \times (1-0.20)^{16-6}+ \binom{16}{7}\times 0.20^{7} \times (1-0.20)^{16-7}$

= $4368 \times 0.20^{5} \times 0.80^{11} +8008 \times 0.20^{6} \times 0.80^{10} +11440 \times 0.20^{7} \times 0.80^{9}$

= 0.1948

Hence, the probability that between 5 and 7 (both inclusive) bulbs from the sample are defective is 0.1948.