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.

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;

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()

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

P(5 X 7) = P(X = 5) + P(X = 6) + P(X = 7)

=

=

= 0.1948

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

step-by-step explanation:

if x = 2 you have to find x to finish the problem

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1