A quality control inspector has drawn a sample of 1616 light bulbs from a recent production lot. Suppose 30%30% of the bulbs in the lot are defective. What is the probability that exactly 44 bulbs from the sample are defective? Round your answer to four decimal places.

0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.

Step-by-step explanation:

For each bulb, there are only two possible outcomes. Either it is defective, or it is not. The probability of a bulb being defective is independent from other bulbs, so we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

A quality control inspector has drawn a sample of 16 light bulbs from a recent production lot.

This means that

What is the probability that exactly 4 bulbs from the sample are defective?

This is P(X = 4).

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[\tex]

[tex]P(X = 4) = C_{16,4}.(0.3)^{4}.(0.7)^{12} = 0.2040[\tex]

0.2040 = 20.40% probability that exactly 4 bulbs from the sample are defective.

3.4 pizzas

step-by-step explanation:

given is a scatter plot showing number of pizzas sold during weeks for different issues of coupon

y represents number of pizzas sold and x number of coupons issued

y is marked verticaly and x horizontally

the trend line is given as

y =3.4x+43

this says that even when no coupon is issued pizzas sold will be 43

and rate of change of pizza sold for 1 unit of coupon = slope = 3.4

i.e. when x increases by 1 unit y increases by 3.4 units

hence for each additional coupon issued 3.4 pizzas more will be sold