A quality control inspector has drawn a sample of 11 light bulbs from a recent production lot. Suppose 20% of the bulbs in the lot are defective. What is the probability that less than 7, but more than 4 bulbs from the sample are defective

0.04845

Step-by-step explanation:

Probability that less than 7 but more than 4 bulbs from the sample are defective :

P(4 < X < 7) = P(x = 5) + P(x = 6)

Using the binomial probability formula :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

From the question :

p = 0.2 ; 1 - p = 0.8 ; n = 11

P(x =5) = 11C5 * 0.2^5 * 0.8^6 = 0.03876

P(x = 6) = 11C6 * 0.2^6 * 0.8^5 = 0.00969

P(4 < X < 7) = P(x = 5) + P(x = 6)

P(4 < X < 7) = 0.03876 + 0.00969

P(4 < X < 7) = 0.04845

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

i dont know

step-by-step explanation: