A quality control inspector has drawn a sample of 15 light bulbs from a recent production lot. Suppose 20% of the bulbs in the lot are defective. What is the probability that exactly 3 bulbs from the sample are defective

0.2501

Step-by-step explanation:

Given : The proportion of the bulbs in the lot are defective = 20%= 0.20

Sample size : n =15

Let x be the binomial variable that represents the defective bulbs.

Binomial probability formula :

The probability that exactly 3 bulbs from the sample are defective : P(X=3)

Required probability = 0.2501

left 3, up 4

step-by-step explanation:

just count the values of x & y of the vertex of the graph and see how many units in the x and y direction will get you from the f(x) graph to the f(x) graph

126 maybe. sorry