A quality control inspector has drawn a sample of 15 light bulbs from a recent production lot. Suppose that 60% of the bulbs in the lot are defective. What is the expected value of the number of defective bulbs in the sample

9

Step-by-step explanation:

Expected number of defective is given as :

N x P = np

Where = number of the sample drawn from the production lot

P= percentage of bulbs In The lot which are defective

N = 15

P = 60% = 0.60

So when we do the multiplication,

We have:

NP = 15x0.60

= 9

So in conclusion the expected value of defective bulbs in the sample is 9.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116