The manufacturing of semiconductor chips produces 2% defective chips. Assume that the chips are independent and that a lot contains 1000 chips. Approximate the following probabilities: .a. More than 25 chips are defective. b. Between 20 and 30 chips are defective.

a) The approximate probability that more than 25 chips are defective is 0.1075.

b)  The approximate probability of having between 20 and 30 defecitve chips is 0.44.

Step-by-step explanation:

Lets call X the total amount of defective chips. X has Binomial distribution with  parameters n=1000, p =0.02. Using the Central Limit Theorem, we can compute approximate probabilities for X using a normal variable with equal mean and standard deviation.

The mean of X is np = 1000*0.2 = 20, and the standard deviation is √np(1-p) = √(20*0.98) = 4.427

We will work with a random variable Y with parameters μ=20, σ=4.427. We will take the standarization of Y, W, given by

The values of the cummmulative distribution function of the standard normal random variable W, which we will denote , can be found in the attached file. Now we can compute both probabilities. In order to avoid trouble with integer values, we will correct Y from continuity.

a)

Hence the approximate probability that more than 25 chips are defective is 0.1075.

b)

As a result, the approximate probability of having between 20 and 30 defecitve chips is 0.44.

based on context clues you gave i'm assuming the answer would be c.

answer: he is 272195

step-by-step explanation: all you have to do is multiple 3535 and 77 and it equals 272195