Question

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.

Answer 1

Answer:

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

$W = \frac{Y-\mu}{\sigma} = \frac{Y-20}{4.427}$

The values of the cummmulative distribution function of the standard normal random variable W, which we will denote $\phi$ , 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)

$P(X > 25) = P(X > 25.5) \approx P(Y>25.5) = P(\frac{Y-20}{4.427} > \frac{25.5-20}{4.427}) =\\P(W > 1.2423) = 1-\phi(1.2423) = 1-0.8925 = 0.1075$

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

b)

$P(20$

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

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