The manufacturing of semiconductor chips produces 2% defective chips. Assume the chips are independent and that a lot contains 1000 chips. (a) Approximate the probability that more than 25 chips are defective.
(b) Approximate the probability that between 20 and 30 chips are defective.

P(X>25)=0.107488

P(20<X<30)= 0.440427

0r 0.53490

Explanation :

Given from the question that the percentage of defective chips produced in the manufacturing of semiconductor is 2%

We are to Assume that the chips are independent and that a lot contains 1000 chips.

Then the following probabilities are to be estimated

a)More than 25 chips are defective

b)Between 20 and 30 chips are defective.

The percentage of defective chips produced in the manufacturing of semiconductor is 2%, therefore, our p=0.02.

The total number of chips, hence our n=1000.

Let X denote the number of defective chips in the manufacturing of semiconductor. So we will be able to calculate our mean

So the mean of X,

np= (10000.02). = 20

The variance of X can also be calculated as

np(1-p)= (10000.020.98). = 19.6

nq= n(1-p)=1000(1-0.2)=980 >5

Since np>5 and n(1-p)>5, then the requirements are satisfied

a) the z- score which is the value decreased by the mean value np and divided by the standard deviation now = np(1-p)

Z=

can be determined here

Z= x-np/√np(1-p)

z= [25.5-1000(0.02)]/√[1000(0.02)(1-0.02)]

z=1.24

if we check the normal distribution table using P(X>25)= P(X>25.5) = P(z>1.24) then we have the value of 0.107488

b) z= 20.5-1000(0.2)/√1000(0.02)(1-0.02)

z= 0.11

z= x-np/√np-(1-np)

z= 29.5-1000(0.2)/√1000(0.02)(1-0.02) = 2.15

CHECK THE ATTACHMENT TO COMPLETE THE SOLUTION