Question

60 % of the chips used by a computer manufacturing company (CMC) are provided by supplier A and the rest by supplier B. The two suppliers have the following faulty chip rates: 0.3% for supplier A and 0.8% for supplier B. Consider a random batch of 20 chips (chips cannot be distinguished by supplier). a. Show that the expected number of faulty chips is 0.1. [15 marks] b. What is the probability of having two or more faulty chips? Justify your answer and the choice of the probability distribution. [15 marks] If the daily production of computer is 1000, and each computer has one chip: c. What is the probability that 8 or more computers do not pass the final quality check due to faulty chips? Justify your approach. [20 marks] The company, CMC, has to decide whether to implement a quality control screening of chips when delivered by the suppliers. In this case, only the chips that pass the screening will be used in the production process. Previous experience of quality control screenings reported that in 95% of the cases it was able to correctly identify faulty chips and in 3% of the cases was giving false negatives (meaning that even though the chip was working correctly the screening identified it as faulty). The screening process will be convenient only if the average number of daily faulty computers will be smaller than 3. d. Should the company implement the quality control screening? [20 marks]

Answer 1

(a) Let us first calcuate the probability of a faulty chip for CMC.

Suppose a random chip is selected, then it is either from supplier A or B accordingly, the probability of being faulty is different.

P[Chip is faulty] = P[Chip is from supplier A] * P[Chip is faulty] + P[Chip is from supplier B] *P[Chip is faulty]

P[Chip is faulty] = 0.6*0.3/100 + 0.4*.8/100 = 0.005

For a random sample of 20 chips, assuming the total population is sufficiently large, we can consider the distribution of number of faulty chips to be binomial with n = 20 and probability of success (being faulty) p = 0.005

Therefore, using the standard formula for mean of binomial , the expected number of faulty chips is given by

E[Chip is faulty] = n*p = 20 * 0.005 = 0.1

(b) What is the probability of having two or more faulty chips?

P[2 or more chips are faulty] from sample of 20 random chips. Again, For this we use the binomial distribution. The number of faulty chips is distributed as a binomial with probability p = 0.005 (as derived in (a) and n = 20.

Assumptions: This is due to assumption that the total population of chips is sufficiently large, and the chips from suppliers A and B are not distinguishable and randomly selected.  Therefore, even for sample of 20 without replacement, we can use the binomial distribution.

P[2 or more chips are faulty] = 1 - P[ less than 2 chips are faulty ] = 1 - { P[faulty chips = 0] + P[faulty chips = 1] }

we can evaluate the 2 probabilities using below formula

where X represents the random variable giving number of faulty chips.

n = 20

p = 0.005

We get P[X=0] = 0.90461

We get P[X=1] = 0.090916

P[2 or more chips are faulty] = 1 - 0.90461 - 0.090916 = 0.004474

(c) If the daily production of computer is 1000, and each computer has one chip: What is the probability that 8 or more computers do not pass the final quality check due to faulty chips?

use the same approach as a and b with same probability of faulty chip p = 0.005. But since each computer has one chip, the sample size n = 1000.

Using these p and n in above formula, we use Excel to evaluate the below

P[ 8 or more computers have faulty chips] = 1 - P[ 7 or less have faulty chips] =

1- {0.0067+0.0334+0.0839+0.1403+0.1757+0.1759+0.1466+0.1046}

which gives

P[ 8 or more computers have faulty chips] = 0.1329

(D) Should the company implement the quality control screening? [20 marks]

Screening process consists of two parts. Consider a randomly selected computer chip. Then probability of it being faulty as discussed above is p =0.005

(1) The chip is really faulty and the screen identifies chip as faulty (True Positive) = 0.005 * 0.95%

or (2) The chip is not faulty, but the screen identifies chip as faulty (False Negative) = (1-0.005)* 3%

Therefore

Prob{Screening identifies faulty] = 0.005*95% + (1-0.005)*3% = 0.0346

Therefore during quality control there is much greater chance of the screening identifying a chip as faulty that the true probability p.

Only the chips that pass the screening will be used in the production process. The screening process will be convenient only if the average number of daily faulty computers will be smaller than 3.

Daily production = 1000

Therefore expected number of daily faulty computers is given by expected value of binomial distribution with n = 1000

and probability of success = 0.0346

By standard formula still using binomial distribution , we get a average number of daily faulty computers = 0.0346*1000

Average = 34.6

Since this number is much larger than that desired by CMC, company should not implement the quality control screening.