Question 05 a a manufacturer sells a certain article in...

Question 05 a) A manufacturer sells a certain article in batches of 5000. By agreement with a customer the following method of inspection is adopted: A sample of 100 items is drawn at random from each batch and inspected. If the sample contains 4 or fewer defective items, then the batch is accepted by the customer. If more than 4 defectives are found, every item in the batch is inspected. If inspection costs are 75 p per hundred articles, and the manufacturer normally produces 2% of defective articles, find the average inspection costs per batch. b) The number of failures occurring in a machine of a certain type in a year has a Poisson distribution with mean 0.4. In a factory there are ten of these machines. i. What is the expected total number of failures in the factory in a year? ii. the probability that there are fewer than two failures in the factory in a year?

Transcript

00:01 In letter a, we need to compute what is the probability that the number of defectives in the batch is at most want.

00:11 I'm using the letter y to express the number of defective in this case items.

00:21 So basically, from the question, we know that in the probability that an item will be defective.

00:30 Is 2%.

00:34 So basically here, to compute this, we're going to use the binomial distribution, because the binomial distribution gives us the probability of a number of successes.

00:49 In this case, the success here is being defective, because it is the one that we have the probability.

00:56 And the one that we are interested.

00:58 So basically, the binomial distribution says that the probability.

01:02 That we are going to observe a little y or small y number of successes is given by the total number of items that we are evaluating, like in this case 10, and we should check how many possible ways can we find wide effective items out of this 10 that were selected.

01:26 Then we need to multiply this by the probability of being defective.

01:32 So in this case, 0 .02 or 2%.

01:35 And since we want like a little y or a small y number of defective items, this means that this event here being defective worker y times.

01:50 So that's why that we are putting here in the exponent the little y.

01:56 Then this means that the rest, which has the probability of not being defective equals to 0 .98, which is 1 minus the probability of being defective, the rest, which is 10 minus y should be, in this case, non -defective.

02:17 So considering this formula, we can open this probability here, the y is less or equal than 1, as the probability of y being zero or y being one.

02:32 So in this case we have that this probability is the same as this, the sum of these two.

02:38 And for each one of this, we should compute this probability here by plugging where we have the small y, the numbers that we have here.

02:50 So basically we are going to do this for zero first.

02:54 So 0 .020 and 0 .9810.

03:00 Then we're gonna do this for the 1.

03:04 So 0 .021 and 0 .989.

03:10 So using a calculator, you're gonna get that this probability is 0 .98.

03:18 Now, for the other part of this question, we want to compute what is the probability? that given that we know that the batch was accepted, which means that the number of defectives in the batch was less or equal than one, what is the probability that we observe only zero, like a no defective items in the batch? so basically, here is what we call the conditional probability.

03:46 So basically, we are going to use the formula for the conditional probability to solve this, which is given by the intersection of these two events...

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