00:01
In this exercise, we make use of the approximation to a binomial random variable for sampling without replacement.
00:09
And this approximation holds when the sample size is small relative to the total population size.
00:16
And so a good rule of thumb for when this approximation is appropriate is if the sample size is at most 5 % of the total population size.
00:26
Now in this exercise, a large batch of components has arrived, and we only want to accept the batch if the proportion of defective components is at most 0 .1.
00:39
Otherwise, we want to reject the batch.
00:43
But since this is a very large batch, the strategy is to draw a sample and then make our decision based on that sample.
00:50
So the strategy is we randomly select 10 components from the batch, and we accept the batch if the number of defective components in this sample of 10 is at most 2.
01:04
So although for each component that we take for a sample, we are not sampling with replacement, this can approximate a binomial based on the reasoning i just provided at the start of the exercise.
01:16
So we can say this that the number of defective components in the sample approximates a binomial based on based on a sample of 10, and the probability of success, p.
01:44
So here, ironically, success is defined as getting a defective component, because we are interested in the number of defective components in the sample.
01:58
For part a, we are asked to find the probabilities that the batch is accepted when the actual proportion of defectives is 0 .01, 0 .05, 0 .1, 0 .2, and 0 .25.
02:10
So we can provide our answers in a table like this.
02:18
So we're looking for the probability that x is at most two, and this is the cumulative probability for two successes based on a binomial random variable with 10 trials and probability of success, p.
02:46
And so i used r to make all these calculations.
02:50
So for the first one, which is p equals 0 .01, we can use the p binome function.
03:02
So we want the cumulative probability up to two successes.
03:06
Sample size is 10, and the probability of success, .01.
03:13
And so we get almost 1.
03:15
So the probability of accepting that batch if we use the strategy is about 1.
03:29
And continuing in this manner, i got .988 .930 .6.
03:44
Then for part b, we are asked to use the results from part a to plot the probability of batch acceptance versus the proportion, the actual proportion of the actual defective proportion of the batch.
04:05
So that is probability of acceptance is this role and the actual defective proportion in the batch is this row.
04:14
So to save some time, i went ahead and did this in r.
04:19
And the plot comes out looking like this.
04:21
So the x -axis, we have the actual defective proportion in the batch.
04:27
And on the y -axis, we have the probability of accepting, given that defective proportion.
04:32
So you can see that at very low probabilities, at very low proportions of defective components in the batch, the probability of accepting the sample is very high.
04:45
Conversely, if the proportion of defective components in the batch is high, then the probability of accepting the batch is low...