00:01
For this exercise, we are told that a company's employees make errorness entries 5 % of the time, so that means a probability of 0 .05, and that an auditor randomly selects 3 to check them.
00:17
So we have a sample of size 3.
00:23
Now, for part a, we are asked to find the probability distribution for why, the number of errors detected by the auditor.
00:32
Now, we can say that each entry check is a bernoulli trial, because each has a probability of success of 0 .05, that should be two possible outcomes.
01:10
And each check is independent from the other checks because they are randomly selected.
01:33
So therefore, y, which is the number of successes out of n -trials, out of ann bernoulli trials, is a binomial random variable.
01:57
And the probability mass function for a binomial random variable is given by this formula.
02:35
And so for the situation at hand, the probability of mass function is 3 choose x times 0 .05 to the exponent x times 0 .95 to the exponent 3 minus x.
02:56
And this is for x is equal to 0, 1, 2, or 3.
03:06
And so if we calculate this for each possible value of x, the first one is 3 -2 -0, 0 .05 to the exponent 0 ,0 times 0 .95 to the exponent 3.
03:24
This is equal to 0 .857.
03:30
If we do the same thing for x equals 1, we get 0 .135.
03:38
X equals 2, we get .007, and for x equals 3 we get .0001.
03:57
So we can represent the distribution in a table like this.
04:15
Alright, so i've got this around here.
04:18
This should be probability of x.
04:41
So this is the probability distribution for the number of errorness entries out of the sample of 3 that the auditor looks at.
04:53
Now, of course, since this is a probability distribution, all of the possible probability masses should add up to 1.
05:02
If you add up these four numbers, you get 0 .991, but that's only happening because these are rounded.
05:19
And one other thing, we've called this random variable y, so i should say y, not x...