00:01
So we know that we have the 90 invoices looked at and using that probability scheme that they discussed, the probability of a 1 or 2 on it, and starting with a 1 or 2 on an invoice is supposed to be 0 .477.
00:14
And so we'll just say that that's probability of success.
00:17
So we would anticipate n times p, this 90 times 0 .477, 90 times 0 .477.
00:27
That expected number is 42 .93 and the n minus p is going to come out to be that 90 times one minus that 0 .477.
00:45
That is going to come out to be 47 .07.
00:49
So both these are greater than are equal to 10.
00:51
So this is an approximately normal setting.
00:55
And so if we look at 90 of these, we would expect the center of that normal distribution to be at a value of 42 .93.
01:07
And our standard deviation would be that square root of n times p times one minus p.
01:13
And let's see, i can take this answer.
01:16
And i already have in my calculator times 0 .477, and then square root that answer.
01:24
And so i get a standard deviation for that random variable of how many have that one or two.
01:30
Being approximately 4 .738...