00:01
So this one's kind of a long one, so bear with me.
00:07
If we're normally distributed and we focus on chicken first, with a mean number of sandwiches of 93 and a standard deviation of 22, don't forget x equals the standard deviation times z plus the mean, z equals x minus the mean, z equals x minus the mean divided by the standard deviation.
00:43
So if we want the probability of some data value k, to be 98 % confident that we're not going to run out of sandwiches, we need to define k.
00:59
Well, the z score that goes with 98 % is 2 .0 -3575.
01:10
Well, that means x is 2 .05375 times 22 plus 933 .3.
01:38
So, you probably have to round that to 139.
01:43
Because you can't have a partial sandwich.
01:46
And also for chicken we need the probability that we'll have at least 150 sandwiches on hand and we want the probability that we're going to need more than 150 well that's the same as the probability that z is greater than 2 .591 or 0 .079.
02:40
So in blue here we're going to do the burgers.
02:45
I want to do a similar thing.
02:47
It has a mean of 313.
02:53
A standard deviation of 57 so we want the probability that we're not going to run out to be 98%.
03:05
We need x as less than some value is 98%.
03:11
Again, z is 2 .05375 so x is the same process except now instead i have an amine of 93.
03:24
We're adding 313 and we get an x of 430 .06 so we probably need 431 and we want to know the probability that will run out on any given day and that's x is greater than 400 well that's the same as the probability that z is greater than 1 .5 263 or 0 .0 .0 .063.
04:18
So those are each individual probabilities of running out of hamburgers and chicken...