0:00
Will solve from problem a.
00:02
So if the budget for next week is 450 cu, we need to calculate the probability that the actual cost will be greater than this amount.
00:12
And we can use the z -score formula, z equals x minus mu over sigma, where z follows the normal distribution with mean 0 and variance 1.
00:27
And the actual value that we have is a x of 450 cu and mean of 400 over the standard deviation of 20, 20 um.
00:43
Okay.
00:45
And so this is, let's calculate this value.
00:50
So this is equal to 5 over 2 and the probability x greater than 450 can be calculated as pz greater than 5 over 2 which is equal to 1 minus p, z less than or equal to 5 over 2 and you can you can use any kind of statistical statistical software or the z table to find this probability and we get the probability of 0 .62 percent and this indicates that it is quite unlikely for the maintenance cost to exceed 450 cu in a given week given that the current distribution given the current distribution of costs now for problem b we want to find the budget that would that would be exceeded only 10 percent of the time.
02:00
And then, so we need to determine the z -score that corresponds to the 90th percentile of the standard normal distribution.
02:11
And this z -score will then be used to find the corresponding x value using the inverse of the z -score formula.
02:19
So x equals z times sigma plus mu.
02:33
And so, okay, and this z corresponds to the value of z such that probability that the standard normal distribution will output the probability of 0 .9.
02:56
And again, you can find this using any kind of statistical software or the z table.
03:13
And this is actually equal to 1 .282.
03:19
Times sigma is 20, plus the mu, which is 400, so that we have a 425 .63 um.
03:36
And so setting the budget at this amount will make it likely that actual cost will not exceed this limit in 90 % of the weeks...