00:01
Suppose the mean length of a time that caller is placed on hold when telephoning a customer service center is 23 .8 seconds with a standard division of 4 .6 seconds.
00:10
So find the probability that the mean length of time on hold in a sample of 1 ,200 calls will be within 0 .5 second of the population mean.
00:20
So our population mean meal is actually equals to 23 .8.
00:26
Population standard division is actually equals to 4 .6.
00:32
Then it was also mentioned that the sample size n is equals to 1 ,200 calls and we are asked to get the probability that the main length of calls on owed in a sample of 1 ,200 calls will be within 0 .2 second of the population mean that so within 0 .5 of the population we simply implies that our x is actually equals to 23 .8 plus or minus 0 .5 that is x is equal to 23 .8 plus 0 .5 or x is equal to 23 .8 minus 0 .5 so let's do the math 23 .8 plus 0 .5 so that gives us 24 .3 or our x is equal to 23 .8 minus 0 .5 that gives us 23 .3 .3 so we are trying to get probability that our mean of call is between 23.
01:35
3 less than x and less than 24 .3.
01:40
So to get this probabilistic value of us, we have to standardize each of the random variable, which is the length of called x is equals to 23 .3 and x is equals to 24 .3.
01:52
And to do so, we use the formula that says as z is equals to x minus mu divided by z sigma divided by the square root of n.
01:59
So let's start at x having a value of 23 .3, we have our z...