00:01
There is given a normal distribution here.
00:02
The population mean denoted by mu that was given as 166 .5 and the population standard division which is 16 .3.
00:10
Let me define a random variable x for this population.
00:14
This is normally distributed, the mean and the standard division.
00:18
So for this one, if we just select a single person or the single item here, we have to use the random variable x and the population random variable x here.
00:27
So i'm going to find x is less than 169 .3.
00:31
To get this probability, i'm going to use the graphing display calculator application normalcdf.
00:35
There is no any lower boundary.
00:37
Put very small number, upper boundary and the mean value and the standard division which is 16 .3.
00:44
So press second variance and the normalcdf lower boundary which is negative 1.
00:48
This is second mean 99.
00:50
The upper boundary is 169 .3.
00:52
So the mean is 166 .5 and the standard division 16 .3.
00:58
So the parameter would be 0 .56 and 82...