00:02
Hi, the mean is given as 1543.
00:07
Mean is given as 1543, right? then we have variation is 304, 304, right? we're to find out the normal probability here.
00:23
P of x greater than 1 .421.
00:27
X greater than 1 .421, right? that's we need to find out.
00:31
So first, we need to find out z score here.
00:33
Z score will be x minus mu.
00:37
Our sigma i will take it as the mu here right that is mu right so x minus mu over sigma but that is coming out to be we have here will be x greater than 1421 right so x is output here 1421 right minus we have mu that is 154 3 must we have right over we have sigma that is given as 3 .04 so i calculate i calculate z from here, right? i calculate z from here.
01:16
So i take z here around to four decimal places, minus 0 .4013, right? now to find out the probability now, probability of z greater than minus 0 .4013.
01:29
Right, so for that, what i can do, i just use here the axle here to just solve this out.
01:39
So, i just put here the formula here.
01:43
In this cell here, i put the formula equals norm distribution, the z value minus 0 .4013.
01:50
The mean is 0, a division is 1, and true for cumulative frequency.
01:55
I find out that it...