00:01
In this question, we say the number of pages printed before replacing the cartridge in a laser printer is normally distributed with a mean of 11 ,500 pages and a standard deviation of 800 pages.
00:14
A new cartridge has just been installed.
00:17
I wanna know what is the probability that the printer produces fewer than 12 ,000 pages before the cartridge is replaced.
00:27
So i have to convert this 12 ,000 into a z -score.
00:31
What do i do? i'm gonna take the 12 ,000.
00:35
From this, i subtract my mean, 11 ,500, and then i divide by the standard deviation, 800.
00:46
I take my value that's given, i subtract my mean, and i divide by the standard deviation in order to get the z -score.
00:57
So this time, 12 ,000 minus 11 ,500, that's 500 over 800.
01:03
That is 0 .625.
01:08
So i'm gonna have a z -score of 0 .625, which we're gonna round to 0 .63.
01:16
And so now, if i go to my z -score table, z equals 0 .63 corresponds with about 0 .735 -ish.
01:29
So it's about 0 .73 because we were between 0 .02 and 0 .03 there with the 0 .025 following the 0 .6.
01:39
So we'll say about 0 .73...