WEBVTT
1
00:00:00.480 --> 00:00:03.399 A:middle L:90%
okay for this problem were just asked to find a
2
00:00:03.399 --> 00:00:07.459 A:middle L:90%
Z score and that we know, uh, were
3
00:00:07.459 --> 00:00:10.289 A:middle L:90%
given for this problem is the other parameters. So
4
00:00:10.289 --> 00:00:16.350 A:middle L:90%
we know the score is just value minus the mean
5
00:00:16.839 --> 00:00:21.149 A:middle L:90%
divided by the standard deviation. And for this problem
6
00:00:21.149 --> 00:00:23.769 A:middle L:90%
So this is just a formula substitute simplify problem.
7
00:00:24.210 --> 00:00:26.170 A:middle L:90%
So for this problem, we were told that did
8
00:00:26.170 --> 00:00:30.170 A:middle L:90%
score was 2 37 the X value forest tracks that
9
00:00:30.179 --> 00:00:35.829 A:middle L:90%
find the difference from the mean of 220. And
10
00:00:35.840 --> 00:00:37.990 A:middle L:90%
by that, but the standard deviation in this case
11
00:00:37.990 --> 00:00:42.409 A:middle L:90%
is 12.3 gives us standardized score. So if we
12
00:00:42.409 --> 00:00:46.649 A:middle L:90%
simplify that, we get 17 divided by 12.3.
13
00:00:51.039 --> 00:00:57.560 A:middle L:90%
So our final answer is 17. Event about 12.3
14
00:00:57.560 --> 00:01:03.969 A:middle L:90%
is 1.38 1.38 Standard deviations away from the mean