00:01
Listed in this chart are the probabilities of job satisfaction scores for both senior executives and middle managers of an information systems company.
00:14
And in part a, we want to determine the expected value of job satisfaction score for senior executives.
00:26
So let's only focus on the senior executives.
00:29
So our job satisfaction scores will be our raw data or our x score, and our probability for the senior executives was 0 .05, 0 .09, 0 .03, 0 .42, and 0 .41.
00:58
Now, to calculate expected value, it's the same as calculating the mean, and when you're working with probability distributions, the mean is the sum of each x times its corresponding probability.
01:17
So that means we're going to add on a column.
01:21
We're going to call it x times p of x.
01:25
And we're going to multiply each job satisfaction score times its probability.
01:31
So when i multiply here, i get 0 .05, then 0 .18.
01:39
0 .09, 1 .68, and 2 .05.
01:49
And if i add up that column, i will get the expected job satisfaction score of 4 .05.
01:59
So that would be the expected score for the senior executives.
02:03
For part b, we're doing the same thing, but this time we're using the data for the middle managers.
02:10
So again, the job satisfaction scores are 1, 2, 3, 4, 5.
02:17
The probabilities for the middle management is 0 .04, 0 .10, 0 .12, 0 .46, and 0 .28.
02:34
Once again, the expected value is the same as the mean, which can be found by 6.
02:43
Summing up x times p of x so we'll create that additional column we'll multiply each x times its corresponding probability so the first one is 0 .04 then 0 .2 .36 1 .84 1 .84 and 1 .40 and when i add those up we get a an average or expected value of job satisfaction score of 3 .84.
03:27
In part c, we want to find the variance for both the senior executives and for middle management.
03:46
So we'll start with the variance for the senior execs, and we'll start with reproducing our information on those senior executives.
03:56
And to talk about variance, we have to know how far each job satisfaction score deviated from the mean before we can talk about that standard deviation or the variance.
04:27
So we're going to create a column called x minus mu, which is our column for the deviation from average.
04:37
So that means we're taking each x value and subtracting the expected value.
04:44
Or the mean to find our deviation.
04:48
So the deviation for a job satisfaction score of 1 is negative 3 .05.
04:56
For 2, it's negative 2 .05.
05:00
For 3, it's negative 1 .05.
05:03
For 3, it's negative 1 .05.
05:04
For 4, it's negative 0 .05.
05:08
And 5 was the only one above the expected value or above the mean, and it was above the mean by 0 .95.
05:17
Now to find the variance, we are going to use the formula, the sum of x minus mu squared times p of x.
05:31
So that means we're going to need to add on an additional column.
05:36
We'll call it x minus mu squared.
05:41
And when i square each of the deviations, when i square negative 3 .05, i get 9 .3025.
05:54
When i square negative 2 .05, we get 4 .2025.
06:02
If you square negative 1 .2025, you'll get 1 .1025.
06:12
If you square negative 0 .05, you'll get a positive 025 .5, you'll get a positive 0 .025.
06:19
And when you square 0 .95, you get 0 .9025.
06:25
Now we have to multiply each of those by their corresponding probabilities.
06:31
So we'll get x minus mu squared times p of x as an additional column...