Urgently Need Help.

[omramos]

I can't seem to figure this problem out.

125 employees in a company. The following probability distribution shows the likelihood that people were absent 0,1,2,3,4,or 5 days last month.

Number of days absent Probability
0 0.60
1 0.20
2 0.12
3 0.04
4 0.04
5 0

What is the mean number of days absent? 1.00
I understand that the mean is the sum of all values divided by the number of values.

What is the variance number of days absent? don't know how to solve.

What is the area under the normal curve between z=0.0 and z= 1.79? don't know how to solve.


Please help me get the right answer and understand it too.

Thank you.

[karynj]

okay try this...

try to do the computational forumala which says EX2 (sum of x squared) -(EX)2 (sum of x) then squared over N so it should be (since i dont have symbols)
EX2-(EX)2
____
N

so
1. make two colums
X and X2
2. list all your X values for column 1, then list all your X values in colum2 squared (so square each x value.) so you should have a list of x scores and al list of squared x scores.
3. find the sum of both colums
4. plug them into your formula
5. do your order of operation
6. your answer should give you your SS (sum of squared deviations)
7. you need the SS to find the variance which the forumla is
sample: SS/n-1
Population: SS/N

and of course to find the Stan. dev you would take the square root of the variance. try this first and we can compare...

[karynj]

x f Efx
__ ___ ____
0.60 0 0
0.20 2 0.4
0.12 3 0.36
0.04 7 0.28
0.05 0 0


i forget to say there was frequency so your formula is

Efx2-(Efx)2
_____
N

so this is your start... but you need to find the squared colums... i didnt do that one for you