Homework Statement
f(x) = (3/4)(-x^2 + 6x - 8) for 2 < x < 4 (0 elsewhere)
A) Find F(x)
integral 2 to 4 ((3/4)(-x^2 + 6x - 8))dx
B) Use F(x) to find P(3 < X < 3.5)
integral 3 to 3.5 ((3/4)(-x^2 + 6x - 8))dx
11/32
C) Use F(x) to find P(X > 3.5)
1-( P(3 < X < 3.5)) = 21/31
D) Find E(X)...
Ok so here is what I have let me know if this looks better
Mean: .21
87(0)+8(1)+3(2)+1(4)+1(3)=21/100
STDev:
E[x^2]-[E[x]]^2
.45-(.21)^2=.4059
sqrt(.4059)
STDEVA: .637
So would it be .45/100%= .0045, which seems also wrong but makes sense. Since the .45 is the number of quarters in the "system" sorta say, and then we accounted for 100% of the population.
So what is throwing me off is I think my calculation of the mean is incorrect. Because more people has 0 quarters then say 2 quarters, the the mean should theoretically be less then 2 correct?
Homework Statement
Mind mean and STDev of number of quarters someone has on them at one time
0 - 87%
1 - 8%
2 - 3%
3- 1%
4 - 1%
The first time I did this I assumed a population but she said that was wrong and not what she was looking for.
Here was my new attempt:
x^2f(x)=
0(.87)
1(.08)
4(.03)...