Question

Probability and Riemann sum question

Answer 1

Is the final answer for part (a) = 0.146854751571888?
Need help with part b

Answer 2

Wolfram says
http://www.wolframalpha.com/input/?i=Left+riemann+sum+1%2Fsqrt(2pi)e%5E(-(x-4)%5E2%2F2)+over+1+to+5+using+10+intervals

Answer 3

May I see how you got 0.146... ?

Answer 4

http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D0%7D%5E9+4%2F10*1%2Fsqrt(2*pi)e%5E(-(1%2B4%2F10*k-4)%5E2%2F2)

Answer 5

Hey, it was just a calculation error. I just double checked.

Answer 6

Any idea about part b?

Answer 7

@ganeshie8

Answer 8

Still here ?

Answer 9

yup

Answer 10

For part B, I think you simply need to integrate the pdf between -inf and +inf

Answer 11

that gives you 1 as any pdf should
so scratch that

Answer 12

lets think a bit hmm

Answer 13

the pdf is of a normal distribution with mean 4 and sd 1

Answer 14

I think you should integrate x*f(x) to get the mean

Answer 15

http://www.wolframalpha.com/input/?i=Integrate%5Bx%2Fsqrt(2pi)e%5E(-(x-4)%5E2%2F2),+%7Bx,-%5Cinfty,+%2B%5Cinfty%7D%5D

Answer 16

\[\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\]

here \(\mu=4\) and \(\sigma=1\)

Answer 17

Nice, that means we could pretty much eyeball mean and sd !

Answer 18

you could also use the hint. it gives the answer away.

Answer 19

yeah it seems they are suggesting to evaluate the integral by using u-substitution and the pdf

Answer 20

They are breaking x*f(x) into two parts

First part can be evaluated using a trivial u substittion
Second pat simply evaluates to 1 as it represents area under the pdf

Answer 21

2nd part is 4

Answer 22

1st part is zero

Answer 23

Looks the first part becomes an odd function after substituting u = x-4, nice

Answer 24

Ok, I think it'll take me a few minutes to get my head around it. I'll write back if I face any issues.

Answer 25

Take your time.
For part B, you just need to integrate x*f(x) between -inf and +inf. You may use the given hint