Question

Math help, typing questing below

Answer 1

The amount of time y (in minutes) that a commuter train is late is a continuous random variable with probabilty density

\[f(y) = \left\{ \frac{3}{500}(25 - y^2) \space \space -5 \le y \le 5\right\} \] and is 0 elsewhere. Find the mean and variance for the amount of time the train is late in minutes, seconds, and hours.

Answer 2

Note: a negative value of y means the train was early

Answer 3

yikes...i honestly don't know man...sorry...

Answer 4

@satellite73 @amistre64

Answer 5

@Tyler1992 this is continuous, right?
Well, if you have a probability density function p(y)
Then the expected value (or mean) is given by:
\[\Large \mathbb{E}(y) = \int\limits_I y \ p(y)dy\]

Answer 6

In this case, your interval I is the interval from -5 to 5.
Just integrate ^_^

Answer 7

oh ok :) and for the variance?

Answer 8

Two ways to get the variance. After you get the mean (let's call it \(\mu\))
\[\Large \mu = \mathbb{E}(y)\]The variance
\[\Large \text{Var}(y) = \mathbb{E}[(y-\mu)^2]\]

Answer 9

In other words, \[\Large \text{Var}(y) = \int\limits_{-5}^5 (y-\mu)^2p(y)dy\]
but there's an easier way :D

Answer 10

lol easy way is always better :D

Answer 11

Agreed. You know that Expected Value is a linear operator, right?
IE
\[\Large \mathbb{E}(x+y) = \mathbb{E}(x) +\mathbb{E}(y) \]\[\Large \mathbb{E}(kx)= k \mathbb{E}(x)\]

Answer 12

confused?

Answer 13

Nope i didnt know that but i know what a linear operator is due to linear algebra

Answer 14

Okay good. Expected Value is a linear operator, given that, we can vastly simplify
\[\Large \text{Var}(y) = \mathbb{E}[(y-\mu)^2]\]

Answer 15

\[\Large = \mathbb{E}[y^2 - 2\mu y +\mu^2]\] right?

Answer 16

yes i see what you did

Answer 17

Okay, let's take advantage of the Linear-ness of expected value...
\[\Large =\mathbb{E}[y^2]-\mathbb{E}[2\mu y]+ \mathbb{E}(\mu^2)\]
right?

Answer 18

correct :D

Answer 19

Okay, you understand that the mean of a constant is that constant, right?

Answer 20

Mean or expected value.
\[\Large \mathbb{E}[k] = k\]

Answer 21

Makes sense, yeah?

Answer 22

makes sense

Answer 23

Okay, so
\[\Large =\color{blue}{\mathbb{E}[y^2]}-\mathbb{E}[2\mu y]+ \mathbb{E}(\mu^2)\]
we leave as is...

This part
\[\Large =\mathbb{E}[y^2]-\color{blue}{\mathbb{E}[2\mu y]}+ \mathbb{E}(\mu^2)\]becomes\[\Large =\mathbb{E}[y^2]-\color{red}{2\mu\mathbb{E}[ y]}+ \mathbb{E}(\mu^2)\]due to linear-ness

And finally, this part\[\Large =\mathbb{E}[y^2]-2\mu\mathbb{E}[ y]+ \color{blue}{\mathbb{E}(\mu^2)}\] becomes \[\Large =\mathbb{E}[y^2]-2\mu \mathbb{E}[ y]+ \color{red}{\mu^2}\] since it was a constant.

Answer 24

Still with me?

Answer 25

yep i gotcha

Answer 26

Okay, we remember that \[\Large \mu = \mathbb{E}(y)\]
So\[\Large \mathbb{E}[y^2]-\color{blue}{2\mu \mathbb{E}[y]}+\mu^2\]
reduces to just
\[\Large \mathbb{E}[y^2]-\color{red}{2\mu^2 }+\mu^2\]

further simplifying, we get the well-known formula for Variance
\[\Large = \mathbb{E}[y^2]-\mu^2\]

Answer 27

I believe this integral
\[\Large \int \limits_{-5}^5 y^2 p(y) dy= \mathbb{E}[y^2]\] is far simpler than THIS integral you would have had to do if you didn't get that formula:
\[\Large \int \limits_{-5}^5 (y-\mu)^2 p(y) dy=\mathbb{E}[(y-\mu)^2]\]

Answer 28

So remember this, you won't go wrong:
\[\Large \text{Var}(y) =\mathbb E[y^2] -\left(\mathbb E [y]\right)^2 \]

Answer 29

I definitely like that integral better lol. And to get the variance in seconds and hours can i just multiply and divide the variance i get for minutes by 60 to get the other two?

Answer 30

whoa, I didn't get the second and hours bit, mind posting your raw question? :D

Answer 31

Oh, wait, I just now got that, yes, that is true, work out the mean and variance first, and then do the conversions :D

Answer 32

Ok thank you very much! If i have more statistics trouble can i ask you for help again?

Answer 33

If I'm online.
A fair warning though, stat isn't my strong point, I just got lucky with this one.

Another warning... Don't be too happy after you evaluate
\[\Large \mathbb{E}[y^2]\] yet, this itself is NOT the variance, you still have to subtract from it the square of the mean :)
\[\Large \text{Var}(y) =\mathbb E[y^2] -\left(\mathbb E [y]\right)^2=\mathbb E[y^2] -\mu^2\]

Variance is "Mean of the square minus the square of the mean"
^_^

Answer 34

Awesome! Thank you very much!!

Answer 35

No problem