Probability of a random variable

**jabba1** · March 29th 2011, 08:20 AM

The reading given by a thermometer calivrated in ice water is a random variable with probablility density function

f(x) { k(1-x) -1

{ 0 otherwise where k is a constant

find the value of k.

What is the probablity that the temperatiure reading is greater than 0C?

What is the mean reading and the standard devation?

**TheEmptySet** · March 29th 2011, 08:28 AM

Originally Posted by jabba1

The reading given by a thermometer calivrated in ice water is a random variable with probablility density function

f(x) { k(1-x) -1

{ 0 otherwise where k is a constant

find the value of k.

What is the probablity that the temperatiure reading is greater than 0C?

What is the mean reading and the standard devation?

Since I can read your density function to find k you need to solve this integral

$\int_{a}^{b}f(x)dx=1$ the limits will be the where $f(x)$ is not zero.

the mean

$\mu=\int_{a}^{b}xf(x)dx$

$\sigma^2=\int_{a}^{b}(x-\mu)^2f(x)dx$

**jabba1** · March 29th 2011, 09:08 AM

i just noticed a mistake...it should say f(x) { k(1-x) -1<x<1

**Plato** · March 29th 2011, 09:15 AM

Originally Posted by jabba1

i just noticed a mistake...it should say f(x) { k(1-x) -1<x<1

Well then use that: $a=-1~\&~b=1$

**TheEmptySet** · March 29th 2011, 09:17 AM

Originally Posted by jabba1

i just noticed a mistake...it should say f(x) { k(1-x) -1<x<1

okay my original post still stands you need to solve this

$\displaystyle \int_{-1}^{1}k(1-x)dx =1 \iff k \int_{-1}^{1}(1-x)dx=1$
$\displaystyle k\left[ -\frac{(1-x)^2}{2}\bigg|_{-1}^{1}\right]=1 \iff 2k=1 \iff k=\frac{1}{2}$

Now $f(x)=\begin{cases} \frac{1-x}{2} , \text{ if }x \in[-1,1] \\ 0, \text{ otherwise }\end{cases}$

Now use this $f(x)$ and the same limits of integration to compute the other 2 integrals.

**jabba1** · March 29th 2011, 04:18 PM

I'm still a little confused on how exactly to use the equation. I think that's my biggest problem

**TheEmptySet** · March 29th 2011, 04:36 PM

Originally Posted by jabba1

I'm still a little confused on how exactly to use the equation. I think that's my biggest problem

$\displaystyle \mu=\int_{-1}^{1}x\left(\frac{1-x}{2} \right)dx=\frac{1}{2}\int_{-1}^{1}x-x^2dx$

Then use this for the 2nd one!

**jabba1** · March 29th 2011, 05:12 PM

thank you! i plugged in the number and for the first one i got .5 and for the second i got .45

Thread: Probability of a random variable

LinkBack

Thread Tools

Display

Probability of a random variable

Similar Math Help Forum Discussions

Probability of a (discrete) random variable

Probability - transformation of random variable

Probability Distribution of random variable X.

random variable probability

probability(mean of a random variable)

Search Tags