# Random Variable (Normal Distribution)

• Jan 19th 2011, 05:25 AM
MattWT
Random Variable (Normal Distribution)
Hello,

Struggling to answer the following question:

The random variable Z has the mean -4 and standard deviation 5. Find:
E[3-2Z]
E[(3-2Z)^2]

Now, I think this is a normal distribution, Z~N(-4, 5^2).

And that E(Z) = -4, but as it has more terms in the E[], i'm a little confused on what to do!

Thanks
• Jan 19th 2011, 08:01 AM
Chris L T521
Quote:

Originally Posted by MattWT
Hello,

Struggling to answer the following question:

The random variable Z has the mean -4 and standard deviation 5. Find:
E[3-2Z]
E[(3-2Z)^2]

Now, I think this is a normal distribution, Z~N(-4, 5^2).

And that E(Z) = -4, but as it has more terms in the E[], i'm a little confused on what to do!

Thanks

By properties of expected value, $\displaystyle E[3-2Z] = 3-2E[Z]$.

Now with regards to $\displaystyle E[(3-2Z)^2]$, there's a quick way to do this using the definition of variances:

$\displaystyle \text{Var}[3-2Z]+(E[3-2Z])^2=E[(3-2Z)^2]$

But then by properties of variances, $\displaystyle \text{Var}[3-2Z] = \text{Var}[-2Z]$. Thus, we can write this as:

$\displaystyle E[(3-2Z)^2]=4\text{Var}[Z]+(E[3-2Z])^2$.

Once you find $\displaystyle E[3-2Z]$, then you should be able to find $\displaystyle E[(3-2Z)^2]$ because we happen to know what $\displaystyle \text{Var}[Z]$ is.

Can you proceed?