Am I right on this proportion

The question is:

The weights of trout in a certain river follow a normal distribution with mean 3 pounds and standard deviation 0.8 pounds. What proportion of trout weigh between 3.2 pounds and 4.8 pounds?

Answers are:

.3891

.9972

.9452

.5865

I think it's either .9972 or .9452 but can't be sure

Re: Am I right on this proportion

If a random variable $\displaystyle X$ has a normal distribution with mean $\displaystyle \mu$ and standard deviation $\displaystyle \sigma$, then the cumulative distribution function (CDF) $\displaystyle F_X(z)=\Phi\left(\frac{z-\mu}{\sigma}\right)$ where $\displaystyle \Phi$ is the CDF of the standard normal distribution (with mean 0 and standard deviation 1) (Wikipedia). You need to find $\displaystyle F_X(4.8)-F_X(3.2)$. The function $\displaystyle \Phi(z)$ can be computed, for example, by Excel function NORMSDIST. There is also a function NORMDIST to compute the CDF of an arbitrary normal distribution.

Re: Am I right on this proportion

Quote:

Originally Posted by

**jay25** The question is:

The weights of trout in a certain river follow a normal distribution with mean 3 pounds and standard deviation 0.8 pounds. What proportion of trout weigh between 3.2 pounds and 4.8 pounds?

Answers are:

.3891

.9972

.9452

.5865

I think it's either .9972 or .9452 but can't be sure

It's the first answer.

Intuitively, if the mean is 3 and your range of values is above that,

the probability of trout being in that weight range must be less than half.

You need to work it out using the Z-tables of the Normal Distribution,

first by converting to Z using

$\displaystyle Z=\frac{x-\mu}{\sigma}$

So the solution is

$\displaystyle P\left(Z\le \frac{4.8-3}{0.8}\right)-P\left(Z\le \frac{3.2-3}{0.8}\right)$