# Am I right on this proportion

• Dec 1st 2011, 11:00 AM
jay25
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?

.3891
.9972
.9452
.5865

I think it's either .9972 or .9452 but can't be sure
• Dec 1st 2011, 11:31 AM
emakarov
Re: Am I right on this proportion
If a random variable $X$ has a normal distribution with mean $\mu$ and standard deviation $\sigma$, then the cumulative distribution function (CDF) $F_X(z)=\Phi\left(\frac{z-\mu}{\sigma}\right)$ where $\Phi$ is the CDF of the standard normal distribution (with mean 0 and standard deviation 1) (Wikipedia). You need to find $F_X(4.8)-F_X(3.2)$. The function $\Phi(z)$ can be computed, for example, by Excel function NORMSDIST. There is also a function NORMDIST to compute the CDF of an arbitrary normal distribution.
• Dec 1st 2011, 11:33 AM
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?

.3891
.9972
.9452
.5865

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

$Z=\frac{x-\mu}{\sigma}$
$P\left(Z\le \frac{4.8-3}{0.8}\right)-P\left(Z\le \frac{3.2-3}{0.8}\right)$