Am I right on this proportion

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December 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?

Answers are:
.3891
.9972
.9452
.5865

I think it's either .9972 or .9452 but can't be sure
December 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.
December 1st 2011, 11:33 AM
Archie Meade

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

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

So the solution is

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