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 has a normal distribution with mean and standard deviation , then the cumulative distribution function (CDF) where is the CDF of the standard normal distribution (with mean 0 and standard deviation 1) (Wikipedia). You need to find . The function 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

So the solution is