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
If a random variablehas 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.
It's the first answer.Originally Posted by jay25
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
-P\left(Z\le \frac{3.2-3}{0.8}\right))
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