Question

ecn 221 The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What lower limit should the State Game Commission set on length if it is desired that 80 percent of the catch may be kept by fishers

Answer 1

Answer:

$a=14 -0.842*3=11.474$

So the value of height that separates the bottom 20% of data from the top 80% is 11.474.  

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the lengths of a population, and for this case we know the distribution for X is given by:

$X \sim N(14,3)$  

Where $\mu=14$ and $\sigma=3$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.8$   (a)

$P(X<a)=0.2$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(z>-0.842)=0.8

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.2$  

$P(z<\frac{a-\mu}{\sigma})=0.2$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.842<\frac{a-14}{3}$

And if we solve for a we got

$a=14 -0.842*3=11.474$

So the value of height that separates the bottom 20% of data from the top 80% is 11.474.