Question

10. Wildlife biologists believe that the weights of adult trout can be described by a normal model. Fishermen report that 22% of all trout caught were thrown back because they weigh below the 2-pound minimum, and only 6% weighed over 5 pounds. What are the mean and standard deviation of the model? 022 P=0.22 P = 0.94 2= -0.77 Z= 1.56 Ź fe 1.56= 5-4 = 20 2 - 5-1.566 = -5tu 1 -0.776. 2-M 0.72 - 0.77 +1.566= 5-M - 0.77 6-2-gu - 2.3363-3 6=273 = 1.29 (1.56) (1.29) = +5-M M=5-2 =3 M=3 lb 6=1. 316 L

Please explain how to solve this type of problem... I'm confused as to where some of the numbers are coming from. Thank you.

Answer 1

Let X is a random variable shows the weight of trout.

Now we need z-score that has 0.22 area to its left. Using excel fucntion "=NORMSINV(0.22)" or z table z-score -0.77 has 0.22 area to its left.

And we need z-score that has 1 -0.06 = 0.94 area to its left. Using excel fucntion "=NORMSINV(0.94)" or z table z-score 1.56 has 0.94 area to its left.

It is given that

P(X <2) = 0.22

So,

$z=\frac{X-\mu}{\sigma}$

2-1 -0.77 =

..............(1)

-0.770 +u = 2

----------------------------------

Also it is given that

P(X > 5) = 0.06

So,

$z=\frac{X-\mu}{\sigma}$

1.56 = 5 - 14

..............(2)

1.56σ +μ = 5

Subtracting equation (1) from (2)

1.56σ +μ + 0.77σ - μ = 5 – 2

2.337 = 3

0 = 3/2.33 = 1.29

-------------------------------

Putting this in equaiton 2 gives

1.56. 1.29 + y = 5

2.0124 + y = 5

u=5 – 2.0124 = 2.9876 3.0