Bass: The bass in Clear Lake have weights that are normally distributed with a m

Question

Bass: The bass in Clear Lake have weights that are normally distributed with a mean of 2.4 pounds and a standard deviation of 0.9 pounds.

(a) Suppose you only want to keep fish that are in the top 5% as far as weight is concerned. What is the minimum weight of a keeper? Round your answer to 2 decimal places.
pounds

(b) Suppose you want to mount a fish if it is in the top 0.5% of those in the lake. What is the minimum weight of a bass to be mounted? Round your answer to 2 decimal places.
pounds

(c) Determine the weights that delineate the middle 95% of the bass in Clear Lake. Round your answers to 2 decimal places.
from  to  pounds

Explanation / Answer

Mean ( u ) =2.4
Standard Deviation ( sd )=0.9
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.6449
P( x-u/ (s.d) > x - 2.4/0.9) = 0.05
That is, ( x - 2.4/0.9) = 1.6449
--> x = 1.6449 * 0.9+2.4 = 3.8804                  
b.
P ( Z > x ) = 0.005
Value of z to the cumulative probability of 0.005 from normal table is 2.5758
P( x-u/ (s.d) > x - 2.4/0.9) = 0.005
That is, ( x - 2.4/0.9) = 2.5758
--> x = 2.5758 * 0.9+2.4 = 4.7182                  
c.
P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.96
P( x-u/s.d < x - 2.4/0.9 ) = 0.025
That is, ( x - 2.4/0.9 ) = -1.96
--> x = -1.96 * 0.9 + 2.4 = 0.636                  
P ( Z > x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is 1.96
P( x-u/ (s.d) > x - 2.4/0.9) = 0.025
That is, ( x - 2.4/0.9) = 1.96
--> x = 1.96 * 0.9+2.4 = 4.164                  

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