Question

Bass: The bass in Clear Lake have weights that are normally distributed with a mean of 2.5 pounds and a standard deviation of 0.6 pounds.

(a) Suppose you only want to keep fish that are in the top 15% as far as weight is concerned. What is the minimum weight of a keeper? Round your answer to 2 decimal places.


(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.

(c) Determine the weights that delineate the middle 99% of the bass in Clear Lake. Round your answers to 2 decimal places.

Answer 1

Solution:-

Given that,

mean = =  2.5

standard deviation = = 0.6

Using standard normal table,

a ) P ( Z > z ) = 15%

1 - P ( Z <  z ) = 0.15

P ( Z <  z ) = 1 - 0.15

P ( Z < 1.04 ) = 0.85

z = 1.04

Using z-score formula,

x = z * +

x = 1.04* 0.6 + 2.5

x = 3.124

The minimum weigh = 3.12

a ) P ( Z > z ) = 0.5%

1 - P ( Z <  z ) = 0.005

P ( Z <  z ) = 1 - 0.005

P ( Z < 2.57 ) = 0.995

z = 2.57

Using z-score formula,

x = z * +

x = 2.57 *0.6+ 2.5

x = 4.042

The minimum weigh = 4.04

c ) P(-z < Z < z) = 99%
P(Z < z) - P(Z < z) = 0.99
2P(Z < z) - 1 = 0.99
2P(Z < z ) = 1 + 0.99
2P(Z < z) = 1.99
P(Z < z) = 1.99 / 2
P(Z < z) = 0.995
z = 2.57 znd z = - 2.57

Using z-score formula,

x = z * +

x = 2.57 *0.6 + 2.5

x = 4.042

The minimum weigh = 4.04

x = z * +

x = - 2.57 *0.6+ 2.5

x = 0.958

The minimum weigh = 0.96