Question

Bass: The bass in Clear Lake have weights that are normally distributed with a mean of 2.7 pounds and a standard deviation of 0.6 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 90% of the bass in Clear Lake. Round your answers to 2 decimal places.
from to pounds

Answer 1

Solution :

mean = = 2.7

standard deviation = = 0.6

Using standard normal table,

(a)

P(Z > z) = 5%

1 - P(Z < z) = 0.05

P(Z < z) = 1 - 0.05

P(Z < 1.645) = 0.95

z = 1.645

Using z-score formula,

x = z * +

x = 1.645 * 0.6 + 2.7 = 3.687 = 3.69

minimum weight of a keeper = 3.69 pounds

(b)

P(Z > z) = 0.5%

1 - P(Z < z) = 0.005

P(Z < z) = 1 - 0.005

P(Z < 2.576) = 0.995

z = 2.576

Using z-score formula,

x = z * +

x = 2.576 * 0.6 + 2.7 = 3.687 = 4.25

minimum weight = 4.25 pounds

(c)

Middle 90% has the two z values : -1.645 and +1.645

x = -1.645 * 0.6 + 2.7 = 3.687 = 1.71

x = 1.645 * 0.6 + 2.7 = 3.687 = 3.69

minimum weight = 4.25 pounds

Weights that delineate the middle 90% of the bass in Clear Lake is : 1.71 pounds to 3.69 pounds .