Question

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? _____ 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? ______pounds (c) Determine the weights that delineate the middle 99% of the bass in Clear Lake. From  _____ to _____ pounds

Answer 1

Solution :

Given that,  

mean = = 2.5

standard deviation = = 0.6

(a)

Using standard normal table ,

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 = 3.12

3.12 pounds

(b)

Using standard normal table ,

P(Z > z) = 0.5%

1 - P(Z < z) = 0.005

P(Z < z) = 1 - 0.005

P(Z < 2.58) = 0.995

z = 2.58

Using z-score formula,

x = z * +

x = 2.58 * 0.6 + 2.5 = 4.05

4.05 pounds

(c)

Middle 99% as the to z values are -2.576 and 2.576

Using z-score formula,

x = z * +

x = -2.576 * 0.6 + 2.5 = 0.95

and

x = 2.576 * 0.6 + 2.5 = 4.05

The middle 99% of the bass in clear lake. from 0.95 and 4.05 pounds