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 10% 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

Answer 1

Solution :

Given that,

mean = = 2.7

standard deviation = = 0.6

a ) P( Z > z) = 10%

P(Z > z) = 0.10

1 - P( Z < z) = 0.10

P(Z < z) = 1 - 0.10

P(Z < z) = 0.90

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 0.6 + 2.7

x = 3.468

The minimum weight = 3.47

b ) P( Z > z) = 0.5%

P(Z > z) = 0.005

1 - P( Z < z) = 0.005

P(Z < z) = 1 - 0.005

P(Z < z) = 0.995

z = 2.58

Using z-score formula,

x = z * +

x = 2.58 * 0.6 + 2.7

x = 4.248

The minimum weight = 4.25

c ) P(-z < Z < z) = 95%
P(Z < z) - P(Z < z) = 0.95
2P(Z < z) - 1 = 0.95
2P(Z < z ) = 1 + 0.95
2P(Z < z) = 1.95
P(Z < z) = 1.95 / 2
P(Z < z) = 0.975
z = *1.96 and z = 1.96

Using z-score formula,

x = z * +

x = 1.96 * 0.6 + 2.7

x =3.876

The minimum weight = 3.88

Using z-score formula,

x = z * +

x = -1.96 * 0.6 + 2.7

x = 1.524

The minimum weight = 1.52