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 2.5 pounds and a standard deviation of 0.5 pounds.

(a) Suppose you only want to keep fish that are in the top 20% 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 99% of the bass in Clear Lake. Round your answers to 2 decimal places.
from ____ to ____ pounds

Explanation / Answer

Given: µ = 2.5, = 0.5

(a) We need to find that P(X > x) = 0.2. Therefore P(X < x) = 1 - 0.2 = 0.8

The Z score at p value of 0.8 = 0.8416

Therefore (X - 2.5)/0.5 = 0.8416. Solving X = (0.8416 * 0.5) + 2.5 = 2.92 lbs

(b) We need to find that P(X > x) = 0.005. Therefore P(X < x) = 1 - 0.005 = 0.995

The Z score at p value of 0.995 = 2.5758

Therefore (X - 2.5)/0.5 = 2.5758. Solving X = (2.5758 * 0.5) + 2.5 = 3.79 lbs

(c) 99% of all weights centered about the mean.

This means that the remaining 1% will be distributed equally to the right and the left = 0.01/2 = 0.005

So the lower p value = 0.005 and the Upper p value = 1 - 0.005 = 0.995

The Z score at p = 0.005 and 0.995 are -2.5758 and +2.5758 respectively.

The Lower value: (X - 2.5)/0.5 = -2.5758. Solving for X, X = -2.5758*0.5 + 2.5 = 1.21 lbs

The Upperer value: (X - 2.5)/0.5 = 2.5758. Solving X = (2.5758 * 0.5) + 2.5 = 3.79 lbs

Therefore 99% will be distributed from 1.21 to 3.79 pounds

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.