Question

The U.S. Fish and Wildlife Service reported that the mean length of six-year-old rainbow trout in the Arolik River in Alaska is 481 millimeters with a standard deviation of 41 millimeters. Assume these lengths are normally distributed. A size limit is to be put on trout that are caught. What should the size limit be so that 15% of six-year-old trout have lengths shorter than the limit

Answer 1

Answer:

The size limit for 15% of of six-year-old trout is 438.52 mm.                

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 481 millimeters

Standard Deviation, σ = 41 millimeters

We are given that the distribution of mean length of six-year-old rainbow trout is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.15

$P( X < x) = P( z < \displaystyle\frac{x - 481}{41})=0.15$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x - 481}{41} = -1.036\\\\x = 438.524$  

Thus, the size limit for 15% of of six-year-old trout is 438.52 mm.

Answer 2

The size limit be so that 15% of six-year-old trout have lengths shorter than the limit is 438.36.

How to calculate the size limit?

From the information given, the mean length is 481 millimeters with a standard deviation of 41 millimeters.

The value from the standard normal table will be:

= P[(z < (x - 481)/41]

= 0.15

The z distribution of the 15% will be:

= (-1.04 × 41) + 481

= 438.36

In conclusion, the size limit is 438.36.

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