A researcher is interested in the lengths of brook trout, which are known to be

Question

A researcher is interested in the lengths of brook trout, which are known to be approximately Normally distributed with mean 80 centimeters and standard deviation 5 centimeters. The probability of catching a brook trout less than 72 centimeters. What proportion of fish are larger than 86 centimeters in length? To ensure that the shortest 8% of the brook trout get thrown back the lower cut off should be set at? The standard is set so all fish shorter than 70.20 centimeters or longer than 89.80 centimeters must be thrown back. What proportion of the population is this?

Explanation / Answer

Let X be the random variable that lengths of brook trout.

X ~ N(mean = 80, sd = 5)

The probability of catching a brook trout less than 72 centimeters.

What proportion of fish are larger than 86 centimeters in length?

Here we have to find P(X >86).

First convert x = 86 into z-score.

z = ( x-mean) / sd

z = (86 - 80) / 5 = 1.2

That is now we have to find P(Z > 1.2)

This probability we can find by using EXCEL.

syntax is,

=1 - NORMSDIST(z)

where z is test statistic value.

P(Z > 1.2) = 0.1151

To ensure that the shortest 8% of the brook trout get thrown back the lower cut off should be set at?

Here we have to find X given that probability is 8% or 0.08.

EXCEL syntax :

=NORMINV(probability, mean, standard_dev)

probability = 0.08

mean = 80

standard_dev = 5

X = 72.975

The standard is set so all fish shorter than 70.20 centimeters or longer than 89.80 centimeters must be thrown back. What proportion of the population is this?

That is here we have to find P(X < 70.20 or X > 89.90).

P(X < 70.20 or X > 89.90) = P(X < 70.20) + P(X > 89.80)

Now find z-score for 70.20 and 89.90.

z = (70.20 - 80) / 5 = -1.96

z = (89.80 - 80) / 5 = 1.96

P(Z < -1.96) + P(Z > 1.96) = 0.0250+0.0250 = 0.05

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