Question

The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm (Ovegard, Berndt & Lunneryd, 2012). Assume the length of fish is normally distributed. What is the length in cm of the longest 15% of Atlantic cod in this area? Round answer to 2 decimal places.

Answer 1

Answer:

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have that:

A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm, so $\mu = 49.9, \sigma = 3.74$.

What is the length in cm of the longest 15% of Atlantic cod in this area?

We have to find the value of X for the value of Z that has a pvalue of 0.85.

Looking at the zscore table, we have that Z = 1.04 has a pvalue of 0.8508. So, we have to find the value of X when $Z = 1.04, \mu = 49.9, \sigma = 3.74$.

So

$Z = \frac{X - \mu}{\sigma}$

$1.04 = \frac{X - 49.9}{3.74}$

$X - 49.9 = 3.8896$

$X = 53.7896$

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.