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 47.8 cm and a standard deviation of 3.72 cm (Overgaard, Berndt & Linneryd, 2012). Assume the length of fish is normally distributed. What is the length in cm of the longest 10% of Atlantic cod in this area

Answer 1

Answer:

The length in cm of the longest 10% of Atlantic cod in this area is at least 53.92 cm.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

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

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

$\mu = 47.8, \sigma = 3.72$

What is the length in cm of the longest 10% of Atlantic cod in this area

At least x cm.

x is the 100 - 10 = 90th percentile, which is X when Z has a pvalue of 0.9. So X when Z = 1.645. Then

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

$1.645 = \frac{X - 47.8}{3.72}$

$X - 47.8 = 1.645*3.72$

$X = 53.92$

The length in cm of the longest 10% of Atlantic cod in this area is at least 53.92 cm.