Question

The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A citation catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, at what weight should the citation designation be established

Answer 1

Answer:

4.844 pounds

Step-by-step explanation:

Problems of normally distributed samples are 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:

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

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 problem, we have that:

$\mu = 3.2, \sigma = 0.8$

Top 2%.

X when Z has a pvalue of 1-0.02 = 0.98. So X when Z = 2.055.

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

$2.055 = \frac{X - 3.2}{0.8}$

$X - 3.2 = 2.055*0.8$

$X = 4.844$

Answer 2

Answer:

The weight of the citation designation should be at 4.8432 pounds.

Explanation:

Given

Mean $= 3.2 pounds.$

Standard deviation$= 0.8 pound.$

Step 1:

Consider 'y' as one of the top weight, that is, $y = 2 \% = 2.054 pounds.$

Let 'x' be the weight of the citation designation.

$y = \frac{x-mean}{standard\ deviation}$

$=2.054 = \frac{x-3.2}{0.8}$

$=2.054\times 0.8 = x-3.2$

$=1.6432 = x-3.2$

$x = 1.6432+3.2$

 $x = 4.8432$

Thus, at 4.8432 pounds citation designation be established.

To learn more about citation, refer:

• https://brainly.com/question/18675815