Question

Catfish weights are normally distributed with a mean of 3.2 pounds and a standard deviation of 0.8 pounds. What is the probability that a randomly selected catfish weighs between 3 and 4 pounds?

A “champion” catfish is one that weighs more than 98% of all catfish. How much must a catfish weigh in order to earn “champion” status?

Answer 1

1) 0.44005 or 44% roughly
2) 4.842 pounds

Explanation:

Question 1

We know the weights are normally distributed according to a mean `mu` of 3.2 pounds and a standard deviation `sigma` of 0.8 pounds. A common way to describe this sort of normal distribution is `N(3.2, 0.8^2)`.

To estimate the probability that a randomly selected catfish weighs between 3 and 4 points requires that we convert the weights to a standard normal distribution `N(0, 1^2)` and then properly use z-scores to determine the area between each respective z-score.

First, I'll denote `z_3^star` as the z-score for the 3 pound lower weight limit, and `z_4^star` as the z-score for the 4 pound upper weight limit. We can convert those weights into z-scores using the formula:

`z^star = (X - mu)/sigma`

(where `X` = catfish weight, `mu` = mean of distribution, and `sigma` = std. deviation of distribution)

Thus, we find:

`z_3^star = (3 - 3.2)/(0.8) = (-0.2)/0.8 = -0.25`

`z_4^star = (4 - 3.2)/0.8 = 0.8/0.8 = +1`

These values make sense in the context of the problem; a weight of 3 pounds is below the mean, thus we expect a negative z-score, while a weight of 4 pounds is exactly 1 standard deviation of weight above the mean, hence a positive z-score of 1.

Now, to find the probability we are seeking we relate this problem to the probability that a z-score `Z` lies between these two z-scores just calculated:

`P(z_3^star < Z < z_4^star) or P(-0.25 < Z < +1)`

Doing this requires a little cleverness on our part and the use of a z-score table (or calculator). We can use tables to determine the left-tail probability `P(Z < +1)`, which would be the probability of a z-score falling anywhere less than a z-score of +1. This would overcount our desired probability, however, because this would include all z-scores < -0.25. Thus, we actually need to subtract from that `P(Z < -0.25)`:

`P(-0.25 < Z < 1) = P(Z < 1) - P(Z < -0.25)`

From tables, we see that `P(Z < 1)` is roughly 0.84134, while #P(Z < -0.25) is roughly 0.40129. Thus:

#P(-0.25 < Z < 1) = 0.84134 - 0.40129 = 0.44005.

There's about a 44% chance of a randomly selected catfish weight falling between 3 and 4 pounds.

Question 2

To find the weight a catfish should be so that it's weight falls in the top 2% of all catfish (and thus 98% of all catfish weights are below it) requires us to work backwards from a z-score table. In this case, we need to look up the z-score `z^star` where the right-tail probability `P(Z > z^star) = 0.02`.

Looking up a z-score table tells us that `z^star` should be roughly 2.053 (to 3 decimal places). This is the z-score where 98% of the normal distribution curve area (aka probability) falls to the left of this z-score.

Using the z-score formula, we can now calculate the weight this corresponds to under our original distribution `N(3.2, 0.8^2)`:

`z^star = (X - mu)/sigma`

`2.053 = (X - 3.2)/0.8`

`2.053*0.8 = X-3.2`

`1.6424 = X - 3.2`

`:. X = 1.6424 + 3.2 = 4.842 " pounds"`