SOLUTION: A fish stall owner at a market place claims that the weight of a catfish he is selling is approximately normally distributed with a mean of 3 pounds and a standard deviation of 0.5

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Question 1207237: A fish stall owner at a market place claims that the weight of a catfish he is selling is approximately normally distributed with a mean of 3 pounds and a standard deviation of 0.5 pounds. A random sample of 20 catfishes is selected.
What is the PROBABILITY that the average weight is at least 2.8 pounds?
What is the PROBABILITY that the average weight is between 2.8 pounds and 3.2 pounds?
What is the AVERAGE WEIGHT of top 10% of catfish?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
population mean is assumed to be 3 pounds.
population standard deviation is assumed to be .5 pounds.

sample size is 20.

you have to use the standard error because you are looking for a distribution of sample means, rather than a distribution of sample elements.

standard error = standard deviation / sqrt(sample size) = .5 / sqrt(30) = .111803 rounded to 6 decimal places.

you are using the z-score because the standard deviation is taken from the population.

What is the PROBABILITY that the average weight is at least 2.8 pounds?

z = (2.8 - 3) / .111803 = -1.7889

area under the normal distribution curve to the right of that z-score is equal to .96318.

probability of getting a z-score greater than that is equal to .96318 rounded to 5 decimal places.

that's the probability of getting a sample with average weight greater than 2.8 pounds.

What is the PROBABILITY that the average weight is between 2.8 pounds and 3.2 pounds?

z-score for 2.8 pounds = (2.8 - 3) / .111803 = -1.7889.
z-score for 3.2 pounds = (3.2 - 3) / .111803 = 1.7889.

area to the left of z-score 0f -1.7889 = .03682 rounded to 5 decimal places.
area to the left of z-score of 1.7889 = .96318 rounded to 5 decimal places.
area in between = larger area minus smaller area = .92636.

that's the probability of getting a z-score greater than -1.7889 and less then 1.7889.

that's the probability of getting a sample with average weight greater than 2.8 and less than 3.2 pounds.

What is the AVERAGE WEIGHT of top 10% of catfish?

look for a z-score that has 10% of the area under the normal distribution curve to the right of it.
that's the same as a z-score with 90% of the area under the normal distribution curve to the left of it.

i get z-score = 1.28188.

use the z-score formula to find the raw score.

1.28188 = (x - 3) / .111803.
solve for x to get x = 3.1433 rounded to 4 decimal places.

the probability of getting a sample with a mean that is in the top 10% of the possible sample means is getting a sample with a mean of 3.1433 rounded to 4 decimal places.

i used an online calculator to verify these answers.
the calculator can be found at

i used the ti-04 plus calculator to get the answers.
i used the calculator at to verify the answers were correct.
the key was to calculate the standard error because you were looking for sample mean rather than a sample element.
everything flowed from there.

here are the calculator results.