Question 1207237
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 <a href = "" target = "_blank"></a>


i used the ti-04 plus calculator to get the answers.
i used the calculator at <a href = "" target = "_blank"></a> 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.


<img src = "http://theo.x10hosting.com/2024/050801.jpg">


<img src = "http://theo.x10hosting.com/2024/050802.jpg">


<img src = "http://theo.x10hosting.com/2024/050803.jpg">


<img src = "http://theo.x10hosting.com/2024/050804.jpg">