Question 1168044: The owner of a fish market determined that the mean weight for a catfish is 3.2 pounds assuming that the weight of a catfish follows a normal distribution and its standard deviation is unknown. He also knew that that probability of a randomly selected catfish that would weigh more than 3.8 pounds is 20% and the probability that a randomly selected catfish that would weigh less than 2.8 pounds is 30%. What is the probability that a randomly selected catfish will weigh between 2.6 and 3.6 pounds?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the mean is 3.2
the weight of a catfish follows a normal distribution and its standard deviation is now known.
the probability of a catfish weighing more than 3.8 pounds is 20%.
the probability of a catfish weighing less than 2.8 pounds is 30%.
as far as i know, you can solve this using z-scores and then approximate the standard deviation from that.
from the z-score calculator, i get.
z-score that has .2 probability greater than it is equal to .8416212335.
z-score that has .3 probability less than it is equal to -.5244005101.
using the z-score formula i can approximate what the standard deviation would be with those scores.
the z-score formula is z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation or the standard error.
in this case s is the standard deviation.
when x = 3.8 and m = 3.2 and z = .8416212335, s = 0.71290977
when x = 2.8 and m = 3.2 and z = -.5244005101, s = 0.76277576
if all catfish came from the same population, s should have been the same, but it's not.
best thing you can do, as far as i know, is take an average of the two standard deviations.
average of the two standard deviations is 0.73784277
use that to solve the rest of your problem.
to find the probability that a random selected catfish weighs between 2.6 and 3.6 pounds, i used a normal distribution calculator.
the one that is used is in the TI-84 Plus.
with a standard deviation of .73784277, i got:
probability of a catfish weighing between 2.6 and 3.6 pounds = .49807604
visually, this looks like this:
the problem that you had here is that, you needed the standard deviation but you didn't have it.
the best you could do was approximate it.
if you were dealing with fish from the same population, then the standard deviation for all problems regarding that same population should have been the same, but it was not.
without knowing which one was right between the two that i calculated, i took the average.
if you redo the problem using the average standard deviation, you would get the following.
a fish weighing 2.8 pounds would have a probability of other fish having less weight than it of .293867331.
that's not .3, but it's pretty close.
a fish weighing 3.8 pounds would have a probability of other fish having more weight than it of .2080566471.
that's not .2, but it's pretty close.
i doubt that any of these probabilities would ever be right on, but the answers do appear reasonable using a standard deviation of .73784277 that was an average of the two standard deviations that i calculated using the z-score formula.
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