Question 1167969: 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?
I know that the answer is 50% or 0.5 but I am having trouble figuring out how to get to that answer. I think it is normal distribution but I don't know where to begin.
Answer by ikleyn(52884)   (Show Source):
You can put this solution on YOUR website! .
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?
I know that the answer is 50% or 0.5 but I am having trouble figuring out how to get to that answer.
I think it is normal distribution but I don't know where to begin.
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I am very glad to find this problem. Usually, in this circle of problems, all problems are
typical and the whole clearing is trampled, so it is a rare event to find something unordinary.
While this one is a truly unordinary (although is not complicated).
Let's draw the coordinate axis and mark there the given numbers.
The coordinate axis represents the weight of catfish.
The mark 3.2 is the mean (the average weight), in pounds.
We are given in the problem that 20% of the fish weigh is on the right of the mark 3.8 pounds,
and 30% of the fish weight is on the left of the mark 2.8 pounds.
Notice that in this problem, the fish weight is always the area under the normal curve
between the given marks, or on the left of the given mark, or on the right of the given mark.
Using the mean, it is the same as to say that 20% of the fish weight is on the right of the mark 3.2+0.6 pounds,
and 30% of the fish weight is on the left of the mark 3.2-0.4 pounds.
So, 50% of the fish weight is in the interval (3.2-0.4,3.2+0.6) pounds.
Now, the question relates to the interval (2.6, 3.6) pounds.
This interval is the same as (3.2-0.6,3.2+0.4) pounds.
Thus, we know that 50% of the fish weight is in the interval (-0.4,+0.6) around the mean,
and the question is what percent of the fish weight is in the interval (-0.6,+0.4) around the mean.
Due to the symmetry of the normal distribution curve, the answer is that
the percent of the fish weight in the interval (-0.6,+0.4) around the mean
is the same 50% as in the interval (-0.4,+0.6) around the mean.
It is the ANSWER to the problem's question.
At this point, the problem is solved in full.
The key to the solution is that the interval under the question
is SYMMETRIC to the interval GIVEN in the problem.
You posted an ELEGANT problem to the forum and get an ELEGANT solution back.
My congratulations ( ! )
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