Question

The owner of a fish market determined that the average weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. A very good in weight catfish should be one of the top 2% in weight. Assuming the weights of catfish are normally distributed, what is the lower limit of these 2%?

Answer 1

Answer:

Mean = 3.2 pounds

Standard deviation = 0.8

z = 2.054 for top 2%

We have z = (x-mean)/standard deviation

Or, 2.054 = (x-3.2)/0.8

Or, x = 3.2+2.054*0.8 = 4.843 pounds

Hence, citation designation should be at 4.843 pounds.

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Answer 2

Given:

The average weight of the catfish = 3.2 pounds

Standard deviation = 0.8 ponds

To Find:

The lower limit of these 2%

Step-by-step explanation:

Hence the Mean = 3.2 pounds

Therefore z = 2.054 ( for that 2% top)

Given z = [(x-mean)/(standard deviation)]

2.054 = [(x-3.2)/(0.8)]

x = 3.2+2.054×0.8

   = 4.843 pounds

Answer- 4.483 pounds  is the lower limit of 2%