The owner of a seafood market determined that the average weight for a crab is 1.6 pounds with a standard deviation of 0.4 pound. What weight is exceeded by 2% of all of the crabs? (Assume the weights are normally distributed.)
A) 0.78 pounds
B) 1.82 pounds
C) 2.42 pounds
D) 4.36 pounds

Question

Quizplus · Accepted Answer

First, we need to find the z-score that corresponds to 2% of the distribution being exceeded. This is equivalent to finding the z-score with an area of 0.02 to the right of it. Using a standard normal distribution table or calculator, we find this z-score to be approximately 2.05.

Next, we can use the formula: 
z = (x - μ) / σ 
where x is the weight we want to find, μ is the mean weight (1.6 pounds), σ is the standard deviation (0.4 pound), and z is the z-score we just found.

Rearranging the formula, we get: 
x = z * σ + μ

Plugging in the values, we get: 
x = 2.05 * 0.4 + 1.6 
x = 2.42 pounds

Therefore, the weight exceeded by 2% of all crabs is 2.42 pounds, so the best choice is C.

The Owner of a Seafood Market Determined That the Average