A machine makes iron rods that are supposed to be 50 inches long. It is known that probability distribution of rods made on this machine is normal with mean of 50 inches and standard deviation of .06 inches. The rods that are either shorter than 49.85 inches or longer than 50.15 inches are discarded. What percentage of rods is discarded?
Added by Reop K.
Step 1
First, we need to find the z-scores for the given lengths (49.85 inches and 50.15 inches). The z-score is calculated as (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation. Show more…
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A machine at Kasem Steel Corporation makes iron rods that are supposed to be 50 inches long. However, the machine does not make all rods of exactly the same length. It is known that the probability distribution of the lengths of rods made on this machine is normal with a mean of 50 inches and a standard deviation of $.06$ inch. The rods that are either shorter than $49.85$ inches or longer than $50.15$ inches are discarded. What percentage of the rods made on this machine are discarded?
Anna D.
A machine at Kasem Steel Corporation makes iron rods that are supposed to be 50 inches long. However, the machine does not make all rods of exactly the same length. It is known that the probability distribution of the lengths of rods made on this machine is normal with a mean of 50 inches and a standard deviation of 0.06 inches. The rods that are either shorter than 49.874 inches or longer than 50.138 inches are discarded. What percentage of the rods made on this machine are discarded?
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