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Chapter 6: Problem 56

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?

Short Answer

Expert verified
Approximately 1.24% of rods produced by the machine are discarded due to size constraints.

Step by step solution

01

Calculate Z-scores

First, calculate the Z-scores for 49.85 inches and 50.15 inches.The Z-score formula is:\[Z = (X - 渭) / 蟽\]where:\(X\) is the value to be standardized (in this case, the lengths 49.85 inches and 50.15 inches),\(渭\) is the mean of the distribution (50 inches), and \(蟽\) is the standard deviation (0.06 inch).To find the Z-score for 49.85 inches:\[Z_1 = (49.85 - 50) / 0.06 = -2.5\]And for 50.15 inches:\[Z_2 = (50.15 - 50) / 0.06 = 2.5\]
02

Find the Probability

Having found the Z-scores, the next step is to find the probability corresponding to these scores. By looking at the standard normal distribution table, the probability (p) for a Z-score of -2.5 is \(0.0062\) or \(0.62\%\), and for 2.5 is \(0.9938\) or \(99.38\%\).Therefore, the percentage of rods within acceptable length is:\[100\% - (0.62\% + 0.62\%) = 98.76\%\]
03

Find the Percentage of Discarded Rods

To find the percentage of rods that are discarded, simply subtract the percentage of acceptable rods from 100\%:\[100\% - 98.76\% = 1.24\%\]

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