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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 inches. The rods that are either shorter than 49.886 inches or longer than 50.150 inches are discarded. What percentage of the rods made on this machine are discarded?

User Ymv
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Final answer:

To find the percentage of discarded rods, calculate the z-scores for the lengths 49.886 and 50.150 inches and use the normal distribution to find the cumulative probabilities. The probabilities of rods being too short or too long add up to give the total percentage discarded.

Step-by-step explanation:

To determine what percentage of iron rods are discarded, we will calculate the probability that a rod is either shorter than 49.886 inches or longer than 50.150 inches, using the normal distribution properties.

The z-score for a given length x is calculated by the formula:

Z = (x - μ) / σ

Where μ is the mean length, and σ is the standard deviation.

First, we calculate the z-scores for the cut-off lengths:

  • Z for 49.886 inches = (49.886 - 50) / 0.06 ≈ -1.900
  • Z for 50.150 inches = (50.150 - 50) / 0.06 ≈ 2.500

The corresponding probabilities can be found using a Z-table or a calculator with normal distribution functions.

The probability of a rod being shorter than 49.886 inches is the cumulative probability up to a Z-score of -1.900.

The probability of a rod being longer than 50.150 inches is found by subtracting the cumulative probability up to a Z-score of 2.500 from 1 (since it's the probability of being in the higher tail).

The total probability of discarding a rod is the sum of these two probabilities.

User Gwenn
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