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Chapter 6: Problem 51

The lengths of 3 -inch nails manufactured on a machine are normally distributed with a mean of \(3.0\) inches and a standard deviation of \(.009\) inch. The nails that are either shorter than \(2.98\) inches or longer than \(3.02\) inches are unusable. What percentage of all the nails produced by this machine are unusable?

Short Answer

Expert verified
The percentage of nails produced by this machine that are unusable is 1.36%

Step by step solution

01

Calculate Z-Scores

To start with this problem, it should be understood that the length of the nails can be represented as a standardized z-Score, which can be calculated using the formula: \( z = \frac{x - \mu}{\sigma} \)\n\nApply this formula for the lengths \(2.98\) and \(3.02\):\n\nFor \(2.98\) inches: \( z_1 = \frac{2.98 - 3.0}{0.009} = -2.22 \) \n\nFor \(3.02\) inches: \( z_2 = \frac{3.02 - 3.0}{0.009} = 2.22 \)
02

Find Probabilities for the Z-scores

Using the z-table, find the probabilities corresponding to these z-values. For both \( z_1 = -2.22 \) and \( z_2 = 2.22 \), the probability is \(0.9864 \) as the z-score table is symmetrical about zero.
03

Calculate usuable and unusable percentages

The probability \(0.9864\) gives us the proportion of nails that are within the usable range of lengths. To get the percentage of unusable nails, subtract the usable percentage from the total, which is \(100\%\). So the percentage of unusable nails is: \(100\% - 98.64\% = 1.36\%\).

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Most popular questions from this chapter

The management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes a. Find the probability that a randomly selected customer will have to wait for less than 3 minutes. b. What percentage of the customers have to wait for 10 to 13 minutes? c. What percentage of the customers have to wait for 6 to 12 minutes? d. Is it possible that a customer may have to wait longer than 16 minutes for service? Explain.

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