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Chapter 7: Problem 78

A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is \(0.500\) in. A bearing is acceptable if its diameter is within \(0.004\) in. of this target value. Suppose, however, that the setting has changed during the course of production, so that the distribution of the diameters produced is well approximated by a normal distribution with mean \(0.499\) in. and standard deviation \(0.002\) in. What percentage of the bearings produced will not be acceptable?

Short Answer

Expert verified
The percentage of the ball bearings that will not be acceptable is 31.47%.

Step by step solution

01

Identify the given parameters

From the problem, we have a normally distributed variable (the diameter of the ball bearings) with mean \(0.499\) in and standard deviation \(0.002\) in. The acceptable range of the diameters is \(0.500 \pm 0.004\) in.
02

Calculate the Z-scores

The Z-score is a measure of how many standard deviations an element is from the mean. For the lower limit of the acceptable range, \(Z_{1} = \frac{0.500 - 0.004 - 0.499}{0.002} = -0.5\). For the upper limit of the acceptable range, \(Z_{2} = \frac{0.500 + 0.004 - 0.499}{0.002} = 2.5\)
03

Calculate the probabilities

Refer to the Z-table (standard normal distribution table) to find the probabilities for these Z-scores. For \(Z = -0.5\), the probability \(P(Z \leq -0.5) = 0.3085\). For \(Z = 2.5\), the probability \(P(Z \leq 2.5) = 0.9938\)
04

Determine the percentage of unacceptable bearings

'Unacceptable' means the diameter is outside the acceptable range. We find this as \(1 - [P(-0.5 \leq Z \leq 2.5)] = 1 - [P(Z \leq 2.5) - P(Z \leq -0.5)] = 1 - [0.9938 - 0.3085] = 0.3147\), or 31.47%

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Most popular questions from this chapter

Consider babies born in the "normal" range of \(37-\) 43 weeks gestational age. Extensive data support the assumption that for such babies born in the United States, birth wcight is normally distributed with mean \(3432 \mathrm{~g}\) and standard deviation \(482 \mathrm{~g}\) ("Are Babies Normal?" The American Statistician [1999]: \(298-302\) ). a. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(4000 \mathrm{~g}\) ? is between 3000 and \(4000 \mathrm{~g}\) ? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than \(2000 \mathrm{~g}\) or greater than \(5000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(7 \mathrm{lb}\) ? (Hint: \(1 \mathrm{lb}=453.59 \mathrm{~g} .)\) d. How would you characterize the most extreme \(0.1 \%\) of all birth weights? e. If \(x\) is a random variable with a normal distribution and \(a\) is a numerical constant \((a \neq 0)\), then \(y=a x\) also has a normal distribution. Use this formula to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from Part (c). How does this compare to your previous answer?

A box contains four slips of paper marked \(1,2,3\), and 4\. Two slips are selected without replacement. List the possible values for each of the following random variables: a. \(x=\) sum of the two numbers b. \(y=\) difference between the first and second numbers c. \(z=\) number of slips selected that show an even number d. \(w=\) number of slips selected that show a 4

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