Question

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

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

Step-by-step solution

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.

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, \(Z_{2} = \frac{0.500 + 0.004 - 0.499}{0.002} = 2.5\)

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 \(P(Z \leq 2.5) = 0.9938\)

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%