The diameters of bearings have a normal distribution with a mean of \(8 \mathrm{~mm}\) and a standard deviation of \(0.04 \mathrm{~mm}\). In a batch of 6000 bearings how many would you expect to have a diameter of (a) more than \(8.03 \mathrm{~mm}\) (b) less than \(7.95 \mathrm{~mm}\) (c) between \(8.01 \mathrm{~mm}\) and \(8.06 \mathrm{~mm}\) (d) more than \(2.5\) standard deviations from the mean?

To calculate z-scores for each diameter range, we use the formula: \(z = \frac{x - \mu}{\sigma}\) where \(x\) is the diameter value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. We need to calculate the z-score for each diameter value (8.03, 7.95, 8.01 and 8.06 mm) using the mean (8 mm) and standard deviation (0.04 mm). For 8.03 mm, the z-score is: \(z = \frac{8.03 - 8}{0.04} = 0.75\) For 7.95 mm, the z-score is: \(z = \frac{7.95 - 8}{0.04} = -1.25\) For 8.01 mm, the z-score is: \(z = \frac{8.01 - 8}{0.04} = 0.25\) For 8.06 mm, the z-score is: \(z = \frac{8.06 - 8}{0.04} = 1.5\)

Now we will use a standard normal distribution table to find the proportions associated with the corresponding z-scores. For a z-score of 0.75, the table value is 0.7734 For a z-score of -1.25, the table value is 0.2119 For a z-score of 0.25, the table value is 0.5987 For a z-score of 1.5, the table value is 0.9332

We will now use the proportions and the total number of bearings (6000) to calculate the expected number of bearings in each of the diameter ranges. (a) To find the expected number of bearings with a diameter more than 8.03 mm: We will subtract the table value of the z-score 0.75 (0.7734) from 1, as we are looking for the proportion of the bearings with a higher diameter, i.e., 1 - 0.7734 = 0.2266. Expected number of bearings = 0.2266 * 6000 = 1359.6 ≈ 1360 bearings (b) To find the expected number of bearings with a diameter less than 7.95 mm: Since we have the table value of the z-score -1.25 (0.2119), we can directly use it as the proportion of the bearings with a lower diameter. Expected number of bearings = 0.2119 * 6000 = 1271.4 ≈ 1271 bearings (c) To find the expected number of bearings with a diameter between 8.01 mm and 8.06 mm: We will subtract the table value of the z-score 0.25 (0.5987) from that of the z-score 1.5 (0.9332) to get the proportion of bearings within this diameter range, i.e., 0.9332 - 0.5987 = 0.3345. Expected number of bearings = 0.3345 * 6000 = 2007 bearings (d) To find the expected number of bearings with a diameter more than 2.5 standard deviations from the mean: Since 2.5 standard deviations from the mean corresponds to diameters less than 7.94 mm and more than 8.06 mm, we will first find the table value of a z-score of -2.5 (0.0062) and use it with the z-score 1.5 (0.9332) as we did before. Proportion of bearings with a diameter more than 2.5 standard deviations from the mean: (1 - 0.9332) + 0.0062 = 0.073. Expected number of bearings = 0.073 * 6000 = 438 bearings So, we can expect: (a) 1360 bearings to have a diameter more than 8.03 mm, (b) 1271 bearings to have a diameter less than 7.95 mm, (c) 2007 bearings to have a diameter between 8.01 and 8.06 mm, and (d) 438 bearings to have a diameter more than 2.5 standard deviations from the mean.