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Chapter 4

Q. 4.21

A process manufactures ball bearings whose diameters are normally distributed with mean 2.505 cm and standard deviation 0.008 cm. Specifications call for the diameter to be in the interval 2.5 ± 0.01 cm. What proportion of the ball bearings will meet the specification?

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Verified Solution

Let X represent the diameter of a randomly chosen ball bearing. Then X ∼ N(2.505, 0.008²). Figure 4.10 presents the probability density function of the N (2.505, 0.008²) population. The shaded area represents P(2.49 < X < 2.51), which is the proportion of ball bearings that meet the specification.
We compute the z-scores of 2.49 and 2.51:

z = \frac{2.49  –  2.505}{0.008} = -1.88

z =\frac{2.51  –  2.505}{0.008} = 0.63

The area to the left of z = -1.88 is 0.0301. The area to the left of z = 0.63 is 0.7357. The area between z = 0.63 \text{ and } z = – 1.88 is 0.7357 – 0.0301 = 0.7056. Approximately 70.56% of the diameters will meet the specification.

fig 4.10