Ball bearings are manufactured with a mean diameter of 5 millimeters (mm). Becau

Question

Ball bearings are manufactured with a mean diameter of 5 millimeters (mm). Because of variability in the manufacturing process, the diameters of the ball bearings are approximately normally distributed, with a standard deviation of 0.02 mm. (a) Any ball bearings that are greater than 5.05 mm and less than 4.95 mm are discarded. What proportion of the ball bearings will be discarded? (b) Using the result from part (a), if 30,000 ball bearings are manufactured, how many should management expect to discard? (c) If an order comes in for 50,000 ball bearings, how many bearings should the plant manufacture if the order states that all ball bearings must be between 4.97 mm and 5.03 mm?

Explanation / Answer

We are given,

Mean = 5

Standard deviation = 0.02

Part a)

Here we need to find the below probability,

1 – P (4.95 < x < 5.05)

z = (x – Mean) / Standard deviation

The z-score for 4.95 = (4.95 – 5)/0.02 = -2.5

The z-score for 5.05 = (5.05 – 5 ) / 0.02 = +2.5

P (4.95 < x < 5.05)

= P (-2.5 < z < 2.5)

= P (z < 2.5) – P (z<-2.5)

From Normal Distribution Table we get,

= 0.9938 – 0.0062

= 0.9876

1 – P (4.95 < x < 5.05) = 1 - 0.9876 = 0.0124

Answer: 0.0124

Part b)

np = 30,000 *0.0124 = 372.6 ~ 373

Answer: 373

Part c)

P (4.97 < x < 5.03)

The z-score for 4.97 = (4.97 – 5)/0.02 = -1.5

The z-score for 5.03 = (5.03 – 5)/0.02 = 1.5

P (-1.5 < z < 1.5)

= P (z<1.5)- P(z<-1.5)

By using Normal Distribution Table we get,

= 0.9332 - 0.0668

= 0.8664

Now we can write

0.8664 = 50000

Total = 50000/ 0.8664

          = 57710.1 ~ 57710

Answer: 57710

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