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) What proportion of ball bearings has a diameter more than 5.03 mm?
(b) Any ball bearings that have a diameter less than 4.95 mm or greater than 5.05 mm are discarded. What proportion of ball bearings will be discarded?
(c) Using the results of part (b), if 30,000 ball bearings are manufactured in a day, how many should the plant manager expect to discard?
(d) If an order comes in for 50,000 ball bearings, how many bearings should the plant manager manufacture if the order states that all ball bearings must be between 4.97 mm and 5.03 mm?

Answer 1

Answer:

a) 0.0668 of ball bearings have a diameter more than 5.03 mm

b) 0.0124 of ball bearings will be discarded

c) The plant manager should expect to discard 372 balls.

d) 57,710 ball bearings should be manufactured.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 5, \sigma = 0.02$

(a) What proportion of ball bearings has a diameter more than 5.03 mm?

This is 1 subtracted by the pvalue of Z when X = 5.03. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.03 - 5}{0.02}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 of ball bearings have a diameter more than 5.03 mm

(b) Any ball bearings that have a diameter less than 4.95 mm or greater than 5.05 mm are discarded. What proportion of ball bearings will be discarded?

Lower than 4.95 or greater than 5.05. Both these proportions are the same, so we find one of them, and multiply by 2.

Less than 4.95:

pvalue of Z when X = 4.95. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.95 - 5}{0.02}$

$Z = -2.5$

$Z = -2.5$ has a pvalue of 0.0062

2*0.0062 = 0.0124

0.0124 of ball bearings will be discarded.

(c) Using the results of part (b), if 30,000 ball bearings are manufactured in a day, how many should the plant manager expect to discard?

0.0124 of 30,000. So

0.0124*30000 = 372

The plant manager should expect to discard 372 balls.

(d) If an order comes in for 50,000 ball bearings, how many bearings should the plant manager manufacture if the order states that all ball bearings must be between 4.97 mm and 5.03 mm?

Proportion of bearings between 4.97 mm and 5.03 mm:

pvalue of Z when X = 5.03 subtracted by the pvalue of Z when X = 4.97. So

X = 5.03

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5.03 - 5}{0.02}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 4.97

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4.97 - 5}{0.02}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9332 - 0.0668 = 0.8664 of ball bearings will have adequate proportions. So 0.8664 of x should be 50,000. then

$0.8664x = 50000$

$x = \frac{50000}{0.8664}$

$x = 57 710$

57,710 ball bearings should be manufactured.