Question

Precision manufacturing: A process manufactures ball

bearings with diameters that are normally distributed
with mean 25.1 millimeters and standard deviation
0.08 millimeter.
a. Find the 60th percentile of the diameters.
b. Find the 32nd percentile of the diameters.
c. A hole is to be designed so that 1% of the ball bearings will
fit through it. The bearings that fit through the hole will be
melted down and remade. What should the diameter of the
hole be?

Answer 1

a. To find the 60th percentile of the diameters, we need to find the diameter that separates the smallest 60% of the diameters from the largest 40%. We can use a standard normal distribution table to find the z-score that corresponds to the 60th percentile:

z = invNorm(0.6) = 0.2533

Now we can use the formula z = (x - mu) / sigma to find the diameter x that corresponds to this z-score:

0.2533 = (x - 25.1) / 0.08

x - 25.1 = 0.020264

x = 25.120264

Therefore, the 60th percentile of the diameters is 25.120264 millimeters.

b. To find the 32nd percentile of the diameters, we need to find the diameter that separates the smallest 32% of the diameters from the largest 68%. We can use a standard normal distribution table to find the z-score that corresponds to the 32nd percentile:

z = invNorm(0.32) = -0.4472

Now we can use the formula z = (x - mu) / sigma to find the diameter x that corresponds to this z-score:

-0.4472 = (x - 25.1) / 0.08

x - 25.1 = -0.035776

x = 25.064224

Therefore, the 32nd percentile of the diameters is 25.064224 millimeters.

c. To find the diameter of the hole, we need to find the diameter that separates the smallest 1% of the diameters from the largest 99%. We can use a standard normal distribution table to find the z-score that corresponds to the 1st percentile:

z = invNorm(0.01) = -2.3263

Now we can use the formula z = (x - mu) / sigma to find the diameter x that corresponds to this z-score:

-2.3263 = (x - 25.1) / 0.08

x - 25.1 = -0.186104

x = 24.913896

Therefore, the diameter of the hole should be 24.913896 millimeters.