Question

A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 millimeters and...

A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 millimeters and standard deviation 0.08 millimeter.

(a) What proportion of the diameters are less than 25.0 millimeters?

(b) What proportions of the diameters are greater than 25.4?

(c) To meet a certain specification, a ball bearing must have a diameter between 25.0 and 25.3 millimeters. What proportions of the ball bearings meet specification?

(d) Find the 95th percentile of the diameters.

Homework Answers

Answer #1

Solution :

Given that ,

mean = = 25.1

standard deviation = =0.08

(a)

P(x < 25.0) = P[(x - ) / < (25.0 - 25.1) / 0.08]

= P(z < -1.25)

= 0.1056

proportion = 0.1056

(b)

P(x > 25.4) = 1 - P(x < 25.4)

= 1 - P[(x - ) / < (25.4 - 25.1) / 0.08)

= 1 - P(z < 3.75)

= 1 - 0.9999

= 0.0001

proportions = 0.0001

(c)

P(25.0 < x < 25.4) = P[(25.0 - 25.1)/ 0.08) < (x - ) /  < (25.3 - 25.1) / 0.08) ]

= P(-1.25 < z < 2.5)

= P(z < 2.5) - P(z < -1.25)

= 0.9938 - 0.1056

= 0.8882

proportions = 0.8882

(d)

Using standard normal table ,

P(Z < z) = 95%

P(Z < 1.65) = 0.95

z = 1.65

Using z-score formula,

x = z * +

x = 1.65 * 0.08 + 25.1 = 25.23

The 95th percentile is 25.23

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