A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 mm and standard deviation 0.08 mm. a) What proportion of the diameters are less than 25.0 mm? b) Find the 10th percentile of the diameters. c) A particular ball bearing has a diameter of 25.2 mm. What percentile is its diameter on? d) To meet a certain specification, a ball bearing must have a diameter between 25.0 and 25.3 millimeters. What proportion of the ball bearings meet the specification?

Question

Anneliese

Mathematics

29 May, 07:10

A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 mm and standard deviation 0.08 mm. a) What proportion of the diameters are less than 25.0 mm? b) Find the 10th percentile of the diameters. c) A particular ball bearing has a diameter of 25.2 mm. What percentile is its diameter on? d) To meet a certain specification, a ball bearing must have a diameter between 25.0 and 25.3 millimeters. What proportion of the ball bearings meet the specification?

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Answers (1)

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Milo Doyle · Answer 1

(a) The proportion of the diameters are less than 25.0 mm is 0.1056.

(b) The 10th percentile of the diameters is 24.99 mm.

(c) The ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d) The proportion of the ball bearings meeting the specification is 0.8881.

Step-by-step explanation:

Let X = diameters of ball bearings.

The random variable X is normally distributed with mean, μ = 25.1 mm and standard deviation, σ = 0.08 mm.

To compute the probability of a Normally distributed random variable we need to first convert the raw scores to z-scores as follows:

z = (X - μ) : σ

(a)

Compute the probability of X < 25.0 mm as follows:

P (X < 25.0) = P ((X - μ) / σ < (25.0-25.1) / 0.08)

= P (Z < - 1.25)

= 1 - P (Z < 1.25)

= 1 - 0.8944

= 0.1056

*Use a z-table for the probability.

Thus, the proportion of the diameters are less than 25.0 mm is 0.1056.

(b)

The 10th percentile implies that, P (X < x) = 0.10.

Compute the 10th percentile of the diameters as follows:

P (X < x) = 0.10

P ((X - μ) / σ < (x-25.1) / 0.08) = 0.10

P (Z < z) = 0.10

z = - 1.282

The value of x is:

z = (x - 25.1) / 0.08

-1.282 = (x - 25.1) / 0.08

x = 25.1 - (1.282 * 0.08)

= 24.99744

≈ 24.99

Thus, the 10th percentile of the diameters is 24.99 mm.

(c)

Compute the value of P (X < 25.2) as follows:

P (X < 25.2) = P ((X - μ) / σ < (25.2-25.1) / 0.08)

= P (Z < 1.25)

= 0.8944

≈ 0.84

*Use a z-table for the probability.

Thus, the ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d)

Compute the value of P (25.0 < X < 25.3) as follows:

P (25.0 < X < 25.3) = P ((25.0-25.1) / 0.08 < (X - μ) / σ < (25.3-25.1) / 0.08)

= P (-1.25 < Z < 2.50)

= P (Z < 2.50) - P (Z < - 1.25)

= 0.99379 - 0.10565

= 0.88814

≈ 0.8881

*Use a z-table for the probability.

Thus, the proportion of the ball bearings meeting the specification is 0.8881.