Question

Based on experience, the Ball Corporation’s aluminum can manufacturing facility in Ft. Atkinson, Wisconsin, knows that the metal thickness of incoming shipments has a mean of 0.2731 mm with a standard deviation of 0.000959 mm.
a. A certain shipment has a diameter of 0.2761. Find the standardized: score of the shipment.
b. Is this an outerlier?

Answer 1

Answer:

a) 3.128

b) Yes, it is an outerlier

Step-by-step explanation:

The standardized z-score for a particular sample can be determined via the following expression:

z_i = {x_i -\bar x}/{s}

Where;

\bar x = sample means

s = sample standard deviation

Given data:

the mean shipment thickness (\bar x) = 0.2731 mm

With the standardized deviation (s) = 0.000959 mm

The standardized z-score for a certain shipment with a diameter x_i= 0.2761 mm can be determined via the following previous expression

z_i = {x_i -\bar x}/{s}

z_i = {0.2761-0.2731}/{ 0.000959}

z_i = 3.128

b)

From the standardized z-score

If [z_i < 2]; it typically implies that the data is unusual

If [z_i > 2]; it means that the data value is an outerlier

However, since our z_i > 3 (I.e it is 3.128), we conclude that it is an outerlier.