A small manufacturing company recently instituted Six Sigma training for its employees. Two methods of training were offered: online and traditional classroom. Management was interested in whether the division in which employees worked affected their choice of method. Below is a table summarizing the data.

Sales Quality Operations Total
Traditional 16 10 8 34
Online 35 23 44 102
Total 51 33 52 136

Required:
a. What is the probability that an employee chose online training?
b. What is the probability that an employee is in the Quality division and chose online training?
c. What is the probability that an employee chose online training given that he/she is in the Sales division?

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MrRoyal MrRoyal · Answer 1 · 2021-07-13T20:44:12+02:00

Answer:

(a) [tex]P(Online\ Training) = 0.750[/tex]

(b) [tex]Pr = 0.169[/tex] --- Quality Division and Online Training

(c) [tex]P(A\ |\ B) = 0.686[/tex] --- Online Training given Sales Division

Step-by-step explanation:

Given

The two-way table

Solving (a): P(Online Training)

The total employee is:

[tex]Total = 136[/tex]

The employees for online training is:

[tex]Online\ Training = 102[/tex]

So, the probability is:

[tex]P(Online\ Training) = \frac{102}{136}[/tex]

[tex]P(Online\ Training) = 0.750[/tex]

Solving (b): P(Quality Division and Online Training)

The number of employees that choose quality Division and online training is 23

So, the probability is:

[tex]Pr = \frac{23}{136}[/tex]

[tex]Pr = 0.169[/tex]

Solving (c): P(Online Training | Sales Division)

This is calculated as:

Let:

[tex]A \to[/tex] Online training

[tex]B \to[/tex] Sales division

So, we have:

[tex]P(A\ |\ B) = \frac{n(A\ n\ B)}{n(B)}[/tex]

From the table:

[tex]n(A\ n\ B) =35[/tex]

[tex]n(B) = 16 + 35 = 51[/tex]

So, the probability is:

[tex]P(A\ |\ B) = \frac{35}{51}[/tex]

[tex]P(A\ |\ B) = 0.686[/tex]