Mathematics, 28.10.2020 14:20 josephmelichar777

A sample of employees of Worldwide Enterprises is to be surveyed about a new health care plan. The employees are classified as follows:
Classification Event Number of Employees
Supervisors A 120
Maintenance B 50
Production C 1,460
Management D 302
Secretarial E 68
(a) What is the probability that the first person selected is:
(i) either in maintenance or a secretary?
(ii) not in management?

(c) Are the events in part (a)(i) complementary or mutually exclusive or both

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Answer from: MayFlowers

Using the probability concept, it is found that:

a) i) There is a 0.059 = 5.9% probability that the first person selected is either in maintenance or a secretary.

ii) There is a 0.849 = 84.9% probability that the first person selected is not in management.

b) The events are neither complementary nor mutually exclusive.

A probability is the number of desired outcomes divided by the number of total outcomes.

Item a:

In this problem, there is a total of 120 + 50 + 1460 + 302 + 68 = 2000 people.

Of those, 50 + 68 = 118 are either in maintenance or a secretary, hence:

$p = \frac{118}{2000} = 0.059$

There is a 0.059 = 5.9% probability that the first person selected is either in maintenance or a secretary.

Item ii:

2000 - 302 = 1698 people are not in management, hence:

$p = \frac{1698}{2000} = 0.849$

There is a 0.849 = 84.9% probability that the first person selected is not in management.

Item b:

If a person is either in maintenance or a secretary, they are also not in management, hence the events are neither complementary nor mutually exclusive.

To learn more about the probability concept, you can take a look at link

Answer from: fatlenny

Following are the solution to the given points:

For point a:

$\to P(B \cup E) = P(B) + P(E) - P( B\cap E)$

Where $P(B \cap E) = 0$ because they are both separate occasions It is impossible to be a member of both departments.

$\to P(B\cup E) =\frac{50}{120 + 50 + 1460 + 302 + 68} +\frac{68}{120 + 50 + 1460+ 302 +68}\\\\$

$=\frac{50}{2000} +\frac{68}{2000}\\\\=\frac{50+68}{2000} \\\\=\frac{118}{2000} \\\\=\frac{ 59}{1000} \\\\=0.059\\$

ii)

$\to P(D^c) = 1 - P(D)$

$=1- (\frac{302}{120+50+1460+302+68})\\\\=1- (\frac{302}{2000})\\\\=1- (\frac{ 151}{1000})\\\\= 1- 0.151 \\\\= 0.849\\\\$

For point b:

Please find the attached file.

For point c:

Events in a. I are mutually incompatible since $P(B \cap E) = 0$ . One would be a part of the both departments.

Learn more:

link

A sample of employees of Worldwide Enterprises is to be surveyed about a new health

care plan. The

Answer from: Quest

the answer i think is a)

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a) |-1|, |3/2|, |-2|, |-3|, |3.5|

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