The employees in the textile industry can be segmented as follows:
Employees Number
Female and union 12,000
Female and nonunion 25,000
Male and union 21,000
Male and nonunion 42,000
a. Determine the probability of each event in this distribution. An event refers to the
combinations
shown in the Employees column. For example, female and union is one event while female and
nonunion is another event.
b. Are the events in this distribution mutually exclusive? Explain.
c. What is the probability that an employee is male?
d. Is this experiment of segmenting employees in the textile industry collectively exhaustive?
Explain.EmployeesNumberFemale and union12,000Female and nonunion25,000Male and
union21,000Male and nonunion42,000
Solution
FU = 12,000/(12,000 + 25,000 + 21,000 + 42,000) = .12
FN = .25
MU = .21
MN = .42
b. yes. male and female are exclusive categories, as are union and non-union
c. MU + MN = .63
d. yes.
A Critique of the Proposed National Education Policy Reform
The employees in the textile industry can be segmented as follows.pdf
1. The employees in the textile industry can be segmented as follows:
Employees Number
Female and union 12,000
Female and nonunion 25,000
Male and union 21,000
Male and nonunion 42,000
a. Determine the probability of each event in this distribution. An event refers to the
combinations
shown in the Employees column. For example, female and union is one event while female and
nonunion is another event.
b. Are the events in this distribution mutually exclusive? Explain.
c. What is the probability that an employee is male?
d. Is this experiment of segmenting employees in the textile industry collectively exhaustive?
Explain.EmployeesNumberFemale and union12,000Female and nonunion25,000Male and
union21,000Male and nonunion42,000
Solution
FU = 12,000/(12,000 + 25,000 + 21,000 + 42,000) = .12
FN = .25
MU = .21
MN = .42
b. yes. male and female are exclusive categories, as are union and non-union
c. MU + MN = .63
d. yes
