Please help asap 25 pts
2 answers:
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Answer:
a) 0.152 = 15.2% probability that this person is a female who engages in physical exercise activities during the lunch hour.
b) 0.248 = 24.8% probability that this person is a female who does not engage in physical exercise activities during the lunch hour.
Step-by-step explanation:
Question a:
20% of employees engage in physical exercise.
This 20% is composed by:
8% of 60%(males)
x% of 100 - 60 = 40%(females).
Then, x is given by:
0.38 = 38%
Probability of being a female who engages in exercise:
40% are female, 38% of 40% engage in exercise. So
0.38*0.4 = 0.152
0.152 = 15.2% probability that this person is a female who engages in physical exercise activities during the lunch hour.
B. If we choose an employee at random from this corporation,what is the probability that this person is a female who does not engage in physical exercise activities during the lunch hour?
40% are female, 100% - 38% = 62% of 40% do not engage in exercise. So
0.62*0.4 = 0.248
0.248 = 24.8% probability that this person is a female who does not engage in physical exercise activities during the lunch hour.