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9966 [12]
3 years ago
12

A company has three different sites: site 1, site 2 and site 3. at site 1 70% of the employees are bu alum, at site 2 20% of the

employees are bu alum, at site 3 10% are bu alum. there are an equal number of employees at each of the three sites. if an employee is randomly selected for employee of the month what is the probability that they are a bu alum
Mathematics
1 answer:
Levart [38]3 years ago
7 0

site 1: 70% = .7

site 2: 20% = .2

site 3: 10% = .1

(.7 + .2 + .1)/3  = 1/3

Answer: 1 out of 3 (1/3)

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Answer:

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And we can find the probability with the complement rule and the normal standard distirbution and we got:

P(z>0.6875)=1-P(z

Step-by-step explanation:

Let X the random variable that represent the time spent reading of a population, and for this case we know the distribution for X is given by:

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We are interested on this probability

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And we can use the z score formula given by:

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Using this formula we got:

P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-49}{16})=P(z>0.6875)

And we can find the probability with the complement rule and the normal standard distirbution and we got:

P(z>0.6875)=1-P(z

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Answer:

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