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yanalaym [24]
3 years ago
14

Sixty seven percent of the employees in a company have managerial positions, and 58 percent of the employees in the company have

MBA degrees.
Also, 67 percent of the managers have MBA degrees.

Using the probability formulas, find the proportion of employees who either have MBAs or are managers. (Round your answer to 2 decimal places.)
Mathematics
1 answer:
Kazeer [188]3 years ago
7 0

Answer: The proportion of employees who either have MBAs or are managers are 0.58.

Step-by-step explanation:

Since we have given that

Probability of employees having managerial positions = 67%

Probability of employees having MBA degrees = 58%

Probability of managers having MBA degrees = 67%

So, using probability formulas, we get that

P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\P(A\cup B)=0.67+0.58-0.67\\\\P(A\cup B)=0.58

Hence, the proportion of employees who either have MBAs or are managers are 0.58.

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Answer:

[1,∞)

Step-by-step explanation:

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A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard devi
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Answer:

Null hypothesis:\mu \leq 3.3  

Alternative hypothesis:\mu > 3.3  

z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768  

Since is a one right tailed test the p value would be:  

p_v =P(z>1.768)=0.039  

Step-by-step explanation:

Data given and notation

\bar X=3.4 represent the sample mean  

\sigma=0.4 represent the population deviation for the sample

n=50 sample size  

\mu_o =3.3 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3.3  

Alternative hypothesis:\mu > 3.3  

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We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

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z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

We can replace in formula (1) the info given like this:  

z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768  

P value

Since is a one right tailed test the p value would be:  

p_v =P(z>1.768)=0.039  

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