A company’s mean salary is $65,000 with a standard deviation of $6,000. What is the probability that an employee makes between $
70,000 and $80,000?
1 answer:
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Answer:
C. (36337.32, 48968.68)
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 42653 - 6315.68 = 36337.32.
The upper end of the interval is the sample mean added to M. So it is 42653 + 6315.68 = 48968.68.
So the correct answer is:
C. (36337.32, 48968.68)
Let's see what to do buddy...
Step-by-step explanation:
_________________________________
Subtract the sides of the equation minus -273.15
Divided the sides of the equation by 59
Subtract the sides of the equation plus 32
And we're done.
Thanks for watching buddy good luck.
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