Respuesta :
The planning value for the population standard deviation is 4330.
Given the confidence interval 20000 and 35000.
We have to find the planning value of standard deviation.
Standard deviation is measuring dispersion of data. Uniform probability distribution has two bounds a and b. The standard deviation is given by :
s=[tex]\sqrt{(b-a)^{2}/12 }[/tex].
Annual starting salaries for college graduates be between 20000 and 35000.
It is uniform in the interval so, a=20000, b=35000.
Now we have to just put the values of a and b in the above formula to get the value of standard deviation.
s=[tex]\sqrt{(35000-20000)^{2} /12}[/tex]
=[tex]\sqrt{(15000)^{2} /12}[/tex]
=[tex]\sqrt{225000000/12}[/tex]
=[tex]\sqrt{18750000}[/tex]
=4330
Hence the planning values for the population standard deviation is 4330.
Learn more about standard deviation at https://brainly.com/question/475676
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Question is incomplete as it should includes the confidence interval of $20000 and $35000.
