Question

professors at a local university earn a salary of $80,000 with a standard deviation of $6,000....

professors at a local university earn a salary of $80,000 with a standard deviation of $6,000. the salary distribution is normally distributed. find the percentage of salaries that are between $62,000 and $98,000?
Math   Statistics-And-Probability   
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Answer #1

Solution :

Given that,

mean = = 80000

standard deviation = = 6000

P (62000 < x < 98000 )

P ( 62000 - 80000 / 6000) < ( x -  / ) < ( 98000- 80000 / 6000)

P ( - 18000 / 6000 < z < - 7 / 6000 )

P (-3 < z < 3 )

P ( z < 3 ) - P ( z < -3 )

Using z table

=0.9987- 0.0013

= 0.9974

Probability = 0.9974

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