The annual salaries of employees in a large company are normally distributed with a mean of $50,000 and a standard deviation of $20,000. What percentage of people earn between $45,000 and $65,000? Round to the second decimal place.
Solution :
Given that ,
mean = = 50000
standard deviation = = 20000
P(45000< x <65000 ) = P[(45000-50000) / 20000< (x - ) / < (65000-50000) /20000 )]
= P( -0.25< Z < 0.25)
= P(Z <0.25 ) - P(Z <-0.25 )
Using z table
= 0.5987 - 0.4013
probability= 0.1974
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