Question

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.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 50000

standard deviation = $\sigma$ = 20000

P(45000< x <65000 ) = P[(45000-50000) / 20000< (x - $\mu$ ) / $\sigma$< (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