Question

Suppose the salaries for recent graduates have a mean of $25,108 with a standard deviation of $1020. What is the minimum percentage of recent graduate who have a salaries between $23,243 and $26,973. I tried to do this question and it gives me a decimal, I don't know if I should use Chebyshev's Theorem or not, please help.

Answer 1

Solution :

Given that ,

mean = = $ 25,108

standard deviation = = $1020

P($ 23,243 < x < $ 26,973) = P[(23,243 - 25,108)/ 1020 ) < (x - ) /  < (26,973 - 25,108) / 1020 ) ]

= P(-1.83 < z < 1.83)

= P(z < 1.83) - P(z < -1.83)

Using z table,

= 0.9664 - 0.0336

= 0.9328

The percentage is = 93.28%