Suppose that the distribution of income in a certain tax bra

Suppose that the distribution of income in a certain tax bracket is approximately normal with a mean of $25,141.52 and a standard deviation of $3,727.774. Approximately 38.62% of households had an income greater than what dollar amount?

Solution

Normal Distribution
Mean ( u ) =25141.52
Standard Deviation ( sd )=3727.774
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P ( Z > x ) = 0.3862
Value of z to the cumulative probability of 0.3862 from normal table is 0.29
P( x-u/ (s.d) > x - 25141.52/3727.774) = 0.3862
That is, ( x - 25141.52/3727.774) = 0.29
--> x = 0.29 * 3727.774+25141.52 = 26218.8467