Research shows that 40 of US adults get no exercise If 250 U

Research shows that 40% of U.S. adults get no exercise. If 250 U.S. adults are selected at random, determine the probabilty that fewer than 90 of them get no exercise.

**Please show work**

Solution

We first get the z score for the critical value:          
          
x = critical value =    89.5      
u = mean = np =    100      
          
s = standard deviation = sqrt(np(1-p)) =    7.745966692      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -1.355544171      
          
Thus, the left tailed area is          
          
P(z <   -1.355544171   ) =    0.087622123 [ANSWER]