Question 933966
mean of 169 days and a standard deviation of 17.5 
 {{{z = blue(x - 169)/blue(17.5)}}}
....
P( x > 310) = P( z > 141/17.5) = normalcdf(8.057,  100) = 3.9x10^(-16)
3.9x10^(-16)of Area to the right of z = 8.057 pretty much says it is NOT going to happen
........
For the normal distribution: Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
Picture the area to the right of z = 8.057
Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50%  to the right
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), green(line(1,0,1,exp(-1^2/2)),line(-1,0,-1,exp(-1^2/2))),green(line(2,0,2,exp(-2^2/2)),line(-2,0,-2,exp(-2^2/2))),green(line(3,0,3,exp(-3^2/2)),line(-3,0,-3,exp(-3^2/2))),green(line( 0,0, 0,exp(0^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}