Question 937336
assume 100 point tests
Maria: p(x = .84) = p(z < .06/.05) = normalcdf(-100,1.2) **** did better
Zoe: p(x = .89) = p(z < .06/.12) = normalcdf(-100, .5)
For the normal distribution: Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
area to the left of z = 1.2 > area to the left of z = .5
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))}}}