The key here is remembering that a given normal distribution's mean and standard deviation, if not 0 and 1, respectively, requires converting the x values into Z values, to be able to use a Z table to look up the correct probabilities.

You also have to determine if the value you looked up matches the area under the curve you seek.  If your table only goes from Z = 0 to, say, 3.5, and you have negative values of Z, you'll have to use the fact that the Z curve is symmetric about Z =0, such that the probability of P(0 < Z < z) is the same as P(-z < Z < 0).  Some tables may run from -3.5 to 3.5 which simplifies things, but remember the probability you look up off this table is for P(Z < z), so if you need P(Z > z) you'll need to subtract the area you look up from 1, i.e., P(Z > z) = 1 - P(Z < z).

z = (x-μ)/σ

a) Using invNorm (.95, 0, 1) = 1.645,

1.645 = (x-75)/6.5

Solve for x,

x = 85.7 <==the lowest score for a full scholarship.

b) z = (63-75)/6.5

P(z < (63-75)/6.5) = 3.24% < 10%

So, the student has to take remedial courses.

c) P (z > (90-75)/6.5) = 1.05% <==Answer