You need to measure two quantities: mean and standard deviation. You also have two bits of data which means you can get a unique answer (which is what you want).
Basically you have P(X > 75) = 0.75 and P(X > 150) = 150 and X ~ N(mu,sigma^2).
Recall that the PDF of a normal is f(x;mu,sigma) = 1/SQRT(2*pi*sigma^2) * e^(-1/2*(x-mu)^2/sigma^2)) and P(X <= x) = Integral (-infinity,x) f(u;mu,sigma)du.
You are given two values of x with a corresponding P(X <= x) that have both the same mu and sigma.
You may need to use a computer to solve this problem and I'd recommend something like R or MATLAB/Octave.