Question 1030196
My scientific calculator is not powerful enough to evaluate {{{(matrix(2,1, 300,96))*0.75^204*0.25^96}}}, which is the true answer.

If you're going to use the Poisson to approximate this, with {{{lambda = np = 300*0.25 = 75}}}, you would get 
{{{p(96) = (75^96/96!)e^-75 = 0.002738}}}, to four significant figures.

I tried approximating using the normal distribution using {{{p(0.315 <= r <= 0.325) = p(2.6 <= z <= 3)}}}, and I got 0.9987-0.9953 = 0.0034, so more or else the real value should be in the close neighborhood.