The demand for newspapers varies over time. We know that on an average, 200 newspapers are sold daily. The standard deviation is 10. Now, the business owner wants to decide how many newspapers does he/she have to print and keep in his/her inventory each day, such that he/she fulfills the demand of the town 95% of the times.

P(X < x) = .95

P[(X - u / std deviation) < (x-u) / std deviation)] = .95

P[ z < ((x-200)/10) = .95

(x - 200)/10 = 1.64

Where does the 1.64 come from?

z = 1.959964

Looking at the Z distribution table z = .4744

I believe the answer is 216.4 = 217 but I am not sure how to get there.

Thanks!