Do not unertand how they found this approximated normal dist

Hi

I recently came upon this and I cant figure it out


P(0.46 < samp pro < 0.52) = P (0.46-0.49)/0.016 < Z < 0.52-0.49)/0.016 = 0.9426

Sorry i forgot to mention the standard deviation is 0.016

I dont know how they figure out this z-score in between.

My exam is tommorow so if anyone can tell me why this happens, it would beb greatly appreciated.

Thanks

Your calculation implies that the mean=0.49.

yes