Elementary Statistics (12th Edition)

We know that the minimum and maximum usual values are 48 and 72, therefore the mean is 60 and the standard deviation is 6. We also know that standard deviation: 6=$\sqrt{n \cdot p \cdot (1-p)}=\sqrt{mean \cdot (1-p)}=\sqrt{60\cdot (1-p)}$, therefore 1-p=0.6, p=0.4. Mean=$n\cdot p$, therefore n=$\frac{mean}{p}=150$.