## Elementary Statistics (12th Edition)

Here, n=50, p=$\frac{1}{38}$. a)Mean=$n\cdot p=50 \cdot \frac{1}{38}=1.32$. Standard deviation: $\sqrt{n \cdot p \cdot (1-p)}=\sqrt{50 \cdot \frac{1}{38} \cdot \frac{37}{38}}=1.13.$ b) If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=1.32-2\cdot1.13=-0.94$ $Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=1.32+2\cdot1.13=3.58$. 0 is between these two bounds, therefore it is not unusually low.