## Elementary Statistics (12th Edition)

Here, n=2600 and p=0.06. a) Mean=$n\cdot p=2600 \cdot 0.06=156$. Standard deviation: $\sqrt{n \cdot p \cdot (1-p)}=\sqrt{2600 \cdot 0.06 \cdot 0.94}=12.11.$ b)If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=156-2\cdot12.11=131.78$ $Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=156-2\cdot12.11=180.22$. 178 is between these value, therefore it is not unusually low or high.