Question 591992
The sample standard deviation is


*[tex \LARGE S_x = \frac{\sigma}{\sqrt{n}}] (n is the sample size)


*[tex \LARGE S_x = \frac{15}{\sqrt{3}} = 8.6602]


The definition for "unusual" varies, but it usually indicates data lying roughly 1.96 standard deviations away from the mean (i.e. outer 5% of data). The corresponding value for z=1.96 is


*[tex \LARGE 100 + (8.6602)(1.96) = 116.97]


115 lies within the middle 95%, so it would not be deemed "unusual."