The data set 10, 12, 14, 16, 90 includes an outlier. By how much does removing the outlier change the mean?

With the outlier, the mean is $\frac{10+12+14+16+90}{5}=\frac{142}{5}=28.4$. Without it, the mean is $\frac{10+12+14+16}{4}=\frac{52}{4}=13$. The mean decreases by $28.4-13=15.4$, choice F. The outlier $90$ pulls the mean far above the other four values, so removing it must lower the mean. Choice G ($15.4$ increase) has the direction wrong. Choice H ($28.4$) is the original mean, not the change. Choice J (stays the same) is wrong because the mean is sensitive to extreme values.