Question 12·Hard·One-Variable Data Distributions; Measures of Center and Spread

Consider separately how multiplying by $2$, subtracting $6$, and adding $4$ affect a mean.

A shift by the same amount does not change standard deviation, but multiplying all values changes their distances from the mean.

Reading the balanced dot plots and tracking the transformations is faster than using Desmos for this question.

For a verification, enter X=[2,4,4,4,6] and Y=[6,8,9,10,12] on separate expressions.

Enter mean(X), mean(Y), stdev(X), and stdev(Y). The displayed results confirm that X has the smaller mean and the smaller standard deviation.

In data set A, the values are balanced around $5$: one value is $4$, three values are $5$, and one value is $6$. Data set B is also balanced around $5$: its values are $2$, $4$, $5$, $6$, and $8$. Thus, both original means are $5$.

Multiplying every value by $2$ multiplies the mean by $2$, and subtracting $6$ then subtracts $6$ from the mean. Therefore, the mean of X is $2(5)-6=4$.

Adding $4$ to every value in B adds $4$ to its mean, so the mean of Y is $5+4=9$.

Data set X has values $2$, $4$, $4$, $4$, and $6$, so three values equal its mean of $4$ and the other two are only $2$ units away.

Data set Y has values $6$, $8$, $9$, $10$, and $12$. Relative to its mean of $9$, two values are $3$ units away and two more are $1$ unit away. Therefore, Y is more spread out, so the standard deviation of X is less than the standard deviation of Y.