Question

the choices below are data sets. in the choices, w is a constant. each choice has the same mean. which choice has the greatest standard - deviation?
o w - 3, w - 1, w, w + 1, w + 3
o w - 2, w - 1, w, w + 1, w + 2
o w - 3, w, w, w, w + 3
o w - 2, w - 2, w, w + 2, w + 2
graphing calculator

Explanation:

Step1: Recall the formula for standard - deviation

The standard - deviation $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\mu)^2}{n}}$, where $\mu$ is the mean and $n$ is the number of data points. Since the mean of all the given data - sets is $w$ (by symmetry), we can simplify the calculation by looking at the squared - differences $(x_i - w)^2$.

Step2: Calculate squared - differences for each option

Option 1: $w - 3,w - 1,w,w + 1,w + 3$

The squared - differences from the mean $w$ are $(w-3 - w)^2=9$, $(w - 1 - w)^2 = 1$, $(w - w)^2=0$, $(w + 1 - w)^2 = 1$, $(w + 3 - w)^2=9$. The sum of squared - differences $\sum=(9 + 1+0 + 1+9)=20$.

Option 2: $w - 2,w - 1,w,w + 1,w + 2$

The squared - differences from the mean $w$ are $(w-2 - w)^2=4$, $(w - 1 - w)^2 = 1$, $(w - w)^2=0$, $(w + 1 - w)^2 = 1$, $(w + 2 - w)^2=4$. The sum of squared - differences $\sum=(4 + 1+0 + 1+4)=10$.

Option 3: $w - 3,w,w,w,w + 3$

The squared - differences from the mean $w$ are $(w-3 - w)^2=9$, $(w - w)^2=0$, $(w - w)^2=0$, $(w - w)^2=0$, $(w + 3 - w)^2=9$. The sum of squared - differences $\sum=(9+0 + 0+0 + 9)=18$.

Option 4: $w - 2,w - 2,w,w + 2,w + 2$

The squared - differences from the mean $w$ are $(w-2 - w)^2=4$, $(w - 2 - w)^2 = 4$, $(w - w)^2=0$, $(w + 2 - w)^2 = 4$, $(w + 2 - w)^2=4$. The sum of squared - differences $\sum=(4 + 4+0 + 4+4)=16$.

Step3: Compare the sums

Since the number of data points $n = 5$ for all options, the data - set with the largest sum of squared - differences from the mean will have the largest standard deviation.

Answer:

$w - 3,w - 1,w,w + 1,w + 3$