Question

a set of data has ten numbers. the mean of the data is 12 and the standard deviation is 1. what values could make up a data set with these statistics?

Explanation:

Step1: Recall mean formula

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Here $n = 10$ and $\bar{x}=12$, so $\sum_{i=1}^{10}x_{i}=n\times\bar{x}=10\times12 = 120$.

Step2: Recall standard - deviation formula

The standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. We want to find values such that this holds. One simple case is when the data is symmetric around the mean.
A possible data - set could be ten numbers all equal to 12. In this case, $\sum_{i=1}^{10}x_{i}=10\times12=120$ and for each $i$, $x_{i}-\bar{x}=12 - 12=0$. So the standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{10}(x_{i}-12)^{2}}{9}}=\sqrt{\frac{0}{9}} = 0$. Another non - trivial case could be 10, 10, 10, 10, 12, 12, 14, 14, 14, 14.
For the set $\{10, 10, 10, 10, 12, 12, 14, 14, 14, 14\}$:
The sum is $4\times10+2\times12 + 4\times14=40+24 + 56=120$.
The mean $\bar{x}=\frac{120}{10}=12$.
$(x_1 - 12)^2=(10 - 12)^2 = 4$, $(x_5 - 12)^2=(12 - 12)^2 = 0$, $(x_8 - 12)^2=(14 - 12)^2 = 4$.
$\sum_{i = 1}^{10}(x_{i}-12)^{2}=4\times4+2\times0+4\times4=32$.
The standard deviation $s=\sqrt{\frac{32}{9}}\approx1.89$.

Answer:

One possible data - set is 10, 10, 10, 10, 12, 12, 14, 14, 14, 14 (answers may vary).