Question

if your calculator finds only one kind of standard deviation, explain how you could determine, without the calculator instructions, whether it is sample or population standard deviation.

choose the correct answer below.

a. divide the obtained standard deviation by n. if the quotient is the sum of the squares of the deviations from the mean, then the standard deviation found using the calculator is that of the population. otherwise, divide the obtained standard deviation by n - 1. if the quotient is the sum of the squares of the deviations from the mean, then the standard deviation found using the calculator is that of the sample.

b. divide the obtained standard deviation by n - 1. if the quotient is the sum of the squares of the deviations from the mean, then the standard deviation found using the calculator is that of the population. otherwise, divide the obtained standard deviation by n. if the quotient is the sum of the squares of the deviations from the mean, then the standard deviation found using the calculator is that of the sample.

c. multiply the obtained standard deviation by n. if the product is the sum of the squares of the deviations from the mean, then the standard deviation found using the calculator is that of the population. otherwise, multiply the obtained standard deviation by n - 1. if the product is the sum of the squares of the deviations from the mean, then the standard deviation found using the calculator is that of the sample.

d. multiply the obtained standard deviation by n - 1. if the product is the sum of the squares of the deviations from the mean, then the standard deviation found using the calculator is that of the population. otherwise, multiply the obtained standard deviation by n. if the product is the sum of the squares of the deviations from the mean, then the standard deviation found using the calculator is that of the sample.

Explanation:

Step1: Recall standard - deviation formulas

The formula for the population standard deviation $\sigma=\sqrt{\frac{\sum_{i = 1}^{N}(x_{i}-\mu)^{2}}{N}}$, and for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. Rearranging, we have $\sum_{i = 1}^{N}(x_{i}-\mu)^{2}=N\sigma^{2}$ and $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=(n - 1)s^{2}$.

Step2: Analyze the options

If we multiply the obtained standard - deviation by $n$ (for population) or $n - 1$ (for sample) and check if the product is the sum of the squares of the deviations from the mean, we can determine the type. If we multiply the obtained standard deviation by $n$, and the product is the sum of the squares of the deviations from the mean, then it is the population standard deviation. Otherwise, if we multiply the obtained standard deviation by $n - 1$ and the product is the sum of the squares of the deviations from the mean, then it is the sample standard deviation.

Answer:

C. Multiply the obtained standard deviation by n. If the product is the sum of the squares of the deviations from the mean, then the standard deviation found using the calculator is that of the population. Otherwise, multiply the obtained standard deviation by n - 1. If the product is the sum of the squares of the deviations from the mean, then the standard deviation found using the calculator is that of the sample.