Question

If the standard deviation for a set of data is​ 0, which of the following must be​ true?

A. All of the data values are negative.
B. All the data values equal 0.
C. All of the data values are identical.
D. None of the above must be true since the standard deviation cannot be equal to 0.

Answer 1

Answer:

C. All of the data values are identical.

Step-by-step explanation:

If the standard deviation for a set of data is​ 0, then all of the data values are identical.

When a standard deviation is 0, then the variance is also 0 and the mean of the deviation that is squared is also 0, thereby making all the values identical to the mean, that is 0.

Answer 2

Standard deviation is a quantity to measure dispersion. If it is 0, then it means we get: Option C: C. All of the data values are identical.

How is standard deviation calculated?

Suppose that the considered data set is $x_i; \: \: i = 1, 2, ... n$

Then, suppose its mean be $\overline{x}$

Then, its standard deviation is the positive square root of the variance of this data set.
Variance of data set is the sum of squared differences of the data values from its mean to the total count of data values.

Thus, we get:

$\sigma = \sqrt{\dfrac{1}{n} \sum_{i=1}^{n} (x_i-\overline{x})^2$

If you focus, we've each single term of that summation as non-negative, as n is always a whole number, and $(x_i - \overline{x})^2 \geq 0$

Thus, $\dfrac{(x_i - \overline{x})^2}{n} \geq 0$
That means, none of the term is going to contribute negative values.

Since it is given that standard deviation of the given data set is 0, that means all of the terms making it must be 0 (since even if one of the value is > 0, no one is there to decrease it to make the sum land on 0).

Thus,

if $\sigma = 0 \implies \dfrac{ (x_i-\overline{x})^2}{n} = 0 \implies x_i = \overline{x} \: \: \forall \: i = 1,2,..,n$

Thus, if standard deviation is 0, all data values are same and they are equal to their mean.

Thus, for the given condition, we get: Option C: C. All of the data values are identical.

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