Question

What are the advantages of squaring difference for calculating variance and standard deviation?
A.Squaring makes each term positive so that values above the means do not cancel below the mean.
B.Squaring adds more weight to the larger differences, and in many cases this extra weight is appropriate since points further from the mean may be more significant.
C.It complicates the calculations
D.All are incorrect

Answer 1

Hint: Here, in this question it is asked the advantages of squaring a difference for calculating standard deviation and variance. For that we need to remember the formula used for standard deviation and variance and the square term is there in the formula from that you will never get negative value because of squaring and also by squaring add more weight to the larger difference due to this deviation of mean will be more accurate and more significant. In this way we can identify the correct option.

Complete step by step solution:
Before we go to the problem first of all we need to understand the concept of standard deviation and variance.
Variance is referred to as the variability to represent the members of the group. It is the distance of observations corresponding to a higher variance value. It is a square measure. Variance is either zero or positive but never negative because of the squared value.
\[\text{variance}={{\sigma }^{2}}_{x}=\dfrac{\sum\limits_{i=1}^{n}{{{({{X}_{i}}-\bar{X})}^{2}}}}{N}\], where, \[\bar{X}=\]mean, \[{{X}_{i}}=\] value of data set
N= number of observations in the given data set.
Now we will come to the standard deviation
A straightforward dispersion measure is the standard deviation. Also, it defines the method of data value spread around the mean in a data set. We can refer to the closeness between the data set values and mean.
The standard deviation always measures according to the original data, and it is still positive and it calculated by \[{{\sigma }_{x}}=\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{({{X}_{i}}-\bar{X})}^{2}}}}{N}}\].
Now, we come to the problem, in this question we have to find the advantages of squaring a difference by calculating the variance and standard deviation.
By definition and formula of variance and standard deviation. We can say that variance and standard deviations are always positive because of the squaring in the formula hence, we can say that Option A is correct.
Now, we come to the option B, by squaring it, more weight is added to the huge disparity as a result of the extra weight. As a result, deviations from the mean may be more accurate and substantial.
So, the correct answer is “Option A and B”.

Note: To solve this problem, it is essential to know the definition and understand the basic concept whenever you need to differentiate two terms in any subject. Most of the students have misconceptions about variance and standard deviation which are similar. But in fact, they both are different. Variance is a numeric and squared value and also defines the variability of observation. Denoted as \[{{\sigma }^{2}}_{x}\] but in case of standard deviation is merely the root square value of the mean square deviation and unit is derived from original data only and it is measured by dispersion in the given data set are standard deviation.