In this article, we’ll explore the key concepts of Variance and Standard Deviation with clear definitions, essential formulas, helpful tips, and important properties. We’ve also included a few solved examples to make it easier for students to understand and apply these concepts effectively. These examples are particularly useful for high schoolers preparing for U.S. competitive exams like the SAT, ACT, and AP Statistics. The most significant difference between Standard Deviation (SD) and Variance is the manner in which they are stated. Standard deviation is stated in the same units as the mean of data, and hence, is more easy to comprehend, whereas variance is stated in units squared and provides a wider measure of the dispersion of the data. Proficiency in these topics is essential for high school math classes as well as competitive exams and will equip students with robust analytical skills required for future academic as well as professional life.

What is Variance and Standard Deviation?

Variance and standard deviation in mathematics can be determined by employing the mean of a group of numbers in question. The mean is defined as the average of a collection of numbers, moreover, the variance estimates the average degree to which each number is different from the mean.

Variance: The variance implies the average of the squared differences from the mean. To calculate the variance, first, determine the difference between each position and the mean; then, square and average the outcomes.

In common terms, variance is a measure of how far a set of data/numbers are dispersed out from their mean/ average value. It is denoted by the symbol \(σ^{2}\).

Standard Deviation: Standard deviation is a statistic that studies how notably from the mean a group of numbers is, by applying the square root of the variance.

The estimation of variance uses squares because it measures outliers more heavily than data closer to the mean. This sort of calculation also limits differences above the mean from cancelling out those below, which would end in a variance of zero.

In simple terms, the spread of statistical data is estimated by the standard deviation. Distribution measures the deviation of data/information from its mean/average state. The degree of dispersion is calculated by the method of estimating the deviation of data points. Standard deviation is expressed by the symbol, ‘σ’.

Variance and Standard Deviation Formula

Variance of the data set is the average square distance between the mean value and each specific data value. Plus standard deviation explains the spread of data values around the mean. With this knowledge let us learn about the standard deviation and variance formula.

Variance Formula

When we have gathered data from every portion of the population then we are interested to get an exact value for population variance.

Population variance is given by the formula:

\(σ^{2}=\frac{1}{N}\sum_{i=1}^N(X_i−μ)^{2}\)

Where:

\(σ^2\)=Population variance

N = Number of observations in the population.

\(X_i\)=ith observation in the population.

\(\mu\)=Population mean or Assumed mean.

When we accumulate data from a sample, the sample variance is applied to make estimates or conclusions about the sample variance.

Sample variance formula is as follows.

\(s^2=\frac{1}{n−1}\sum_{i=1}^n(x_i−\overline{x})^2\)

Where:

\(s^2\)=Sample Variance.

n= Number of observations in the sample.

\(x_{i}\)=ith observation in the sample.

\(\overline{x}\)=Sample mean or Arithmetic mean.

The extent of the variance corresponds to the dimension of the overall range of numbers; meaning the variance is higher when there is a broader range of numbers in the group, and the variance is smaller when there is a narrower array of numbers.

Standard Deviation Formula

Below is the formula for standard deviation regarding population and sample:

Population standard deviation is given by the formula:

\(σ=\sqrt{\frac{1}{N}\sum_{i=1}^N(X_i−μ)^2}\)

Where:

\(\sigma\)= Population standard deviation

With samples, we practice ‘n – 1’ in the formula because applying ‘n’ would provide us with a biased estimate that consistently minimizes variability. The sample variance would be lower than the actual variance of the population.

Sample standard deviation formula is as follows.

\(s=\sqrt{\frac{1}{n−1}\sum_{i=1}^n(x_i−\overline{x})^2}\)

Where:

s = Sample standard deviation

Reducing the sample ‘n’ to ‘n – 1’ gets the variance artificially large, providing you with an unbiased estimate of variability. We practice such a model as it is better to overestimate rather than underestimate variability in samples given.

How to calculate the Variance and Standard Deviation?

Now that you know the variance is the square of standard deviation as per the formula and definition, let us step towards how to calculate variance and standard deviation and also learn how to find standard deviation from variance with a basic example.

Consider the given sample space where n = 5 and the data set is given as = { 1,2,3,4,5}.

• As per the formula first, obtain the mean for the set of data.
• To get the mean, add up all the data, then divide them by the number of data i.e 5.
• Mean, \(\overline{x}\)=\(\frac{\left(1+2+3+4+5\right)}{5}=\frac{15}{5}=3\).
• Next, find each value’s deviation from the mean i.e subtracts the mean from each value to get the deviations from the mean.
• Mean=3
• Therefore the deviation from the mean is; { 1-3, 2-3, 3-3, 4-3, 5-3}={-2, -1, 0, 1, 2}
• Once the deviations from the mean are obtained multiply every deviation from the mean by itself. This will appear in positive numbers.
• {4, 1, 0, 1, 4}
• The summation of squares by adding up all of the squared deviations is; {4+1+ 0+ 1+ 4}=10.
• In the next step, divide the sum of squares by n – 1(for a sample) or N (for a population) depending on the data type.
• \(s^2=\frac{1}{n−1}\sum_{i=1}^n(x_i−\overline{x})^2\), As we are working with a sample, we will use n – 1, where n = 5.
• \(s^2=\frac{1}{n−1}\sum_{i=1}^n(x_i−\overline{x})^2=\frac{1}{5−1}\left(10\right)=\frac{10}{4}=2.5\)
• The respective sample standard deviation=\(s=\sqrt{\frac{1}{n−1}\sum_{i=1}^n(x_i−\overline{x})^2}\)
• s=1.5811.
• Now the population variance=\(σ^{2}=\frac{1}{N}\sum_{i=1}^N(X_i−μ)^{2}\)
• As we are working with population, we will use N, where N = 5.
• Population variance= \(\frac{10}{5}=2\)
• The corresponding population standard deviation=\(σ=\sqrt{\frac{1}{N}\sum_{i=1}^N(X_i−μ)^2}\)
• =\(\sqrt{2}\)=1.4142.

Variance and Standard Deviation Properties

Standard deviation and variance are 2 different mathematical theories that are both closely correlated to one another. The variance is required to calculate the standard deviation. These numbers help dealers and investors define the volatility of an expense and hence allow them to make well-informed trading decisions.

Properties of Variance

• The output of the final variance is always a non-negative value because every term in the variance sum is squared and hence the result is either positive or zero.
• The variance of the given set of data always possesses squared units. For example, the variance of a set of loads estimated in kilograms will be presented in kg squared. As the population variance is squared.
• We cannot compare variance directly with the mean or the data themselves due to the difference in the unit.

Properties of Standard Deviation

• It represents the square root of the mean of the squares of every value in a data set and therefore is also named the root-mean-square deviation.
• The least value of the standard deviation is zero since it cannot be a negative value.
• Meanwhile, if the data values of a group are comparable, then the standard deviation will be extremely low or even close to zero. However when the data values differ from each other, then the standard variation is large or even far from zero.
• The above point can be understood in this way also; if the points are further from the mean, there exists a higher deviation within the date. However, if they are closer to the mean, there is a lower deviation.
• The more spread out the group of numbers is, the greater the standard deviation obtained.

Relationship between Variance and Standard Deviation

Variance is equivalent to the average squared deviations from the mean, while standard deviation implies the number’s square root. The standard deviation is a square root of the obtained variance.

Both models exhibit variability in distribution, although their units vary: i,e the standard deviation is represented in the same units as the primary values, whereas the variance is represented in squared units.

Important Points on Variance and Standard Deviation

Some important points on variance and standard deviation are as follows:

• Both these concepts are helpful and significant for traders, who practice them to measure market volatility.
• When we have “N” data values then the population is divided by N when calculating the variance. Whereas for a sample we divide by n-1 when calculating variance.
• The variance is obtained by applying the sample data gives the sample variance. On the other hand, variance is calculated by applying the entire data, giving population variance. Also, a sample (a selection chosen from a bigger population), gives sample variance.

Solved Examples of Variance and Standard Deviation

Let’s learn how to calculate variance and standard deviation with some solved examples from the examination viewpoint as well.

Solved Example 1: The standard deviation of 4, 6, 10, 5, 10, is?

Solution:

Here, n = 5,

Mean (x̅) =\(\sum_{i=1}^5\frac{x_i}{5}=\frac{\left(4+6+10+5+10\right)}{5}=\frac{35}{5}=7\)

Variance

\((σ^2)=\sum_{i=1}^n\frac{\left(x_i−\mu\right)^2}{n}=\frac{(4−7)^2+(6−7)^2+(10−7)^2+(5−7)^2+(10−7)^2}{5}\)

\(=\frac{9+1+9+4+9}{5}=6.4\)

Standard deviation \((σ) = \sqrt{σ^2} = \sqrt{6.4}\)

Solved Example 2: If the standard deviation of a distribution is 29. Find the variance?

Solution:

\(\text{Variance}=(S.D)^2\)

\(\text{Variance}=(29)^2=841\)

Solved Example 3: For the given data set 1, 1, 2, 3, 1, 1, 12, 1, 5 find the value of variance?

Solution:Variance=\(σ^{2}=\frac{1}{N}\sum_{i=1}^N(X_i−μ)^{2}\)

To start with obtaining the mean first;

Mean = (1+ 1+ 2+ 3+ 1+ 1+ 12+ 1+ 5)/9 = 27/9 = 3

Mean Deviation (M.D) about mean= Subtract mean from each given data set values.

\(\left(X_i−μ\right)^2=(2^2+2^2+1^2+0^2+2^2+2^2+9^2+2^2+2^2)=106\)

Substituting the values in the formula:

The value of Variance = \(\frac{106}{9}=11.77\)

Solved Example 4: If the mean and the coefficient variation of distribution is 25% and 35% respectively, find variance.

Solution: The relation between mean, coefficient of variation and the standard deviation is as follows:

\(\text{Coefficient of variation}=\frac{\text{S.D}}{\text{Mean}}\times100\)

⇒\(35=\frac{\text{S.D}}{25}\times100\)

⇒ S.D = 8.75

\(\text{Variance}=(SD)^2\)

= \((8.75)^2 = 76.5625\)

Conclusion

In short, variance and standard deviation are fundamental to the mastery of statistical concepts, most particularly for high school students gearing up for competitive examinations such as the SAT, ACT, and AP Statistics. These two give useful information about the spread and variability of data, where variance is the squared deviations from the mean, and standard deviation provides a more intuitive measure in the same units as the data. Proficiency in these aspects not only improves analytical abilities but also prepares students for the critical thinking skills necessary for academic achievement and intelligent decision-making in future professional life. By learning these formulas and their uses, students can face statistical problems with confidence and achieve excellence in their exams.