The variance is the mean squared deviation from the mean. The variance for population data is denoted by \(\sigma^2\) (read as sigma squared) and the variance calculated for sample data is denoted by \(s^2\).

\[ \sigma^2 = \frac{\sum_{i=1}^n (x_i-\mu)^2}{N} \]

and

\[ s^2 = \frac{\sum_{i=1}^n (x_i-\bar x)^2}{n-1} \]

where \(\sigma^2\) is the population variance and \(s^2\) is the sample variance. The quantity \(x_i-\mu\) or \(x_i-\bar x\) in the above formulas is called the deviation of the \(x_i\) value (\(x_1, x_2,...,x_n\)) from the mean (Mann 2012).

The standard deviation is the most-used measure of dispersion. The value of the standard deviation tells us how closely the values of a data set are clustered around the mean. In general, a lower value of the standard deviation indicates that the values of the data set are spread over a relatively smaller range around the mean. In contrast, a larger value of the standard deviation for a data set indicates that the values of that data set are spread over a relatively larger range around the mean (Mann 2012).

The standard deviation is obtained by taking the square root of the variance. Consequently, the standard deviation calculated for population data is denoted by \(\sigma\) and the standard deviation calculated for sample data is denoted by \(s\).

\[ \sigma = \sqrt{\frac{\sum_{i=1}^N (x_i-\mu)^2}{N}} \]

and

\[ s = \sqrt{\frac{\sum_{i=1}^n (x_i-\bar x)^2}{n-1}} \]

where \(\sigma\) is the standard deviation of the population and \(s\) is the standard deviation of the sample.

As an exercise we compute the mean, the median, the variance and the standard deviation for some numerical variables of interest in the students data set, and present them in a nice format.

students <- read.csv("https://userpage.fu-berlin.de/soga/data/raw-data/students.csv")
quant_vars <- c("name", "age", "nc.score", "height", "weight")
students_quant <- students[quant_vars]
head(students_quant, 10)
##                    name age nc.score height weight
## 1   Gonzales, Christina  19     1.91    160   64.8
## 2        Lozano, T'Hani  19     1.56    172   73.0
## 3        Williams, Hanh  22     1.24    168   70.6
## 4           Nem, Denzel  19     1.37    183   79.7
## 5       Powell, Heather  21     1.46    175   71.4
## 6        Perez, Jadrian  19     1.34    189   85.8
## 7         Clardy, Anita  21     1.11    156   65.9
## 8  Allen, Rebecca Marie  21     2.03    167   65.7
## 9         Tracy, Robert  18     1.29    195   94.4
## 10       Nimmons, Laura  18     1.19    165   66.0
# mean
students_quant_mean <- apply(students_quant[, !(colnames(students_quant) == "name")], 2, mean)
# median
students_quant_median <- apply(students_quant[, !(colnames(students_quant) == "name")], 2, median)
# variance
students_quant_var <- apply(students_quant[, !(colnames(students_quant) == "name")], 2, var)
# standard deviation
students_quant_sd <- apply(students_quant[, !(colnames(students_quant) == "name")], 2, sd)
# concatenate the vectors and round to 2 digits
students_quant_stats <- round(cbind(
  students_quant_mean,
  students_quant_median,
  students_quant_var,
  students_quant_sd
), 2)
# rename column names
colnames(students_quant_stats) <- c("mean", "median", "variance", "standard deviation")
students_quant_stats
##            mean median variance standard deviation
## age       22.54  21.00    36.79               6.07
## nc.score   2.17   2.04     0.66               0.81
## height   171.38 171.00   122.71              11.08
## weight    73.00  71.80    74.57               8.64

Use of the Standard Deviation

By using the mean and standard deviation we can find the proportion or percentage of the total observations that fall within a given interval around the mean.

Chebyshev’s Theorem

Chebyshev’s theorem gives a lower bound for the area under a curve between two points that are on opposite sides of the mean and at the same distance from the mean.

For any number \(k\) greater than 1, at least \((1-1/k^2)*100\) % of the data values lie within \(k\) standard deviations of the mean.

Let us use R to gain some intuition for Chebyshev’s theorem.

k <- seq(1, 4, by = 0.1)
auc <- 1 - (1 / k^2)
auc_percent <- round(auc * 100)
cbind(k, auc_percent)
##         k auc_percent
##  [1,] 1.0           0
##  [2,] 1.1          17
##  [3,] 1.2          31
##  [4,] 1.3          41
##  [5,] 1.4          49
##  [6,] 1.5          56
##  [7,] 1.6          61
##  [8,] 1.7          65
##  [9,] 1.8          69
## [10,] 1.9          72
## [11,] 2.0          75
## [12,] 2.1          77
## [13,] 2.2          79
## [14,] 2.3          81
## [15,] 2.4          83
## [16,] 2.5          84
## [17,] 2.6          85
## [18,] 2.7          86
## [19,] 2.8          87
## [20,] 2.9          88
## [21,] 3.0          89
## [22,] 3.1          90
## [23,] 3.2          90
## [24,] 3.3          91
## [25,] 3.4          91
## [26,] 3.5          92
## [27,] 3.6          92
## [28,] 3.7          93
## [29,] 3.8          93
## [30,] 3.9          93
## [31,] 4.0          94

To put it in words: Let us pick a value for \(k\): \(k= 3\). This means that at least 89% of the data values lie within 3 standard deviations of the mean.

Let us plot Chebyshev’s theorem with R:

plot(k,
  auc_percent,
  col = "blue",
  pch = 19,
  xlab = "k",
  ylab = "percent",
  main = "Chebyshev's theorem"
)

The theorem applies to both sample and population data. Note that Chebyshev’s theorem is applicable to a distribution of any shape. However, Chebyshev’s theorem can be used only for \(k>1\). This is so because when \(k = 1\), the value of \((1-1/k^2)\) is zero, and when \(k<1\), the value of \((1-1/k^2)\) is negative (Mann 2012).

Empirical Rule

While Chebyshev’s theorem is applicable to any kind of distribution, the empirical rule applies only to a specific type of distribution called a bell-shaped distribution or normal distribution. There are 3 rules:

For a bell-shaped distribution, approximately

  1. 68% of the observations lie within one standard deviation of the mean.
  2. 95% of the observations lie within two standard deviations of the mean.
  3. 99.7% of the observations lie within three standard deviations of the mean.

Since we have sufficient coding abilities by now, we will try to test if the three rules are valid. (1) First, we will explore the rnorm function in R to generate normally distributed data and (2) second, we will go back to our students data set and validate those rules on that data set.

The normal distribution belongs to the family of continuous distributions. In R there are a lot of probability distributions available (here). To generate data from a normal distribution one may use the rnorm() function, which is a random variable generator for the normal distribution.

We can sample n values from a normal distribution with a given mean (default is 0) and standard deviation (default is 1) using the rnorm() function: rnorm(n=1, mean=0, sd=1). Let us give it a try:

rnorm(n = 1, mean = 0, sd = 1)
## [1] 0.3024223
rnorm(n = 1, mean = 0, sd = 1)
## [1] 0.1607108
rnorm(n = 1, mean = 0, sd = 1)
## [1] -0.5589824
rnorm(n = 1, mean = 0, sd = 1)
## [1] -0.0868194

As we can see, the rnorm() function returns (pseudo-)random numbers. We can fairly easily ask the function to draw hundreds or thousands or even more (pseudo-)random numbers:

rnorm(n = 10, mean = 0, sd = 1)
##  [1]  0.305534 -1.449842  1.124206 -1.250876  1.520984  1.292920 -0.506880
##  [8] -1.755317  1.399782 -2.066587
rnorm(n = 100, mean = 0, sd = 1)
##   [1] -0.95408954 -0.36226697  0.81396156  0.53698581  1.57695272 -0.51980926
##   [7] -0.34556678 -0.06558639 -0.94565297 -0.20274276  1.00864513  1.26060728
##  [13]  1.52324542 -0.81329537 -0.04527783 -0.73954000  0.33078526 -1.61404629
##  [19] -0.75955544  0.44716717 -0.43674032 -1.38611811  1.11213934 -1.91651982
##  [25]  0.64893972  1.67199076 -0.56596237  0.15142877 -0.08794681 -0.91326167
##  [31]  0.81229043 -1.62428413  0.14954278 -0.95877422  0.26793841  0.80352371
##  [37] -0.20803389  0.71325008  1.60517143  0.39996411  0.15522763 -0.83475926
##  [43]  2.40961970 -0.49223334  0.66283628 -1.30077324  0.46532720 -1.04467400
##  [49] -1.73866438  0.26044254  2.10025571 -1.85331303 -0.88941184  0.33500572
##  [55]  0.07303722 -2.07474575  0.42468379 -1.06423040  0.36929172 -0.07309428
##  [61]  0.42295545  0.70401118  0.10954828 -0.72429535 -0.50137046 -0.33520835
##  [67]  0.56041721 -0.33140872  0.62048804  1.34317852  0.24645206  0.30737213
##  [73] -0.17031416 -0.48449711 -1.20424081 -0.97730543  0.07287897 -0.31357427
##  [79] -0.13024810  0.51945538  0.95817660 -1.79337310 -0.63578623  1.43155929
##  [85]  1.29249493 -0.98061237  0.46659636  1.04861061  0.19487780 -0.89221195
##  [91]  2.12972550  0.62830767 -1.17800505  0.42219325  0.21551023  1.70225208
##  [97] -0.98230728  0.13922632 -0.25711721 -0.18386610
y_norm <- rnorm(n = 100000, mean = 0, sd = 1)

If we plot a histogram of those numbers, we see the eponymous bell shaped distribution.

hist(y_norm, breaks = 100, main = "Normal distribution", xlab = "")

We already know the mean and the standard deviation of the y_norm vector, as we explicitly called the function rnorm() with mean = 0 and sd = 1. So, we just have to count those numbers of the y_norm vector bigger than 1, and respectively smaller than -1, bigger than 2, respectively -2, and 3, respectively -3, and relate them to the length of the vector, in our case 100,000, to validate the three rules claimed above.

sd1 <- sum(y_norm > -1 & y_norm < 1) / length(y_norm) * 100
sd2 <- sum(y_norm > -2 & y_norm < 2) / length(y_norm) * 100
sd3 <- sum(y_norm > -3 & y_norm < 3) / length(y_norm) * 100

cbind(c("1sd", "2sd", "3sd"), c(sd1, sd2, sd3))
##      [,1]  [,2]    
## [1,] "1sd" "68.107"
## [2,] "2sd" "95.323"
## [3,] "3sd" "99.735"

Perfect match! The three empirical rules are obviously valid. To visualize our findings we re-plot the histogram and add some annotations. Please note that in the hist() function we set the argument freq = F, which is the same as freq = FALSE. As a consequence, the resulting histogram does not show counts on the y-axis anymore, but the density values (normalized count divided by bin width), which means that the bar areas sum to 1.

h <- hist(y_norm, breaks = 100, plot = F)
cuts <- cut(h$breaks, c(-Inf, -3, -2, -1, 1, 2, 3, Inf), right = F) # right = False;
                                                                    # sets intervals to be open on the right closed on the left
plot(h,
  col = rep(c("white", "4", "3", "2", "3", "4", "white"))[cuts],
  main = "Normal distribution",
  xlab = "",
  freq = F,
  ylim = c(0, 0.6)
)

lwd <- 3
# horizontal lines
lines(x = c(2, -2), y = c(0.48, 0.48), type = "l", col = 3, lwd = lwd)
lines(x = c(3, -3), y = c(0.55, 0.55), type = "l", col = 4, lwd = lwd)
lines(x = c(1, -1), y = c(0.41, 0.41), type = "l", col = 2, lwd = lwd)
# vertical lines
lines(x = c(1, 1), y = c(0, 0.41), type = "l", col = 2, lwd = lwd)
lines(x = c(-1, -1), y = c(0, 0.41), type = "l", col = 2, lwd = lwd)
lines(x = c(2, 2), y = c(0, 0.48), type = "l", col = 3, lwd = lwd)
lines(x = c(-2, -2), y = c(0, 0.48), type = "l", col = 3, lwd = lwd)
lines(x = c(3, 3), y = c(0, 0.55), type = "l", col = 4, lwd = lwd)
lines(x = c(-3, -3), y = c(0, 0.55), type = "l", col = 4, lwd = lwd)
# text
text(0, 0.44, "68 %", cex = 1.5, col = 2)
text(0, 0.51, "95 %", cex = 1.5, col = 3)
text(0, 0.58, "99.7 %", cex = 1.5, col = 4)

Well, now let us work on our 2nd task: Validate the three empirical rules on the students data set. For this, we have to check whether any of the numeric variables in the students data set are normally distributed. We start by extracting numeric variables of interest from the students data set. Then we check the data set by calling the function head().

cont_vars <- c("age", "nc.score", "height", "weight", "score1", "score2", "salary")
students_quant <- students[, cont_vars]
head(students_quant, 10)
##    age nc.score height weight score1 score2 salary
## 1   19     1.91    160   64.8     NA     NA     NA
## 2   19     1.56    172   73.0     NA     NA     NA
## 3   22     1.24    168   70.6     45     46     NA
## 4   19     1.37    183   79.7     NA     NA     NA
## 5   21     1.46    175   71.4     NA     NA     NA
## 6   19     1.34    189   85.8     NA     NA     NA
## 7   21     1.11    156   65.9     NA     NA     NA
## 8   21     2.03    167   65.7     58     62     NA
## 9   18     1.29    195   94.4     57     67     NA
## 10  18     1.19    165   66.0     NA     NA     NA

To get an overview of the shape of the distribution of each particular variable, we apply the histogram() function of the lattice package. If the lattice package is not yet installed on your computer, type install.packages("lattice")in your console. The coding is a little bit different than for standard histograms.

library(lattice)
histogram(~ height + age + weight + nc.score + score1 + score2 + salary,
  breaks = 50,
  type = "density",
  xlab = "",
  ylab = "density",
  layout = c(4, 2),
  scales = list(relation = "free"),
  col = "black",
  data = students_quant
)

We immediately realize that some variables are positively skewed, thus we exclude them and keep those that appear to be normally distributed.

histogram(~ height + salary,
  breaks = 50,
  type = "density",
  xlab = "",
  ylab = "density",
  layout = c(2, 1),
  scales = list(relation = "free"),
  col = "black",
  data = students_quant
)

Both the height and the salary variable seem to more or less follow a normal distribution. So, the choice which to pick for further analysis is a matter of taste. For now, we stick to the salary variable and check, if the three empirical rules claimed above are valid. Let us switch to R and validate those rules by calculating the mean and the standard deviations first. Please note, that the salary variable includes empty cells marked by NA. Thus, we first exclude all NA values by applying the na.omit() function.

salary_vector <- na.omit(students$salary)

# calculate mean
salary_vector_mean <- mean(salary_vector)

# calculate standard deviations
salary_vector_sd1_pos <- salary_vector_mean + sd(salary_vector)
salary_vector_sd2_pos <- salary_vector_mean + sd(salary_vector) * 2
salary_vector_sd3_pos <- salary_vector_mean + sd(salary_vector) * 3

salary_vector_sd1_neg <- salary_vector_mean - sd(salary_vector)
salary_vector_sd2_neg <- salary_vector_mean - sd(salary_vector) * 2
salary_vector_sd3_neg <- salary_vector_mean - sd(salary_vector) * 3

As in the generic example from above, we count the number of values, which are bigger than +1 s.d., respectively smaller than -1 s.d., and +2 s.d., respectively -2 s.d., and +3 s.d., respectively -3 s.d., and relate them to the length of the vector, in our case 1753.

salary_sd1 <- 100 - sum(salary_vector > salary_vector_sd1_pos | salary_vector < salary_vector_sd1_neg) /
  length(salary_vector) * 100
salary_sd2 <- 100 - sum(salary_vector > salary_vector_sd2_pos | salary_vector < salary_vector_sd2_neg) /
  length(salary_vector) * 100
salary_sd3 <- 100 - sum(salary_vector > salary_vector_sd3_pos | salary_vector < salary_vector_sd3_neg) /
  length(salary_vector) * 100
cbind(c("1sd", "2sd", "3sd"), c(round(sd1), round(sd2), round(sd3, 1)), c(salary_sd1, salary_sd2, salary_sd3))
##      [,1]  [,2]   [,3]              
## [1,] "1sd" "68"   "67.0849971477467"
## [2,] "2sd" "95"   "95.6075299486594"
## [3,] "3sd" "99.7" "99.7718197375927"

Wow, quite close! Obviously the salary variable shows a strong tendency to support the so called empirical rule. We plot the histogram of the salary variable to confirm our impression. We colorize the standard deviations for a better visual impression. There are several ways to pick a particular color in R. In this code example we use the color name “white” (type colors() in your console to see a full list of colors) and combine it with numbers 2, 3, 4, a shortcut for accessing the colors of the standard color palette (type palette() in your console to see the list of colors in the standard color palette).

h <- hist(salary_vector, breaks = 100, plot = F)
x_vector2plot <- seq(salary_vector_sd3_neg * 0.9, salary_vector_sd3_pos * 1.1, 1)

cuts <- cut(h$breaks, c(
  -Inf,
  salary_vector_sd1_neg,
  salary_vector_sd2_neg,
  salary_vector_sd3_neg,
  salary_vector_sd1_pos,
  salary_vector_sd2_pos,
  salary_vector_sd3_pos,
  Inf
))
plot(h,
  col = rep(c("white", "4", "3", "2", "3", "4", "white"))[cuts],
  main = "Normal distribution",
  xlab = "Annual Salary in EUR",
  freq = F
)

# add legend
legend(
  x = min(x_vector2plot) * 1.1,
  y = max(h$density) * 0.9,
  legend = c("1 s.d.", "2 s.d.", "3 s.d."),
  col = c(2, 3, 4),
  pch = 15
)

We can now extend our visualization approach by plotting the empirical density estimate, using the density() function, and checking its shape. We display the empirical density estimate as a dashed line by setting the line type argument lty = 2 with a line width of 3 (argument lwd = 3).

h <- hist(salary_vector, breaks = 100, plot = F)
x_vector2plot <- seq(salary_vector_sd3_neg * 0.9, salary_vector_sd3_pos * 1.1, 1)

cuts <- cut(h$breaks, c(
  -Inf,
  salary_vector_sd1_neg,
  salary_vector_sd2_neg,
  salary_vector_sd3_neg,
  salary_vector_sd1_pos,
  salary_vector_sd2_pos,
  salary_vector_sd3_pos,
  Inf
))
plot(h,
  col = rep(c("white", "4", "3", "2", "3", "4", "white"))[cuts],
  main = "Normal distribution",
  xlab = "Annual Salary in EUR",
  freq = F
)

# add empirical density
lines(density(salary_vector), col = "black", lty = 2, lwd = 3)

# add legend
legend(
  x = min(x_vector2plot) * 1.1,
  y = max(h$density) * 0.9,
  legend = c("1 s.d.", "2 s.d.", "3 s.d.", "emp. density"),
  col = c(2, 3, 4, "black"),
  lty = c(NA, NA, NA, 2),
  lwd = c(NA, NA, NA, 2),
  pch = c(15, 15, 15, NA),
  cex = 0.7
)

Finally, we compare our empirical density estimate to the theoretical probability density function based on the actual mean and standard deviation of the data (salary_vector). For a better visual comparison, we switch back to a non-colorized histogram plot.

h <- hist(salary_vector, breaks = 100, plot = F)
x_vector2plot <- seq(salary_vector_sd3_neg * 0.9, salary_vector_sd3_pos * 1.1, 1)

plot(h,
  main = "Normal distribution",
  xlab = "Annual Salary in EUR",
  freq = F
)

# add pdf
lines(x_vector2plot,
  # dnorm() function, returns the probability density function (pdf) of the normal distribution
  dnorm(
    x = x_vector2plot,
    mean = salary_vector_mean,
    sd = sd(salary_vector)
  ),
  lwd = 3,
  col = "red"
)

# add empirical density
lines(density(salary_vector), col = "black", lty = 2, lwd = 3)

# add legend
legend(
  x = min(x_vector2plot) * 1.1,
  y = max(h$density) * 0.9,
  legend = c("pdf", "emp. density"),
  col = c("red", "black"),
  lty = c(1, 2),
  lwd = 2,
  cex = 0.7
)

We may conclude, that the salary variable in the students data set is roughly normally distributed. However, the plot indicates that the distribution of the salary variable is slightly skewed to the left. We can see that by the small deviation between the empirical density estimate and the probability density function.

Citation

The E-Learning project SOGA-R was developed at the Department of Earth Sciences by Kai Hartmann, Joachim Krois and Annette Rudolph. You can reach us via mail by soga[at]zedat.fu-berlin.de.

Creative Commons License
You may use this project freely under the Creative Commons Attribution-ShareAlike 4.0 International License.

Please cite as follow: Hartmann, K., Krois, J., Rudolph, A. (2023): Statistics and Geodata Analysis using R (SOGA-R). Department of Earth Sciences, Freie Universitaet Berlin.