In this lab we’ll investigate the probability distribution that is most central to statistics: the normal distribution. If we are confident that our data are nearly normal, that opens the door to many powerful statistical methods. Here we’ll use the graphical tools of R to assess the normality of our data and also learn how to generate random numbers from a normal distribution.

The Data

This week we’ll be working with measurements of body dimensions. This data set contains measurements from 247 men and 260 women, most of whom were considered healthy young adults.

download.file("http://www.openintro.org/stat/data/bdims.RData", destfile = "bdims.RData")
load("bdims.RData")

Let’s take a quick peek at the first few rows of the data.

head(bdims)
##   bia.di bii.di bit.di che.de che.di elb.di wri.di kne.di ank.di sho.gi
## 1   42.9   26.0   31.5   17.7   28.0   13.1   10.4   18.8   14.1  106.2
## 2   43.7   28.5   33.5   16.9   30.8   14.0   11.8   20.6   15.1  110.5
## 3   40.1   28.2   33.3   20.9   31.7   13.9   10.9   19.7   14.1  115.1
## 4   44.3   29.9   34.0   18.4   28.2   13.9   11.2   20.9   15.0  104.5
## 5   42.5   29.9   34.0   21.5   29.4   15.2   11.6   20.7   14.9  107.5
## 6   43.3   27.0   31.5   19.6   31.3   14.0   11.5   18.8   13.9  119.8
##   che.gi wai.gi nav.gi hip.gi thi.gi bic.gi for.gi kne.gi cal.gi ank.gi
## 1   89.5   71.5   74.5   93.5   51.5   32.5   26.0   34.5   36.5   23.5
## 2   97.0   79.0   86.5   94.8   51.5   34.4   28.0   36.5   37.5   24.5
## 3   97.5   83.2   82.9   95.0   57.3   33.4   28.8   37.0   37.3   21.9
## 4   97.0   77.8   78.8   94.0   53.0   31.0   26.2   37.0   34.8   23.0
## 5   97.5   80.0   82.5   98.5   55.4   32.0   28.4   37.7   38.6   24.4
## 6   99.9   82.5   80.1   95.3   57.5   33.0   28.0   36.6   36.1   23.5
##   wri.gi age  wgt   hgt sex
## 1   16.5  21 65.6 174.0   1
## 2   17.0  23 71.8 175.3   1
## 3   16.9  28 80.7 193.5   1
## 4   16.6  23 72.6 186.5   1
## 5   18.0  22 78.8 187.2   1
## 6   16.9  21 74.8 181.5   1

You’ll see that for every observation we have 25 measurements, many of which are either diameters or girths. A key to the variable names can be found at http://www.openintro.org/stat/data/bdims.php, but we’ll be focusing on just three columns to get started: weight in kg (wgt), height in cm (hgt), and sex (1 indicates male, 0 indicates female).

Since males and females tend to have different body dimensions, it will be useful to create two additional data sets: one with only men and another with only women.

mdims <- subset(bdims, sex == 1)
fdims <- subset(bdims, sex == 0)
  1. Make a histogram of men’s heights and a histogram of women’s heights. How would you compare the various aspects of the two distributions?

Let’s draw frequency histograms for male and female distributions.

hist(mdims$hgt, main = "Male Height Distribution", breaks = 10, xlab = "Height (cm)")

hist(fdims$hgt, main = "Female Height Distribution", breaks = 10, xlab = "Height (cm)")

Male height distribution is more bell-shaped than female height distribution. Male height distribution is pretty symmetric. The mean and the median are almost the same. Female height distribution is a little bit right-skewed. It means that the mean is a little bit greater than the median.
Male height distribution is closer to normal distribution than female distribution.

The normal distribution

In your description of the distributions, did you use words like bell-shaped or normal? It’s tempting to say so when faced with a unimodal symmetric distribution.

To see how accurate that description is, we can plot a normal distribution curve on top of a histogram to see how closely the data follow a normal distribution. This normal curve should have the same mean and standard deviation as the data. We’ll be working with omen’s heights, so let’s store them as a separate object and then calculate some statistics that will be referenced later.

fhgtmean <- mean(fdims$hgt)
fhgtsd   <- sd(fdims$hgt)

Next we make a density histogram to use as the backdrop and use the lines function to overlay a normal probability curve. The difference between a frequency histogram and a density histogram is that while in a frequency histogram the heights of the bars add up to the total number of observations, in a density histogram the areas of the bars add up to 1. The area of each bar can be calculated as simply the height times the width of the bar. Using a density histogram allows us to properly overlay a normal distribution curve over the histogram since the curve is a normal probability density function. Frequency and density histograms both display the same exact shape; they only differ in their y-axis. You can verify this by comparing the frequency histogram you constructed earlier and the density histogram created by the commands below.

hist(fdims$hgt, probability = TRUE,ylim = c(0, 0.06))
x <- 140:190
y <- dnorm(x = x, mean = fhgtmean, sd = fhgtsd)
lines(x = x, y = y, col = "blue")

After plotting the density histogram with the first command, we create the x- and y-coordinates for the normal curve. We chose the x range as 140 to 190 in order to span the entire range of fheight. To create y, we use dnorm to calculate the density of each of those x-values in a distribution that is normal with mean fhgtmean and standard deviation fhgtsd. The final command draws a curve on the existing plot (the density histogram) by connecting each of the points specified by x and y. The argument col simply sets the color for the line to be drawn. If we left it out, the line would be drawn in black.

The top of the curve is cut off because the limits of the x- and y-axes are set to best fit the histogram. To adjust the y-axis you can add a third argument to the histogram function: ylim = c(0, 0.06).

  1. Based on the this plot, does it appear that the data follow a nearly normal distribution?

The data is not perfectly bell-shaped but data is not perfectly symmetric and a little bit right-skewed (mean is not exactly equals median). It’s not clear that 68% of values are within 1 standard deviation of the mean. So that, it’s hard to say ether the data follows nearly normal distribution by just looking at the graph.

Evaluating the normal distribution

Eyeballing the shape of the histogram is one way to determine if the data appear to be nearly normally distributed, but it can be frustrating to decide just how close the histogram is to the curve. An alternative approach involves constructing a normal probability plot, also called a normal Q-Q plot for “quantile-quantile”.

qqnorm(fdims$hgt)
qqline(fdims$hgt)

A data set that is nearly normal will result in a probability plot where the points closely follow the line. Any deviations from normality leads to deviations of these points from the line. The plot for female heights shows points that tend to follow the line but with some errant points towards the tails. We’re left with the same problem that we encountered with the histogram above: how close is close enough?

A useful way to address this question is to rephrase it as: what do probability plots look like for data that I know came from a normal distribution? We can answer this by simulating data from a normal distribution using rnorm.

sim_norm <- rnorm(n = length(fdims$hgt), mean = fhgtmean, sd = fhgtsd)

The first argument indicates how many numbers you’d like to generate, which we specify to be the same number of heights in the fdims data set using the length function. The last two arguments determine the mean and standard deviation of the normal distribution from which the simulated sample will be generated. We can take a look at the shape of our simulated data set, sim_norm, as well as its normal probability plot.

  1. Make a normal probability plot of sim_norm. Do all of the points fall on the line? How does this plot compare to the probability plot for the real data?
qqnorm(sim_norm)
qqline(sim_norm)

The points in the middle mostly fall on the line. However, there are some outlines in the begging and the end of the line. Both plots have outlines. Although, the plot of sim_norm has more points that fall on the line. The plot of fdims$hgt is more step-wise in the middle and has less pints that fall on the line.

Even better than comparing the original plot to a single plot generated from a normal distribution is to compare it to many more plots using the following function. It may be helpful to click the zoom button in the plot window.

qqnormsim(fdims$hgt)

  1. Does the normal probability plot for fdims$hgt look similar to the plots created for the simulated data? That is, do plots provide evidence that the female heights are nearly normal?

    The normal probability plot for fdims$hgt looks very similar with some of the plots created for the simulated data. I think that the female heights are nearly normal because some of the simulated plots contains more points that don’t fall on the line that the plot of real data.

  2. Using the same technique, determine whether or not female weights appear to come from a normal distribution.

  3. Let’s calculate the mean and the standard deviation of the weight.

fwgtmean <- mean(fdims$wgt)
fwgtsd   <- sd(fdims$wgt)
  1. Let’s build density histogram for fdims$wgt.
hist(fdims$wgt, probability = TRUE,ylim = c(0, 0.06))
x <- 40:110
y <- dnorm(x = x, mean = fwgtmean, sd = fwgtsd)
lines(x = x, y = y, col = "blue")

The data is not symmetric and right-skewed. It means that the mean is greater than the median. Moreover, normal distribution curve touches x-axis. If data is normally distributed a normal distribution curve doesn’t taus x-axis.

Let’s check wether mean is greater than the median.

summary(fdims$wgt)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##    42.0    54.5    59.0    60.6    65.6   105.2

Mean is a little bit greater than the median.

Now, let’s compare actual weight data with simulated normally distributed weight data.

3.Build a normal Q-Q plot for "quantile-quantile".
qqnorm(fdims$wgt)
qqline(fdims$wgt)

  1. Simulate data from a normal distribution using rnorm.
sim_norm_wgt <- rnorm(n = length(fdims$wgt), mean = fwgtmean, sd = fwgtsd)
  1. Make a normal probability plot of sim_norm_wgt.
qqnorm(sim_norm_wgt)
qqline(sim_norm_wgt)

The normal probability plot of sim_norm_wgt has much more points that fall on the line than the probability plot for the real data.

qqnormsim(fdims$wgt)

The normal probability plot for fdims$wgt looks very similar with the top left plot. However, it looks worser than all other plots. I think that the female wight might not be normally distributed because almost all of the simulated plots contains more points that fall on the line that the plot of real data.

Normal probabilities

Okay, so now you have a slew of tools to judge whether or not a variable is normally distributed. Why should we care?

It turns out that statisticians know a lot about the normal distribution. Once we decide that a random variable is approximately normal, we can answer all sorts of questions about that variable related to probability. Take, for example, the question of, “What is the probability that a randomly chosen young adult female is taller than 6 feet (about 182 cm)?” (The study that published this data set is clear to point out that the sample was not random and therefore inference to a general population is not suggested. We do so here only as an exercise.)

If we assume that female heights are normally distributed (a very close approximation is also okay), we can find this probability by calculating a Z score and consulting a Z table (also called a normal probability table). In R, this is done in one step with the function pnorm.

1 - pnorm(q = 182, mean = fhgtmean, sd = fhgtsd)
## [1] 0.004434387

Note that the function pnorm gives the area under the normal curve below a given value, q, with a given mean and standard deviation. Since we’re interested in the probability that someone is taller than 182 cm, we have to take one minus that probability.

Assuming a normal distribution has allowed us to calculate a theoretical probability. If we want to calculate the probability empirically, we simply need to determine how many observations fall above 182 then divide this number by the total sample size.

sum(fdims$hgt > 182) / length(fdims$hgt)
## [1] 0.003846154

Although the probabilities are not exactly the same, they are reasonably close. The closer that your distribution is to being normal, the more accurate the theoretical probabilities will be.

  1. Write out two probability questions that you would like to answer; one regarding female heights and one regarding female weights. Calculate the those probabilities using both the theoretical normal distribution as well as the empirical distribution (four probabilities in all). Which variable, height or weight, had a closer agreement between the two methods?

    Question 1. What is the probability that a randomly chosen young adult female is lower than 5 feet 7 inches (about 170.18 cm)?

pnorm(q = 170, mean = fhgtmean, sd = fhgtsd)
## [1] 0.7833331
sum(fdims$hgt < 170) / length(fdims$hgt)
## [1] 0.7538462

Theoretical probability is 78.33% while the empirical probability is 75.38%. Two numbers are pretty close.

Question 2. What is the probability that a randomly chosen young adult female weights more than 130 lbs?

1-pnorm(q = 59, mean = fwgtmean, sd = fwgtsd)
## [1] 0.5660925
sum(fdims$wgt > 59) / length(fdims$wgt)
## [1] 0.4961538

Theoretical probability is 56.61% while the empirical probability is 49.62%. Two numbers are not as close as in case of female height.

Theoretical probability is what is expected to happen while experimental probability is what actually happened. Since female weight deviates from normal distribution much more that female height the difference between theoretical and empirical probability is greater for weight that for height.

On Your Own

qqnorm(fdims$kne.di)
qqline(fdims$kne.di)

The pattern on the plot suggests that knee diameter data is right-skewed.

If data is right-skewed mean should be greater that the median. Let’s check.

summary(fdims$kne.di)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##    15.7    17.3    18.0    18.1    18.7    24.3

Mean is a liitle bit greater than the median.

Let’s calculate the mean the standard deviation.

knedimean <- mean(fdims$kne.di)
knedisd <- sd(fdims$kne.di)

Let’s build density histogram with normal distribution curve.

hist(fdims$kne.di, probability = TRUE,ylim = c(0, 0.4))
x <- 14:26
y <- dnorm(x = x, mean = knedimean, sd = knedisd)
lines(x = x, y = y, col = "blue")

This is a product of OpenIntro that is released under a Creative Commons Attribution-ShareAlike 3.0 Unported. This lab was adapted for OpenIntro by Andrew Bray and Mine Çetinkaya-Rundel from a lab written by Mark Hansen of UCLA Statistics.