Normal Distribution

Course Overview

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Bell curve (AKA normal distribution)

Parametric statistics often compare values to a normal distribution of expected data, based on the estimated mean and SD. Lets start by showing a (made up) normal distribution of heights in centimeters:

So lets say the average person’s height is 150cm, and the standard deviation of height across the population is 10cm. The data would look something like:

R

# Plot a normal distribution of heights
population_heights_x <- seq(
  120,    # min
  180,    # max
  by = 1  
)
population_heights_y <- dnorm(
  population_heights_x,
  mean = 150,
  sd   = 10
)
plot(
  population_heights_x,
  population_heights_y,
  xlab = "height",
  ylab = "frequency"
)
# Add line to show mean and median
abline(
  v=150,  # where the line for the mean will be 
  lwd=5
)

Python

# Plot a normal distribution of heights
import numpy as np
from scipy.stats import norm
import matplotlib.pyplot as plt
from matplotlib.ticker import FormatStrFormatter

# vector for the x-axis
population_heights_x = [x for x in range(120, 181, 1)]

# vector for the y-axis
population_heights_y = norm.pdf(population_heights_x, loc=150, scale=10)

fig, ax = plt.subplots(figsize =(7, 5))
ax.yaxis.set_major_formatter(FormatStrFormatter('%.2f'))

# plot
plt.scatter(population_heights_x, population_heights_y, color='w', edgecolors='black')

# Add line to show mean and median
plt.axvline(x=150, color='black', ls='-', lw=5)

# add title on the x-axis
plt.xlabel("height")

# add title on the y-axis
plt.ylabel("Frequency")

# show the plot
plt.show()

# show plot
plt.show()

Normal Distribution

Excel

Download and open the normal.xlsx file in this repository to see data being used to create the below figure:

(note that making a vertical line to reflect the mean didn’t seem as easy in Excel as other languages, so this hasn’t been added)

Normal Distribution using Excel

JASP

Download and open height.csv in JASP, and then complete the following steps to generate the figure below:

• Click on the Descriptives panel

• add height as a variable

• open the Basic plots interface and then select Distribution plots and Display density

SPSS

Download and open height.sav in SPSS and then complete the following steps to generate the figure below:

• Select Analyze –> Descriptive Statistics –> Explore…

• Move the height variable to the dependent list, then select Plots… and choose Histogram as a descriptive figure

You can see that the above fits a bell-curve, and the middle represents both the mean and the median as the data is symmetrical. In reality, almost no data is a perfect bell-curve, but there are ways to test if the data isn’t sufficiently normal to use parametric tests with.

Next, we will look at how normal distributions allow you to transform your data to z-scores to compare to a z-distribution.

Z-scores and the z-distribution

A z-score is a standardised value that captures how many standard deviations above or below the mean an individual value is. Thus, to calculate the z-score

\[ Z = \frac{individualScore-meanScore}{StandardDeviation} \]

Or in formal terminology:

\[ Z = \frac{x-\bar{x}}{\sigma} \]

The calculated score can then be applied to a z-distribution, which is parametric/normally distributed. Lets have a look at a z-distribution:

R

# vector for the x-axis
z_score_x <- seq(
  -3,    # min
  3,    # max
  by = .1  
)

# vector for the y-axis
z_score_y <- dnorm(
  z_score_x,
  mean = 0,
  sd   = 1
)

plot(
  z_score_x,
  z_score_y,
  xlab = "z-score (SDs from the mean)",
  ylab = "frequency"
)

Python

import numpy as np
from scipy.stats import norm
import matplotlib.pyplot as plt
from matplotlib.ticker import FormatStrFormatter

# vector for the x-axis
z_score_x = np.arange(-3.0, 3.1, 0.1)

# vector for the y-axis
z_score_y = norm.pdf(z_score_x, loc=0, scale=1)

fig, ax = plt.subplots(figsize =(7, 5))
ax.yaxis.set_major_formatter(FormatStrFormatter('%.1f'))

# plot
plt.scatter(z_score_x, z_score_y, color='w', edgecolors='black')

# add title on the x-axis
plt.xlabel("z-score (SDs from the mean)")

# add title on the y-axis
plt.ylabel("frequency")

# show plot
plt.show()

Z-score Distribution

Excel

Download and open the normal_z_scores.xlsx file in this repository to see data being used to create the below figure. To get the colors you need to use the “Select Data…” option when right clicking on the chart:

Normal Distribution Z - scores

JASP

It’s not clear that generating a z-distribution like this would be helpful to do with JASP, so the Excel figure is included below:

Normal Distribution Z - scores

SPSS

It’s not clear that generating a z-distribution like this would be helpful to do in SPSS, so the Excel figure is included below:

Normal Distribution Z - scores

If you compare the height distribution above to the z-score distribution, you should see that they are identically distributed. This is useful, as we know what percentage of a population fits within each standard deviation of a normal distribution:

R

library(ggplot2)
# https://stackoverflow.com/a/12429538
norm_x<-seq(-4,4,0.01)
norm_y<-dnorm(-4,4,0.0)

norm_data_frame<-data.frame(x=norm_x,y=dnorm(norm_x,0,1))


shade_50 <- rbind(
  c(0,0), 
  subset(norm_data_frame, x > 0), 
  c(norm_data_frame[nrow(norm_data_frame), "X"], 0))

shade_34.1 <- rbind(
  c(1,0), 
  subset(norm_data_frame, x > 1), 
  c(norm_data_frame[nrow(norm_data_frame), "X"], 0))


shade_13.6 <- rbind(
  c(2,0), 
  subset(norm_data_frame, x > 2), 
  c(norm_data_frame[nrow(norm_data_frame), "X"], 0))


p<-qplot(
  x=norm_data_frame$x,
  y=norm_data_frame$y,
  geom="line"
)

Warning: `qplot()` was deprecated in ggplot2 3.4.0.

 p +
   
   geom_polygon(
     data = shade_50,
     aes(
       x,
       y,
       fill="50"
      )
    ) +
   geom_polygon(
     data = shade_34.1,
     aes(
       x,
       y,
       fill="34.1"
      )
    ) +
   geom_polygon(
     data = shade_13.6, 
     aes(
       x, 
       y,
       fill="13.6"
      )
    ) +
   
   annotate(
     "text", 
     x=0.5, 
     y=0.01, 
     label= "34.1%"
   ) + 
   annotate(
     "text", 
     x=1.5, 
     y=0.01, 
     label= "13.6%"
   ) + 
   annotate(
     "text", 
     x=2.3, 
     y=0.01, 
     label= "2.3%"
   ) +
   xlab("Z-score") +
   ylab("Frequency") +
   theme(legend.position="none")

Python

norm_x = np.arange(-4.0, 4.1, 0.1)

norm_y = norm.pdf(norm_x, loc=0, scale=1)

fig, ax = plt.subplots(figsize =(8, 5))
ax.yaxis.set_major_formatter(FormatStrFormatter('%.1f'))

ax.plot(norm_x, norm_y, color='black', alpha=1.00)
ax.fill_between(norm_x, norm_y, where= (0 <= norm_x)&(norm_x <= 1), color='blue', alpha=.4)
ax.fill_between(norm_x, norm_y, where= (1 < norm_x)&(norm_x <= 2), color='green', alpha=.4)
ax.fill_between(norm_x, norm_y, where= (norm_x > 2), color='red', alpha=.4)

plt.text(2.05, 0.001, '2.3%', fontsize = 10, color='black',weight='bold')
plt.text(1.1, 0.001, '13.6%', fontsize = 10, color='black',weight='bold')
plt.text(0.1, 0.001, '34.1%', fontsize = 10, color='black',weight='bold')

# add title on the x-axis
plt.xlabel("Z-score")

# add title on the y-axis
plt.ylabel("Frequency")

plt.show()

Z-score Distribution

Excel

Download and open the normal_z_scores_0_plus.xlsx file in this repository to see data being used to create the below figure. To get the colors you need to use the “Select Data…” option when right clicking on the chart:

Normal Distribution Z-scores with percentages for each boundary

JASP

It’s not clear that generating a distribution like this would be helpful to do with JASP, so the Excel figure is included below:

Normal Distribution Z-scores with percentages for each boundary

SPSS

It’s not clear that generating a distribution like this would be helpful to do in SPSS, so the Excel figure is included below:

Normal Distribution Z-scores with percentages for each boundary

The above visualises how 34.1% of a population’s scores will be between 0 and 1 standard deviation from the mean, 13.6% of the population’s scores will be between 1 to 2 standard deviations above the mean, and 2.3% of the population will be more then 2 standard deviations above the mean. Remember that the normal distribution is symmetrical, so we also know that 34.1% of the population’s score will be between 0 to 1 standard deviations below the mean (or 0 to -1 SDs), 13.6% of the population’s score will be between -2 to -1 standard deviations from the mean, and 2.3% of the population’s score will be more negative than -2 standard deviations from the mean. Lets look at this cumulative distribution:

R

# https://stackoverflow.com/a/12429538
norm_x<-seq(-4,4,0.01)
norm_y<-dnorm(-4,4,0.0)

norm_data_frame<-data.frame(x=norm_x,y=dnorm(norm_x,0,1))


shade_2.3 <- rbind(
  c(-8,0), 
  subset(norm_data_frame, x > -8), 
  c(norm_data_frame[nrow(norm_data_frame), "X"], 0))

shade_13.6 <- rbind(
  c(-2,0), 
  subset(norm_data_frame, x > -2), 
  c(norm_data_frame[nrow(norm_data_frame), "X"], 0))

shade_34.1 <- rbind(
  c(-1,0), 
  subset(norm_data_frame, x > -1), 
  c(norm_data_frame[nrow(norm_data_frame), "X"], 0))

shade_50 <- rbind(
  c(0,0), 
  subset(norm_data_frame, x > 0), 
  c(norm_data_frame[nrow(norm_data_frame), "X"], 0))

shade_84.1 <- rbind(
  c(1,0), 
  subset(norm_data_frame, x > 1), 
  c(norm_data_frame[nrow(norm_data_frame), "X"], 0))


shade_97.7 <- rbind(
  c(2,0), 
  subset(norm_data_frame, x > 2), 
  c(norm_data_frame[nrow(norm_data_frame), "X"], 0))


p<-qplot(
  x=norm_data_frame$x,
  y=norm_data_frame$y,
  geom="line"
)

 p +
   geom_polygon(
     data = shade_2.3,
     aes(
       x,
       y,
       fill="2.3"
      )
    ) +
   geom_polygon(
     data = shade_13.6,
     aes(
       x,
       y,
       fill="13.6"
      )
    ) +
   geom_polygon(
     data = shade_34.1,
     aes(
       x,
       y,
       fill="34.1"
      )
    ) +
   geom_polygon(
     data = shade_50,
     aes(
       x,
       y,
       fill="50"
      )
    ) +
   geom_polygon(
     data = shade_84.1,
     aes(
       x,
       y,
       fill="84.1"
      )
    ) +
   geom_polygon(
     data = shade_97.7, 
     aes(
       x, 
       y,
       fill="97.7"
      )
    ) +
   xlim(c(-4,4)) +
   
   annotate("text", x=-2.3, y=0.01, label= "2.3%") + 
   annotate("text", x=-1.4, y=0.01, label= "15.9%") + 
   annotate("text", x=-0.4, y=0.01, label= "50%") + 
   annotate("text", x=0.5, y=0.01, label= "84.1%") + 
   annotate("text", x=1.5, y=0.01, label= "97.7%") + 
   annotate("text", x=2.3, y=0.01, label= "100%") +

   xlab("Z-score") +
   ylab("Frequency") +
   theme(legend.position="none")

Python

norm_x = np.arange(-4.0, 4.1, 0.1)

norm_y = norm.pdf(norm_x, loc=0, scale=1)

fig, ax = plt.subplots(figsize =(8, 5))
ax.yaxis.set_major_formatter(FormatStrFormatter('%.1f'))

ax.plot(norm_x, norm_y, color='black', alpha=1.00)
ax.fill_between(norm_x, norm_y, where= (norm_x < -2), color='gold', alpha=.4)
ax.fill_between(norm_x, norm_y, where= (-2 <= norm_x)&(norm_x <= -1), color='red', alpha=.4)
ax.fill_between(norm_x, norm_y, where= (-1 <= norm_x)&(norm_x <= 0), color='green', alpha=.4)
ax.fill_between(norm_x, norm_y, where= (0 <= norm_x)&(norm_x <= 1), color='cyan', alpha=.4)
ax.fill_between(norm_x, norm_y, where= (1 < norm_x)&(norm_x <= 2), color='blue', alpha=.4)
ax.fill_between(norm_x, norm_y, where= (norm_x > 2), color='purple', alpha=.4)

plt.text(2.05, 0.001, '100%', fontsize = 10,color='black',weight='bold')
plt.text(1.2, 0.001, '97.7%', fontsize = 10,color='black',weight='bold')
plt.text(0.2, 0.001, '84.1%', fontsize = 10,color='black',weight='bold')
plt.text(-0.8, 0.001, '50%', fontsize = 10,color='black',weight='bold')
plt.text(-1.85, 0.001, '15.9%', fontsize = 10,color='black',weight='bold')
plt.text(-2.7, 0.001, '2.3%', fontsize = 10,color='black',weight='bold')

plt.yticks(np.arange(0, 0.5, step=0.1))

# add title on the x-axis
plt.xlabel("Z-score")

# add title on the y-axis
plt.ylabel("Frequency")


plt.show()

Z-score Distribution

Excel

Download and open the normal_z_scores_percentages.xlsx file in this repository to see data being used to create the below figure. To get the colors you need to use the “Select Data…” option when right clicking on the chart:

Normal Distribution Z-scores and percentages

JASP

It’s not clear that generating a distribution like this would be helpful to do with JASP, so the Excel figure is included below:

Normal Distribution Z-scores and percentages

SPSS

It’s not clear that generating a distribution like this would be helpful to do in SPSS, so the Excel figure is included below:

Normal Distribution Z-scores and percentages

The above figure visualises how 13.6% of the population have score that is more negative than -2 standard deviations from the mean, 34.1% of the population have a standard deviation that is more negative than -1 standard deviations from the mean (this also include all the people who are more than -2 standard deviations from the mean), etc.

We can now use the above information to identify which percentile an individual is within a distribution.

For example, let’s imagine that an individual called Jane wants to know what percentile she’s at with her height. Lets imagine she is 170cm tall, the mean height of people 150cm, and the SD 10cm. That would make her z-score:

\[ Z_{score} = \frac{170 - 150}{10} = 2 \]

As we can see from the figure above, that puts her above 97.7% of the population, putting her in the top 2.3%.

Consolidation questions

Question 1

rand_maths_score = 40 + Math.round(Math.random() * 60);
mean_maths_score = 70
sd_maths_score   = 10

Jamie has just completed a mathematics test, where the maximum score is 100%. Their score was , the mean maths score was and the SD was . What is their Z-score?

viewof normal_question_1_response = Inputs.number([-7,3], {label: "Z-score", step:.1});
correct_z_score = (rand_maths_score - mean_maths_score)/sd_maths_score;

normal_question_1_result = { 
  if(normal_question_1_response == correct_z_score){
    return "Correct! (" + rand_maths_score + " - " + mean_maths_score + ")/" + sd_maths_score + " = " + correct_z_score;
  } else {
    return "Missing or incorrect. Remember that how Z is calculated by dividing the difference between a value and the mean value by the SD."
  } 
}

Your answer is… .

Question 2

Using the above value, which percentile group would you put Jamie’s score into?

normal_question_2_correct = {
  if(correct_z_score < -2){
    return "bottom 2.3%";
  } else if(correct_z_score < -1){
    return "bottom 15.9%";
  } else if(correct_z_score < 0){
    return "bottom 50%";
  } else if(correct_z_score < 1){
    return "top 50%";
  } else if(correct_z_score < 2){
    return "top 15.9%";
  } else {
    return "top 2.3%";
  }
}

viewof normal_question_2_response = Inputs.radio([
  "bottom 2.3%", 
  "bottom 15.9%",
  "bottom 50%",
  "top 50%",
  "top 15.9%",
  "top 2.3%", 
  ], {label: "", value: "A"});
normal_question_2_result = { 
  if(normal_question_2_response == ""){
    return "awaiting your response";
  } else if(normal_question_2_correct == normal_question_2_response){
    return "Correct!";
  } else {
    return "Missing or Incorrect - have a look at the plots above to help you find the correct answer. Note, the distributions are symmetrical, so the pattern for the top half will mirror that for the bottom half.";
  }
}

Your answer is… .

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