Download presentation

Published byGabrielle Patterson Modified over 4 years ago

1
Normal Distribution This lecture will give an overview/review of normal distribution. 2005

2
**Objectives Learning Objective Performance Objectives**

- To understand the topic on Normal Distribution and its importance in different disciplines. Performance Objectives At the end of this lecture the student will be able to: Draw normal distribution curves and calculate the standard score (z score) Apply the basic knowledge of normal distribution to solve problems. Interpret the results of the problems. Tripthi M. Mathew, MD, MPH

3
**Types of Distribution Frequency Distribution**

Normal (Gaussian) Distribution Probability Distribution Poisson Distribution Binomial Distribution Sampling Distribution t distribution F distribution There are several types of distribution: frequency distribution, normal distribution, probability and sampling distributions. This lecture will focus on normal distribution. Tripthi M. Mathew, MD, MPH

4
**What is Normal (Gaussian) Distribution?**

The normal distribution is a descriptive model that describes real world situations. It is defined as a continuous frequency distribution of infinite range (can take any values not just integers as in the case of binomial and Poisson distribution). This is the most important probability distribution in statistics and important tool in analysis of epidemiological data and management science. The Normal distribution is also known as the Gaussian Distribution and the curve is also known as the Gaussian Curve, named after German Mathematician-Astronomer Carl Frederich Gauss. Tripthi M. Mathew, MD, MPH

5
**Characteristics of Normal Distribution**

It links frequency distribution to probability distribution Has a Bell Shape Curve and is Symmetric It is Symmetric around the mean: Two halves of the curve are the same (mirror images) Tripthi M. Mathew, MD, MPH

6
**Characteristics of Normal Distribution Cont’d**

Hence Mean = Median The total area under the curve is 1 (or 100%) Normal Distribution has the same shape as Standard Normal Distribution. Tripthi M. Mathew, MD, MPH

7
**Characteristics of Normal Distribution Cont’d**

In a Standard Normal Distribution: The mean (μ ) = and Standard deviation (σ) =1 Tripthi M. Mathew, MD, MPH

8
**Z Score (Standard Score)3**

Z = X - μ Z indicates how many standard deviations away from the mean the point x lies. Z score is calculated to 2 decimal places. σ The relationship between the normal variable X and Z score is given by the Z score or standard score. Mu (μ) is the mean and sigma (σ) is the standard deviation of the population. Tripthi M. Mathew, MD, MPH

9
**Tables Areas under the standard normal curve (See Normal Table)**

The value of z can be calculated using the Z score. The z value can also be found in tables on standard normal curve or normal distribution curve which can be found in the appendices of most statistics or modelling textbooks. Tripthi M. Mathew, MD, MPH

10
**Diagram of Normal Distribution Curve (z distribution)**

33.35% 13.6% 2.2% 0.15 μ This is the diagram of a normal distribution curve or z distribution. Note the bell shape of the curve and that its ends/tail don’t touch the horizontal axis below. As I mentioned earlier, the area under the curve equals 1 or 100%. Therefore, each half of the distribution from the center (that is from the mean is equal to 50%. Thus, the area from/above the mean up to 1 standard deviation is equal to 33.35%, area above +1 standard deviation is equal to 13.6%, the area above +2 standard deviation is equal to 2.2% and area above +3 standard deviations is equal to 0.1%. Since the other half is a mirror image, the percentage/proportion of area above -1 standard deviation is the same as the area above + 1 standard deviation i.e. it is 33.35%. And -2 standard deviation=+2 standard deviation and so forth…. Modified from Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

11
**Distinguishing Features**

The mean ± 1 standard deviation covers 66.7% of the area under the curve The mean ± 2 standard deviation covers 95% of the area under the curve The mean ± 3 standard deviation covers 99.7% of the area under the curve Therefore, we can see that the mean +/- SD contains 66.7% of the area under the curve. i.e. total area of + 1SD and -1SD is equal to 66.7% (33.35% %). Similarly, the mean +/- 2 SD contains 95% of the area under the normal curve and the mean +/- 3 standard deviations contains 99.7% of the area under the normal curve. Tripthi M. Mathew, MD, MPH

12
**Skewness Positive Skewness: Mean ≥ Median**

Negative Skewness: Median ≥ Mean Pearson’s Coefficient of Skewness3: = 3 (Mean –Median) Standard deviation As I mentioned earlier the Mean = Median in the normal distribution curve. However, when the Mean is greater than the median there is a positive skewness (that is the tail is to the right) to the normal distribution curve. And when the Median is greater then the mean, the tail is skewed to the left, that there is negative skewness. The skewness can be calculated using the Pearson’s Coefficient of Skewness. Tripthi M. Mathew, MD, MPH

13
**Positive Skewness (Tail to Right)**

This diagram illustrates positive skewness (a move to the right). Tripthi M. Mathew, MD, MPH

14
**Negative Skewness (Tail to Left)**

This diagram depicts negative skewness (a move to the left) Tripthi M. Mathew, MD, MPH

15
Exercises Assuming the normal heart rate (H.R) in normal healthy individuals is normally distributed with Mean = 70 and Standard Deviation =10 beats/min The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

16
**Exercise # 1 Then: 1) What area under the curve is above 80 beats/min?**

Now we know, Z =X-M/SD Z=? X=80, M= 70, SD=10 . So we have to find the value of Z. For this we need to draw the figure…..and find the area which corresponds to Z. Modified from Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

17
**Diagram of Exercise # 1 13.6% 0.15 -3 -2 -1 μ 1 2 3 33.35% 2.2% 0.159**

μ 0.159 Since M=70, then the area under the curve which is above 80 beats per minute corresponds to above + 1 standard deviation. The total shaded area corresponding to above 1+ standard deviation in percentage is 15.9% or Z= 15.9/100 = Or we can find the value of z by substituting the values in the formula Z= X-M/ standard deviation. Therefore, Z= 70-80/ /10= is the same as The value of z from the table for 1.00 is How do we interpret this? This means that 15.9% of normal healthy individuals have a heart rate above one standard deviation (greater than 80 beats per minute). The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

18
**Exercise # 2 Then: 2) What area of the curve is above 90 beats/min?**

As in question, proceed by drawing the normal distribution curve and calculate the z value……. The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

19
**Diagram of Exercise # 2 13.6% 0.15 -3 -2 -1 μ 1 2 3 33.35% 2.2% 0.023**

μ 0.023 The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

20
**Exercise # 3 Then: 3) What area of the curve is between**

50-90 beats/min? The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

21
**Diagram of Exercise # 3 13.6% 0.15 -3 -2 -1 μ 1 2 3 33.35% 2.2% 0.954**

μ 0.954 The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

22
**Exercise # 4 Then: 4) What area of the curve is above 100 beats/min?**

The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

23
**Diagram of Exercise # 4 13.6% 0.15 -3 -2 -1 μ 1 2 3 33.35% 2.2%**

μ 0.015 The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

24
Exercise # 5 5) What area of the curve is below 40 beats per min or above 100 beats per min? The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

25
**Diagram of Exercise # 5 13.6% 0.15 -3 -2 -1 μ 1 2 3 33.35% 2.2%**

μ In this question, we need to calculate Z1 and Z2. Therefore, Z1 =70-40/10, which is equal to 3. The z value of 3 is Similarly, the value of Z2 is /10 which is equal to -3. Thus, the value of z is 0.015 And so, Z1+Z2 is equal to 0.3%. But how do we interpret this value? Please see the solution/answer #5 slide for its interpretation. 0.015 0.015 The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

26
**Solution/Answers 1) 15.9% or 0.159 2) 2.3% or 0.023 3) 95.4% or 0.954**

Calculation of the problems and Interpretation of results: For calculation of exercise # 1 see earlier slide. The result of exercise # 1 is 15.9%. This means that 15.9% of normal healthy individuals have a heart rate above one standard deviation (greater than 80 beats per minute). 2) Calculation for exercise #2 Z = X- μ z = = 20/10 = If we look at the normal distribution tables, then the z value of 2.00 corresponds to or 2.3% σ This means that 2.3% of normal healthy individuals have a heart rate above two standard deviation (greater than 90 beats per minute). 3) Calculation for exercise # 3 Z = X- μ Z1 = = -20/10 = and Z2= 90-70/10 =2.00. The area between -2 standard deviations and +2 standard deviations from The z tables is or 95.4%. This means that 95.4% have a heart rate between -2 and +2 standard deviations (between beats per minute). The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

27
**Solution/Answers Cont’d**

4) 0.15 % or 0.015 5) 0.3 % or (for each tail) 4) Calculation for exercise #4 Again, z = X- μ = = 30/10 = From the z tables the value of 3.00 corresponds to or 0.15% σ This means that only 0.15% have a heart rate above 3 standard deviations (greater than 100 beats per minute). 5) For calculations for this question, please see the earlier slide on this problem and diagram. The answer is 0.3%. This means that only 0.3% have a heart rate either below or above 3 standard deviations (less than 40 or greater than 100 beats per minute). The exercises are modified from examples in Dawson-Saunders, B & Trapp, RG. Basic and Clinical Biostatistics, 2nd edition, 1994. Tripthi M. Mathew, MD, MPH

28
**Application/Uses of Normal Distribution**

It’s application goes beyond describing distributions It is used by researchers and modelers. The major use of normal distribution is the role it plays in statistical inference. The z score along with the t –score, chi-square and F-statistics is important in hypothesis testing. It helps managers/management make decisions. Tripthi M. Mathew, MD, MPH

Similar presentations

OK

Problem: Assume that among diabetics the fasting blood level of glucose is approximately normally distributed with a mean of 105mg per 100ml and an SD.

Problem: Assume that among diabetics the fasting blood level of glucose is approximately normally distributed with a mean of 105mg per 100ml and an SD.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google