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Normal distribution

The document provides an overview of normal distribution, including its properties and significance in statistics, often represented as a bell curve or Gaussian curve. It elaborates on the normal and standard normal distributions, explaining how to calculate z-scores and probabilities for normally distributed data. Additionally, the document includes practical examples to apply these concepts in determining abnormal values and analyzing statistical data.

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Normal Distribution Jagdish D. Powar Statistician cum Tutor Community Medicine SMBT, IMSRC, Nashik JDP-CM-SMBT 1
Competency & Learning objectives Competency Learning Objectives CM6.4, CM6.2 Describe and discuss the principles and demonstrate the methods of collection, classification, analysis, interpretation and presentation of statistical data The student should be able to  Define what is normal distribution and normality limits  Define and explain Properties of normal distribution  Find out probability or percentage for normal variables from normal distribution for given examples JDP-CM-SMBT 2
The Normal Distribution A symmetrical probability distribution where most results are located in the middle and few are spread on both sides. The normal distribution is the most important and most widely used distribution in statistics. It is sometimes called the "bell curve," It is also called the "Gaussian curve" after the mathematician Karl Friedrich Gauss. 50%50% mean = median = mode Normal Distribution JDP-CM-SMBT 3
Normal Distribution properties A continuous random variable X is said to follow Normal distribution if it satisfies following property a) It is bell shaped curve. b) The curve is symmetrical and asymptotic (i.e. it never touches to x-axis) c) All measures of central tendency are equal and stable on the highest peak axis mean = median = mode d) Total area under the curve is equal to 1. e) On the basis of mean, µ and standard deviation, σ the area of normal curve is distributed as: µ ± 1*σ=68.27% (µ - 1σ, µ -1σ) µ ± 2*σ=95.45% (µ - 2σ, µ + 2σ) µ ± 3*σ=99.73% (µ -3*σ, µ +3*σ)JDP-CM-SMBT 4
Normal Curve Population Mean(µ) and SD(σ) Sample mean(x̅) and SD JDP-CM-SMBT 5
Standard Normal Distribution Standard Normal Variable(z-score) Any normally distributed data can be converted to the standardized form using the formula: Z = (𝑥−µ) σ If we take x = µ, then Z = 0 x = µ -1σ, then Z = -1 x = µ + 1σ, then Z = +1 x = µ ± 2*σ, then Z=-2, +2 x = µ ± 3*σ, then Z= -3, +3 JDP-CM-SMBT 6
Properties of Standard Normal Distribution a) It is bell shaped curve. b) The curve is symmetrical and asymptotic (i.e. it never touches to x-axis) c) All measures of central tendency are equal and stable on the highest peak axis mean = median = mode = 0 d) Total area under the curve is equal to 1. e) On the basis of mean, µ= 0 and standard deviation, σ=1 the area of normal curve is distributed as: 68% of observations lie within -1 to +1 95% of observations lie within -2 to +2 99% of the observation lie within -3 to +3 JDP-CM-SMBT 7
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Use of Normal Curve Normality limits 95 % confidence interval/ 5% level of significance  x̅ ± 2*SD / x̅ ± 1.96* SD Z- Score:- The value of the z-score tells you how many standard deviations you are away from the mean. To find the probability or percentage by using normal curve and normal table Z = (𝑥−µ) 𝜎 = (𝑥−x̅ ) 𝑆𝐷 JDP-CM-SMBT 9
1) The average fasting blood sugar level of adult male in some area is 90 mg% with SD of 5 mg%. If fasting blood sugar levels are normally distributed, then would you regard-Fasting blood sugar of 95 mg% as abnormal? a) Fasting blood sugar of 105 mg% as abnormal? Ans:- Let us consider variable X- fasting blood sugar level(FBSL). Given values x̅ = mean FBSL= 90 mg%, SD=5% Normality limits x̅ ± 2*SD = (90-2*5, 90+2*5) = (80, 100) a) Fasting blood sugar of 95 mg% as abnormal? 95 lie within normality limits hence it is not abnormal. b) Fasting blood sugar of 105 mg% as abnormal? 105 lie outside of normality limits hence it is abnormal JDP-CM-SMBT 10
3) Mean height of 500 students is 160 cm and SD is 5 cm. If height follows normal distribution then calculate a) What is the chance of height above 168 cm? b) What percentage of students will have height above 168 cm? c) How many of students will have height between 150 cm to 175cm? Ans- Here variable X-height Mean Height = x̅ = 160 cm SD = 5 cm n= number of students =500 JDP-CM-SMBT 11
a) What is the chance of height above 168 cm? Z = (𝑥−x̅ ) 𝑆𝐷 = (168−160) 5 =1.60 A(X > 168) =A(Z > 1.6) = 0.5- A(Z=0 to Z=1.6) =0.5-0.4452 =0.0548 b) What percentage of students will have height above 168 cm? =A(X >168) *100 =0.0548*100 =5.48% JDP-CM-SMBT 12
c) How many of students will have height between 150 cm to 175cm? Z1 = (𝑥1−x̅ ) 𝑆𝐷 = (150−160) 5 =-2 Z2 = (𝑥2−x̅ ) 𝑆𝐷 = (175−160) 5 =3 A(150< X< 175) =A(-2< Z <3) =A(Z=-2 to Z=0) + A(Z=0 to Z=3) =0.4987+0.4772 =0.9759 Number of students will have height between 150 cm to 175cm =A(150< X< 175) *500 =0.9759*500 =487.95≈ 488 Students JDP-CM-SMBT 13
2) The mean weight of 200 students 50 kg and SD is 4 kg. If weight is normally distributed then find the percentage of students having weight [ A(Z=2)= 0.4772, A(Z=1.25)=0.3944 ] a) Above 58 kg c) Above 50 kg b) Less than 45 kg d) Below 58 kg Ans:- Here variable X- weight. Given values x̅ = mean weight = 50 kg, SD=4 kg a) Percentage of students will have weight above 58 kg Z = (𝑥−x̅ ) 𝑆𝐷 = (58−50) 4 =2 A(Z > 2) =0.5-0.4772 =0.0228 %of students weight above 58 kg =0.0228*100= 2.28% JDP-CM-SMBT 14
b) Percentage of students will have weight less than 45 kg Z = (𝑥−x̅ ) 𝑆𝐷 = (45−50) 4 =1.25 % of students having weight less than 45 kg A(Z < 1.25) =A(Z < 0) +A(Z=0 to Z=1.25) =0.5+0.3944 =0.8944 Percentage =0.8944*100 =89.44% JDP-CM-SMBT 15
C) Percentage of students will have weight above 50 kg Z = (𝑥−x̅ ) 𝑆𝐷 = (50−50) 4 =0 A( Z > 0) =0.5 Percentage =0.5*100 =50% JDP-CM-SMBT 16
d) Percentage of students will have weight below 58 kg Z = (𝑥−x̅ ) 𝑆𝐷 = (58−50) 4 =2 A(Z < 2) =A(Z<0 )+A(Z=0 to Z=2) =0.5+0.4772 =0.9772 %of students weight above 58 kg =0.9772*100= 97.72% Another Method:- %( students below 58 kg) =100- %( students above 58 kg) =100- 2.28 =97.72% JDP-CM-SMBT 17
JDP-CM-SMBT 18 Thank You