270 likes | 431 Vues
This document provides an insightful overview of normal distribution, a crucial concept in statistics. It differentiates between parameters, which are fixed values related to populations, and statistics, which are derived from sample data. The characteristics of normal distribution, such as its bell-shaped curve and the significance of mean (µ) and standard deviation (σ), are discussed. Practical examples illustrate how to calculate probabilities and interpret z-scores. This resource is essential for anyone studying statistics, particularly in health and community medicine.
E N D
Normal Distribution Nojomi, MD, MPH Department of Community Medicine IUMS
Introduction • Parameters and Statistics • Value or characteristic associated with population • Is called : Parameter • Value or characteristic calculated from a sample • Is called : Statistics Normal Disrtibution
Introduction • Parameters are fixed • Parameters such as : µ , σ • Single sample fixed statistic • Statistics such as : x , SD Normal Disrtibution
Frequency Distribution • Is a tabulation , by table or graph , of the frequency at which values occur in a set . Normal Disrtibution
Frequency distribution of cholinestrase in RBC Normal Disrtibution
Frequency Distribution • RBC – Cholinestrase ? • Numeric Continuous variable • Graph ? • Histogram ! Normal Disrtibution
Histogram Normal Disrtibution
Histogram Normal Disrtibution
The Normal Distribution • The most important frequency distribution in statistic. • It is for numeric variable . • Shape : smooth, bell-shaped, symmetric • It has two parameter : µ , σ Normal Disrtibution
68% of data : µ ± 1SD 95% of data : µ ± 2SD 99% of data : µ ± 3SD The Normal Distribution Normal Disrtibution
Determining Probabilities Using the Normal Distribution • Example 1: • What is the probability that an individual randomly selected from population will have an RBC- choliestrase value between 11.95 to 13.95? • zi = (xi- µ) / σ xi zi ~ N(0 , 1) • zi= (11.95 – 11) / 2 = 0.48 • zi = (13.95 – 11) / 2 = 1.48 • P (.48 < z <1.48) Normal Disrtibution
Determining Probabilities Using the Normal Distribution Normal Disrtibution
Determining Probabilities Using the Normal Distribution • Area beyond z = .48 = .31 • Area beyond z = 1.48 = .06 • .31 - .06 = .25 Normal Disrtibution
Example 2 • After prolonged contact with a sample of cholinomimetic compound, a 14-year old boy is brought to Emergency room, with vomiting, headache, bradycardia, ….. Lab test : • RBC-cholinestrase 20 µmol/ml. How unusual is this ? • Solution : • z = (xi - µ) / σ • = (20 – 11) / 2 = 4.5 Normal Disrtibution
Solution • Area beyond z for z = 4.5 ? Normal Disrtibution
Example 3 • Between which two RBC- cholinestrase values, 95% of the values in G.P be expected to fall ? • It means, 95% of the mid of the curve ! • So, the rest is : 2.5% on the right , and 2.5% on the left of curve . • We have to know, area beyond which z is .025? • The curve is symmetric,… Normal Disrtibution
Solution • If the area beyond z = .025 • The z value is +1.96 on the right, and -1.96 on the left . • z = (xi - µ) / σ • +1.96 = (xi – 11) / 2 • xi =3.92 + 11 = 14.92 • - 1.96 = (xi +11) / 2 • xi = -3.92 + 11= 7.08 Normal Disrtibution
Solution • Is there an easier alternative way ? • What is the SD ? • Ok, the SD is equal 2, in normal distribution : • 95% of curve is between : µ± 2SD • So, the mean is : 11 ± 2 × 2 = 11± 4 • 95% of general population have RBC- choliestrase between 7 and 15 . Normal Disrtibution
Problems • In a normal distribution the mean and standard deviation are 40 and 5 respectively. Calculate : • A. Area under curve on the left of xi = 34 ? • xi < 34 ? • z = (xi- µ) /σ • z = (34 – 40) / 5 = -1.2 • Area beyond z = -1.2 ? • = .11 • 11% of the subjects have score less than 34 . Normal Disrtibution
Problems • B. the point on which, 95% of the curve is the left . • It means, .05 is beyond that point . • Z point that .05 is beyond it , is ? • 1.64 • 1.64 = (xi – 40) / 5 • xi = 48.25 Normal Disrtibution
Problems • Distribution of serum glucose is normal with mean and SD 100 and 2.5 respectively. • If select a person randomly, what is the probability of over than 105 for her/his serum glucose? • Z= (105-100) / 2.5 = 2 • Z > 2 ? • .02 = 2% Normal Disrtibution
Problems • نمرات درس آمار دانشجویان پزشکی از توزیع نرمال با میانگین 70 و انحراف معیار 10 برخوردار است. در یک نمونه 100 نفری از دانشجویان مذکور، تقریبأ چند نفر نمره بین 60 و 80 می گیرند؟ • 68 Normal Disrtibution
Problems • اگر Z متغیر نرمال استاندارد باشد، و P(0<Z<Z1)= a باشد آنگاه P(Z<Z1) برابر است با : • 1/2 + a Normal Disrtibution
Problems • کمیت تصادفی X دارای توزیع دو جمله ای با میانگین 20 و انحراف معیار 4 است. پارامتر n چقدر است ؟ • np= 20 • npq=16, np × 0.8= 18, p=0.2 • n × 0.2 = 20 • n=100 Normal Disrtibution