Day4 Normal Curve

8/3/2019 Day4 Normal Curve

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Statistics in MedicalResearch



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Curves of Frequency Distributions

skewed to the rightA=mode B=median C=mean

skewed to the leftA=mean B=median C=mode



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Normal Distributions



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| 68.26 |---------95.45%----

--------------99.74%--------- _______|___|___|___|___|___|________

3sd 2sd 1sd 1sd 2sd 3sdMo Md



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Properties of the normal distribution 1. The normal curve is a theoretical distribution defined

by two parameters, the mean µ and the standarddeviation σ.

2. It has the appearance of a symmetrical bell-shapedcurve extending infinitely in both directions. It issymmetrical about the mean.

3. The total area under the curve is 1 or 100%. Whetherthe mean or standard deviation is large or small, the

relative areas between any two points are always thesame. With µ±1σ the area under the curve is 68.26 or 68.26% of the observations fall within µ±1σ, withµ±2σ, the area is 95.45% and with µ±3σ the area is99.74%

4. The mean, median and mode are equal.



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Standard Normal Curve The normal distribution is a large family of

distributions corresponding to many different

values of parameters µ and σ.

Application of the normal curve involves computationsof the areas of its segments. Every time it is needed, tocompute for a specific area given µ and σ, we have togo back to the formula which is very time consuming.

Statisticians have computed extensively all the possible

members of the family or areas under the curve andpresented in the so called the standardized normalcurve with mean 0 and standard deviation of 1. Astandardized Z score have been tabulated.



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1 -(x-)2 f(x) = ---------- exp -----------), - x

(2

2

) 22

where: = 3.1416 e = 2.71

= mean of the distribution = standard deviation of the

distribution



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Application of the normal distribution

1. Computation of proportion, percentages orprobabilities of x values falling in a given range

Ex. What is the proportion of students who got 60% inthe examination or what is the probability that astudent will get 60% in the exam?

2. Determining the bounding x values given theproportion or probability

Ex. Determining the weights of people belonging to

the lowest 10% of the group



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Formula for transformation:

x-µz = -------- Where:

σ z = standard normaldeviate

x = value of the parameteror measurement

µ = population meanσ = population standard

deviation



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Normal DistributionsExample:

A group of subjects is known to have a mean weight of135+/-20lbs.

Application #1a. find the proportion of individuals with weights

above 150lbs.x-µ 150-135 15z = -------- = -------------- =- -------- = 0.75

σ 20 20

Referring to the table of values of the standardnormal curve, the area corresponding to 0.75 is0.2266. Multiplying 0.2266 by 100, the proportionof subjects with weights of 150lbs is 22.66%.



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0.2266x100=22.66%------------׀-----------------------------------------------

135 150 x0 0.75 z



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b. find the proportion of subjects with weightsbelow 110 lbs

x-µ 110-135

z = --------- = ------------- = -1.25σ 20

z of –1.25 corresponds to the area 0.1056 in

the normal curve.

Therefore the proportion of subjects with

weights below 110lbs is 10.56%



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0.1056X100 =10.56%--------------------------------------׀------------

x 110 135

z -1.25 0



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c. find the proportion of subjects with weights between130 to 160 lbs

when x1 = 130, when x2 = 160

130-135 160-135 z = ----------- = -0.25 z = ---------- = 1.25 20 20

area of z = -0.25 is –0.4013, z = 1.25 is 0.1056 area = 1 – (0.4013+0.1056) = 0.4931 therefore the proportion of subjects between

130 to 160lbs is 49.31%



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-0.4013 0.4931 0.1056------------------|--------------------|------------

x 130 135 160

z -0.25 0 1.25



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Application # 2d. find the weight of the heaviest 10% in the group

In the table of the standard deviate, the nearest corresponding z value of 10% or 0.1000 is 1.28 (0.1003) Using the transformation formula: x - 135

10% or 0.1000 = ------------ 20

X - 135

1.28 = ------------ 20 1.28 (20) = X – 135 X = 128 (20) + 135 X = 160.6 lbs



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160.6

| | 10% ------------׀-----------------------------------------------

135 160.6 x0 1.28 z



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Normal Distributionse. find the weights of the middle 50%

Divide the other half or the other 50% into two,corresponding to the two tail end since what isasked is the middle 50%. Therefore the two tails willhave an area of 0.25 each, -0.25 to the left and +0.25to the right. The corresponding Z scores are –0.67

and +0.67 respectively

Lower limit: Upper limit: x-135 x-135

-0.67 = ------------ 0.67 = --------- 20 20

x-135 = -0.67(20) x-135 = 0.67(20)

x = -0.67(20) +135 x = 0.67(20)+135 x = 121.6 x = 148.4



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| | 50% |

------------׀--------------------|---------------------------121.6 135 148.4 x-0.67 0 0.67 z

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