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SADC Course in Statistics The normal distribution (Session 08)

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SADC Course in Statistics The normal distribution (Session 08)
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Page 1: SADC Course in Statistics The normal distribution (Session 08)

SADC Course in Statistics

The normal distribution

(Session 08)

Page 2: SADC Course in Statistics The normal distribution (Session 08)

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Learning Objectives

At the end of this session you will be able to:

• describe the normal probability distribution

• state and interpret parameters associated with the normal distribution

• use a calculator and statistical tables to calculate normal probabilities

• appreciate the value of the normal distribution in practical situations

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The Normal Distribution• In the previous two sessions, you were

introduced to two discrete distributions, the Binomial and the Poisson.

• In this session, we introduce the Normal Distribution – one of the commonest distributions followed by a continuous random variable

• For example, heights of persons, their blood pressure, time taken for banana plants to grow, weights of animals, are likely to follow a normal distribution

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Example: Weights of maize cobs

Graph shows histogram of 100 maize cobs.

Data which follows the bell shape of this histogram are said to follow a normal distribution.

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Frequency definition of probabilityIn a histogram, the bar areas correspond to frequencies.

For example, there are 3 maize cobs with weight < 100 gms, and 19 maize cobs with weight < 120 gms.

Hence, using the frequency approach to probability, we can say that

Prob(X<120) = 19/100 = 0.019

The areas under the curve can be regarded as representing probabilities since the curve and edges of histogram would coincide for n=.

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Probability Distribution Function

The mathematical expression describing the form of the normal distribution is

f(x) = exp(–(x–)2/22)/(22)

0.4

0.2

0.0

f(x)

x

Two parameters associated with the normal distribution, its mean and variance 2.

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Properties of the Normal Distribution

• Total area under the curve is 1

• characterised by mean & variance: N(,2 )

• symmetric about mean ()

• 95% of observations lie within ± 2 of mean

0.4

0.2

0.0

f(x)

x

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The Standard Normal Distribution

This is a distribution with =0 and =1, shown below in comparison with N(0,2), =3.

-6 -4 -2 0 +2 +4 +6

0.4 f(x)

0.2

x

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The Standard Normal DistributionThis is a distribution with =0 and =1.

Tables give probabilities associated with this distribution, i.e. for every value of a random variable Z which has a standard normal distribution, values of Pr(Z<z) are tabulated.

P

z 0

In graph on right, P=Pr(Z<z).

Symmetry means any area (prob) can be found.

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Calculating normal probabilitiesAny random variable, say X, having a normal distribution with mean and standard deviation , can be converted to a value (say z) from the standard normal distribution.

This is done using the formula

z = (X - ) /

The z values are called z-scores. The z scores can be used to compute probabilities associated with X.

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An exampleThe pulse rate (say X) of healthy individuals is expected to have a normal distribution with mean of 75 beats per minute and a standard deviation of 8. What is the chance that a randomly selected individual will have a pulse rate < 65?

We need to find Pr(X < 65)

i.e. Pr(X - 75 < 65 - 75)

= Pr[ (X – 75/8) < (-10/8) ]

= Pr(Z < -1.28) Pr(Z<-1.3) = 0.0968

(using tables of the standard normal distn)

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A practical applicationMalnutrition amongst children is generally measured by comparing their weight-for-age with that of a standard, age-specific reference distribution for well-nourished children.

A child’s weight-for-age is converted to a standardised normal score (an z-score), standardised to and of the reference distribution for the child’s gender and age.

Children whose z-score<-2 are regarded as being underweight.

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A Class Exercise

Similarly to the above, height-for-age is used as a measure of stunting again converted to a standardised z score (stunted if z-score<-2).

Suppose for example, the reference distribution for 32 months old girls has mean 91 cms with standard deviation 3.6 cms.

What is the probability that a randomly selected girl of 32 months will have height between 83.8 and 87.4 cms?

Graph below shows the area required. A class discussion will follow to get the answer.

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Depicting required probability as an area under the normal curve

2.986.940.914.878.83

Answer =

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Is a child stunted?

Suppose a 32 month old girl has height-for-age value = 82.1

Would you consider this child to be stunted?

Discuss this question with your neighbour and write down your answer below.

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Cumulative normal distribution

a x

In example above, the shaded area is 0.6,

the value of a from tables of the standard normal distribution is 0.726.

Cumulative distribution is given by the function

F(x) = P(X ≤a)

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ba

P(a<X<b) = F(b)-F(a) is the area under the cumulative normal curve between points a and b.

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Practical work follows …


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