Probability Theory

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Lecture 9
Uniform Distribution.
Normal
(Gaussian Distribution)
Distributions.

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Contents
Uniform distribution – the general view
The Normal Distribution

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Uniform Distribution: general view
The random variable X is said to have uniform distribution on , if its probability density function is
constant for ,
and is equal to 0 for

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To find the constant C:
We use the property of continuous random variable
(this property sometimes is called normalizing condition)

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So we have in the general terms:
For example, if we are interesting in probability density function for uniform distribution on

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To find the expectation of the uniformly distributed random variable:
We remember the definition of expectation for continuous random variable
Substituting
we obtain

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To find the expectation of the uniformly distributed random variable:
For the uniform distribution

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To find the variance of the uniformly distributed random variable:
We remember the definition of variance for continuous random variable

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To find the variance of the uniformly distributed random variable:
Substituting the value of and
probability density function
for uniform distribution
We have

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To find the variance of the uniformly distributed random variable:

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So the variance of the uniformly distributed random variable:
It could be calculated as

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To find the cumulative distribution function for uniform distribution
We use the definition of cumulative distribution function
So we have for 3 different intervals
1. For

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To find the cumulative distribution function for uniform distribution
2. For
3. For

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Finally we obtain the cumulative distribution function for uniform distribution
We have

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The Normal Distribution
We introduce now a continuous distribution that plays a central role in a very large body of statistical analysis.
For example, suppose that a big group of students takes a test. A large proportion of their scores are likely to be concentrated about the mean, and the numbers of scores in ranges of a fixed width are likely to "tail off” away from the mean.

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The Normal Distribution
If the average score on the test is 60, we would expect to find, for instance, more students with scores in the range 55-65 than in the range 85-95
These considerations suggest a probability density function that peaks at the mean and tails off at its extremities. One distribution with these properties is the normal distribution, whose probability density function is shown below.
As can be see this density function is bell-shaped.

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Probability Density Function
of the Normal Distribution
The shape of the probability density function is a symmetric bell-shaped curve centered on the mean , that peaks at the mean and tails off at its extremities

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Probability Density Function
of the Normal Distribution
If the random variable X has probability density function
where and are any number such that and
and where and are physical constants,
= 2.71828 ... and = 3.14159 ...,
then X is said to follow a normal distribution.

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Comments
It can be seen from the definition that there is not a single normal distribution but a whole family of distributions, resulting from different specifications of and .
These two parameters have very convenient interpretations

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Some Properties of the Normal Distribution
Suppose that the random variable X follows a normal distribution with parameters and . The following properties hold:
(i) The mean of the random variable is ;
that is
(ii) The variance of the random variable is ;
that is

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Some Properties of the Normal Distribution
(iii) The shape of the probability density function is a symmetric bell-shaped curve centered on the mean .

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Comments & Notation
It follows from these properties that given the mean and variance of a normal random variable, an individual member of the family of normal distributions is specified.
This allows use of a convenient notation.
If the random variable X follows a normal distribution with mean and variance , we write

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Comments
Now,
the mean of any distribution provides a measure of central location,
while the variance gives a measure of spread or dispersion about the mean.
Thus, the values taken by the parameters and have different effects on the probability density function of a normal random variable.

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Comments
We shows probability density functions for two normal distributions with a common variance but different means.
It can be seen that increasing the mean while holding the variance fixed shifts the density function but does not alter its shape.

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Comments
The two density functions are of normal random variables with a common mean but different variances.
Each is symmetric about the common mean, but that with the larger variance is more disperse

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Cumulative Distribution Function of the Normal Distribution
An extremely important practical question concerns the determination of probabilities from a specified normal distribution.
As a first step in determining probabilities, we introduce the cumulative distribution function

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Cumulative Distribution Function of the Normal Distribution
Suppose that X is a normal random variable with mean and variance
that is,
Then the cumulative distribution function
This is the area under the probability density function to the left of
As for any proper density function, the total area under the curve is 1; that is

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Cumulative Distribution Function of the Normal Distribution
The shaded area is the probability that X does not exceed for a normal random variable

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Cumulative Distribution Function of the Normal Distribution
There is no simple algebraic expression for calculating the cumulative distribution function of a normally distributed random variable.
That is to say that the integral
does not have a simple algebraic form

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Cumulative Distribution Function
of the Normal Distribution
The general shape of the cumulative distribution function is shown below

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Range Probabilities
for Normal Random Variables
We have already seen
that for any continuous random variable, probabilities can be expressed in terms of the cumulative distribution function

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Range Probabilities
for Normal Random Variables
Let X be a normal random variable with cumulative distribution function ,
and let a and b be two possible values of X,
with a < b.
Then

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Range Probabilities
for Normal Random Variables
Any required probability can be obtained from the cumulative distribution function.
However, a crucial difficulty remains because there does not exist a convenient formula for determining the cumulative distribution function.
In principle, for any specific normal distribution, probabilities could be obtained by numerical methods using an electronic computer.

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Range Probabilities
for Normal Random Variables
However, it would be enormously tedious if we had to carry out such an operation for every normal distribution we encountered.
Fortunately, probabilities for any normal distribution can always be expressed in terms of probabilities for a single normal distribution for which the cumulative distribution function has been evaluated and tabulated.

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The Standard Normal Distribution
We now introduce the particular distribution that is used for this purpose
Let Z be a normal random variable with mean 0 and variance 1; that is
Then Z is said to follow the standard normal distribution.

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The Standard Normal Distribution
If the cumulative distribution function of this random variable is denoted ,
and a* and b* are two numbers with a* < b*,
then
The cumulative distribution function of the standard normal distribution is tabulated.

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The table of Normal Distribution
This table gives values of
for nonnegative values of z.
For example
Thus, the probability is 0.8944 that the standard normal random variable takes a value less than 1.25

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Values of the cumulative distribution function for negative values of z can be inferred from the symmetry of the probability density function
Let be any positive number, and suppose that we require
The density function of the standard normal random variable is symmetric about its mean, 0, the area under the curve to the left of is the same as the area under the curve to the right of ; that is

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Probability density function for the standard normal random variable Z;
the shaded areas, which are equal, show the probability that Z does not exceed and the probability that Z is greater than

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Moreover, since the total area under the curve is 1:
Hence, it follows that
For example

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Example
If Z is a standard normal random variable, find
The required probability is
Then, using Table we obtain

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How can probabilities for any normal random variable be expressed in terms of those for the standard normal random variable?
Let the random variable X be normally distributed with mean and variance .
We know that subtracting the mean and dividing by the standard deviation yields a random variable Z that has mean 0 and variance 1.
It can also be shown that if X is normally distributed, so is Z.
Hence, Z has a standard normal distribution

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How can probabilities for any normal random variable be expressed in terms of those for the standard normal random variable?
Suppose, then, that we require the probability that X lies between the numbers a and b.
This is equivalent to lying between and ,
so that the probability of interest is

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Finding Range Probabilities
for Normal Random Variables
Let X be a normal random variable with mean and variance
Then the random variable has a standard normal distribution; that is,
It follows that if a and b are any numbers
with a < b, then
where Z is the standard normal random variable and denotes its cumulative distribution function

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Probability density function for normal random variable X with mean 3 and standard deviation 2; shaded area is probability that X lies between 4 and 6
Figure shows the probability density function of a normal random variable X with mean and standard deviation

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Probability density function for normal random variable X with mean 3 and standard deviation 2; shaded area is probability that X lies between 4 and 6
The shaded area shows the probability that X lies between 4 and 6.
This is the same as the probability that a standard normal random variable lies between
and
that is, between 0.5 and 1.5.
This probability is the shaded area under the standard normal curve

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Probability density function for standard normal random variable Z; shaded area is probability that Z lies between 0.5 and 1.5 and is equal to shaded area in the previous slide

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Example
If X ~ N(15, 16), find the probability that X is larger than 18.
This probability is
From the Table we have
so

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Example
If X is normally distributed with mean 3 and standard deviation 2, find P(4 < X < 6).
We have

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Example
A company produces lightbulbs whose lifetimes follow a normal distribution with mean 1,200 hours and standard deviation 250 hours.
If a lightbulb is chosen randomly from the company's output, what is the probability that its lifetime will be between 900 and 1,300 hours?
Let X represent lifetime in hours

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Example.
Then
Hence, the probability is approximately 0.54 that a lightbulb will last between 900 and 1,300 hours

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Example
A very large group of students obtains test scores that are normally distributed with mean 60 and standard deviation 15.
What proportion of the students obtained scores between 85 and 95?
Let X denote the test score.

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Example
Then we have
That is, 3.76% of the students obtained scores in the range 85 to 95.

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Example
For the test scores of the previous Example, find the cutoff point for the top 10% of all students.
We have previously found probabilities corresponding to cutoff points. Here we need the cutoff point corresponding to a particular probability.
The position is illustrated in Figure (next slide), which shows the probability density function of a normally distributed random variable with mean 60 and standard deviation 15.

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Example
The probability is 0.10 that the random variable X exceeds the number b;
Here X is normally distributed, with mean 60 and standard deviation 15

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Example
Let the number b denote the minimum score needed to be in the highest 10%. Then, the probability is 0.10 that the score of a randomly chosen student exceeds the number b.
This probability is shown as the shaded area in Figure.
If X denotes the test scores, then the probability that X exceeds b is 0.1,

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Example
So
Hence, it follows that
Now, from Table, if then z = 1.28.

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Example
Therefore, we have
So
The conclusion is that 10% of the students obtain scores higher than 79.2

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Comments
In Examples, if the scores awarded on the test were integers, the distribution of scores would be inherently discrete.
Nevertheless, the normal distribution can typically provide an adequate approximation in such circumstances.
We will see later that the normal distribution can often be employed as an approximation to discrete distributions.
As a preliminary, we introduce in the next lecture a result that provides strong justification for the emphasis given to the normal distribution.