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Lec Uniform Disttribution

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complete description on the topic of uniform distribution. Everything is fully explained with examples.
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'Uniform Distribution'
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Page 1: Lec Uniform Disttribution

'Uniform Distribution'

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Uniform Distribution

• In statistics, a type of probability distribution in which all outcomes are equally likely.

• There are two types of uniform distributions: discrete and continuous.

• The possible results of rolling a die provide an example of a discrete uniform distribution

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Examples: discrete uniform

• have a countable number of outcomes

• Coin p=1/2• Dice p=1/6• So probability= 1/x in case of x outcomes

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Continuous uniform random variable

• Infinite values but have equally probable intervals

• Eg• Height ranges• Baby weights• Waiting time at bus stop• Late from office• Blood pressures/ sugar levels

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Probability functions• A probability function maps the possible values of

x against their respective probabilities of occurrence, p(x)

• p(x) is a number from 0 to 1.0.• The area under a probability function is always 1.• Example: rolling dice sum=1• Defined at points only

x

p(x)

1/6

1 4 5 62 3

xall

1 P(x)

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Probability mass function (pmf)

x p(x)1 p(x=1)=1

/62 p(x=2)=1

/63 p(x=3)=1

/64 p(x=4)=1

/65 p(x=5)=1

/66 p(x=6)=1

/61.0

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Cumulative distribution function (CDF)

x

P(x)

1/6

1 4 5 62 3

1/31/22/35/61.0

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Cumulative distribution functionx P(x≤A)1 P(x≤1)=1/6

2 P(x≤2)=2/6

3 P(x≤3)=3/6

4 P(x≤4)=4/6

5 P(x≤5)=5/6

6 P(x≤6)=6/6

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Continuous case The probability function that accompanies

a continuous random variable is a continuous mathematical function that integrates to 1.

The probabilities associated with continuous functions are just areas under the curve (integrals!).

Probabilities are given for a range of values, rather than a particular value (e.g.,heights between 5.9-6 feet).

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examples

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• For continuous data we don’t have equally spaced discrete values so instead we use a curve or function that describes the probability density over the range of the distribution.

• The curve is chosen so that the area under the curve is equal to 1.

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probability density function

• The probability that a continuous random variable will be between limits a and b is given by an integral, or the area under a curve.

• The probability that the • continuous random variable, X, is between a and

b corresponds to the area under the curve representing the probability density function between the limits a and b.

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For uniform continuous variable• It would be a rectangle/ straight line • At height 1/n in case of n equal divisions/intervals• Area under curve= area of rectangle• Sum of total area =1•

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Density and cumulative distt

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Developing the prob function• Say a plane flying to karachi from islamabad takes

anywhere from 120 to 140 minutes• Say the distribution of flying time is uniform over the

interval 120 to 140• This basically means the probability of each 1 minute

interval between 120 and 140 has the same probability of occurring.

• In other words, the plane has the same chance of landing in the 120 to 121 interval as the 131 to 132 interval and so on.

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17

x = minutes of flight time.120 140

a b

1/20

0

f(x)

1/(b – a)

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18

On the previous screen you see a rectangle. It is on the interval 120 to 140 because this is the number of minutes a plane would take.

In general we say the interval is from a to b. Then the base of the rectangle is b – a = 140 –120 = 20 in our example.

Now, here is an important point based on rules of probability. Remember that the sum of all the probabilities of the outcomes must sum to 1. Since we want the area to represent probability the area of the rectangle must equal =1.

We do this by making the height 1/(b – a) = 1/(140 – 20) = 1/20.

Base time height is then (20)(1/20) = 1

Now we can answer more questions dealing with the probability of an event when we have a uniform distribution.

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19

•What is the probability the plan will land between 122 and 124 minutes?

•We just want to find the area inside the rectangle between 122 and 124. The base is 124 – 122 = 2 and the height is 1/20, so we have a probability 2(1/20) = .1.

•You might see the same problem asked in the following way:

•P(122 <= x <= 124), or what is the probability x is greater than or equal to 122, but less than or equal to 124?

•Special case

•P(x = 125)? What is the probability the plane lands in 125 minutes? Well 125 – 125 = 0 and 0(1/20) = 0. Since probabilities are areas for continuous variable, the probability for any one value is zero. We can only do probabilities over an interval.

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x = minutes of flight time.120 122 124 140

a b

1/20

0

f(x)

1/(b – a)

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Example 2• A student arrives at a bus stop and waits for the

bus. He knows that the bus comes every 15 minutes (which we will assume is exact), but he doesn’t know when the next bus will come. Let’s assume the bus is as likely to come in any one instant as in any other within the next 15 minutes. Let the time the student has to wait for the bus be x minutes.

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Mean , Variance

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The expected value for a uniform variable is

E(x) = (a + b)/2 and in our example equals

(120 + 140)/2 = 130.

The variance is

Var(x) = (b – a)2/12 and in our example equals

(140 – 120)2/12 = 400/12 = 33.33

prove

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