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Concepts of Probability Distribution Prepare by: Rina Isanan

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Concepts of Probability Distribution

P r e p a r e b y :

R i n a I s a n a n

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Random Variable

= is a function or rule that assigns a number to each outcome of an experiment;

=symbol is X; =outcomes are numbers.

Two types of random variables

1. Discrete Random variable = one that can take on a countable number of values.

2. Continuous Random variable = one that values are uncountable; x is continuous

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Probability Distribution

= a table, formula or graph that describes the values of a random variable and the probability associated with these values.

X= random variable name; x = random variable value

P ( X = x )P ( x )

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Number of Color Televisions Number of Households (thousands)

0 1,218

1 32,379

2 37,961

3 19,387

4 7,714

5 2,842

Total 101,501

DISCRETE PROBALITY DISTRIBUTION

Probabilities of the values of a discrete random variable may be derived by means of probability tools.

Example 7.1 Develop the probability distribution of the random variable defined as the number of color televisions per household.

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X P(x)0 1,218/101,501 = .0121 32,379/101,50 =.3192 37,961/101,50 =.3743 19,387/101,501=.1914 7,714/101,501 =.0765 2,842/101,50 =1.028

Total 1.000

S o l u t i o n

The probability of each value of x, the number of color televisions per household, is computed as the relative frequency. We divide the frequency for each value of X by the total number of households, producing the following probability distribution.As you can see, the requirements are satisfied. Each probability lies between 0 and 1 and the total is 1.

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The importance of probability distribution derives from their use as representative of population. Probability distribution

represents populations.

Population mean = weighted average of all of it’s value; expected value of x and is represented by E (x).

E(x)== all x xP(x)Population Variance

V(x)=2=all x (x-u)2 P(x)Standard deviation

=2

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Example 7.3

Find the mean, variance and standard deviation for the population of the number of color television per household.

SolutionThe mean of X is;

E(X) = = all x xP(x) = 0P (0) + 1P(1)+ 2P(2)+ 3P(3)+ 4P(4)+ 5P(5) = 0(.012)+ 1(.319)+ 2(.374)+ 3(.191)+ 4(.076)+

5(.028) =2.084

The variance of X is;V(x)=2=all x (x-u)2 P(x) = (0-2.084)2(.012) +(1-2.084)2(.319) +(2-2.084)2(.374) +(3-2.084)2(.191) +(4-2.084)2(.076) +(5-2.084)2(.028)

= 1.107

The standard deviation is=2= 1.107= 1.052

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One of the Most Important Discrete Distribution

POISSON DISTRIBUTION

Poisson Random Variable*Is the number of occurrences of events which we’ll continue to call successes.*The number of successes in an interval of time or specific region of space.*No. of successes that occur in a period of time or an interval of space in a poisson experiment.

Examples:1.The number of cars arriving at a service station in 1 hour ( the interval of time is 1 hr.)2.The number of flaws in a bolt of cloth (the specific region is a bolt of cloth)3.The number of accidents in 1 day on a particular stretch or highway. (the interval is defined by both time, 1 day and space, the particular stretch of highway)

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Poisson Experiment

1.The number of successes that occur in any interval is independent of the number of successes that occur in any other interval.

2.The probability of a success in an interval is the same for all equal-size intervals.

3.The probability of a success in an interval is proportional to the size of the interval.

4.The probability of more than one success in an interval approaches 0 as the interval becomes smaller.

As a general rule, poisson random variable is the number of occurrences of a relatively rare event that occurs randomly and independently.

Example: The no. of hits on an active website is not poisson random variable because hits are not rare.

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POISSON PROBABILITY DISTRIBUTION

P(x)= e- x = for x= 0,1,2….

X!

= mean number of successes in the interval or region and e is the base of the natural

logarithm (approx. 2.71828)

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Example. 7.12 An instructor has observed that the number of typographical errors in new editions of textbooks varies considerably from book to book. After some analysis, he concludes that the number of errors is Poisson distributed with a mean of 1.5 per 100 pages. The instructor randomly selects 100 pages of a new book. What is the probability that there are no typographical errors?

Determine the probability that a Poisson random variable with a mean of 1.5 is equal to 0. Thus, we substitute x=0 and =1.5 into the formula for the

Poisson distribution.P(x)= e-1.5 1.50 = (2.71828)-1.5(1)

0! 1=.2231

The probability that in the 100 pages selected there are no typographical errors is .2231.

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CONTINUOUS RANDOM VARIABLE

One that can assume an uncountable number of values.Because this type of random variable is so different from a discrete

variable we need to treat it completely different.We cannot list the possible values because there are an infinite

number of them.Because there is an infinite number of values the probability of

each individual value is virtually 0.

*probability of a range of values only.

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NORMAL DISTRIBUTION

= considered to be the most important of all probability distributions because of its crucial role in statistical inference.

Figure 8.7

= notice that the curve is symmetric about its mean and the random variable ranges/ bell-shaped

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Figure 8.8

Normal distribution is described by two parameters. The mean () and standard deviation ().-increasing shifts the curve to the right-decreasing shifts the curve to the left* same variance but different means.

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Figure 8.9

Effect of , larger values of widen the curve and smaller ones narrow it. Same mean but different standard deviation.

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Calculating Normal Probabilities

We need to reduce the no. of tables needed to one by standardizing the random variable. We standardize a random variable by subtracting its mean and dividing by its standard deviation.

When the variable is normal, the transformed variable is called a Standard Normal Random variable and denoted by .

= x-

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Example : Supposed that the amount of time to assemble a computer is

normally distributed with a mean of 50 minutes and a standard deviation of 10 mins. We would like to know the probability that a computer is assembled in a time between 45 and 60 minutes.

To find the probability;P (45<x<60)

Figure 8.10Describes a normal curve with mean 50 and standard deviation 10, and the area we want to find.

45 50 60x

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We need to standardize X and if we perform any operations on x, we must perform the same on 45 and 60.

P(45<x<60)=P(45-50 < x- < 60-50) 10 10=P (-.5 <z<1)

= notice that variable x was transformed into z.= notice that variable 45 was transformed into -.5.= notice that variable 60 was transformed into 1.

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Figure 8.11

Conclude P(45<x<60) is identical to P (-.5 <z<1)The area has not changed. The value of z specify the location of the corresponding value of x.= a value of z = -.5 corresponds to a value of x that is ½ a standard deviation below the mean and that is x=45, is 5 mins. Below the mean and the standard deviation is 10.= a value of z = 1 corresponds to a value of x that is 1 standard deviation above the mean.= normal curve is symmetric about its mean and the total area under the curve is 1.P(z>o)= P (z<0)=0.05

-.5 0 1z

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Table

8.1Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40151.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.41771.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.43191.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.44411.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.45451.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.46331.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.47061.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.47672.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.48172.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.48572.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.48902.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.49162.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.49362.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.49522.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.49642.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.49742.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.49812.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.49863.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990

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Example

The probability P( 0<Z<2.00) is found by finding 2.0 in the left margin and under the heading .00, finding .4778.

The probability that we seek is actually the sum of two probabilities:

P (-.5 <Z<1)= P (0 <Z<1) + P (-.5 <Z<0)The first probability on the right is easily determined from the table.

P (0 <Z<1) = .3413P (-.5 <Z<0) =.1915P (-.5 <Z<1)= .3413 + .1915 = .5328

Therefore the probability that a randomly selected computer takes between 45 and 60 min. to assemble is .5328.

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Example 8.12

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Finding values of Z

= there is a family of problems that requires us to determine the value of z given a probability.

ZA = represent the value of z such that the area to its right under the standard normal curve is A. The value of a standard normal random variable.P( Z > ZA)= A

Figure 8.14

To find ZA for any value of A requires us to use the standard normal table backwards. To use the table backward, we need to specify a probability and then determine the z value associated with it.

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Let’s find Z.025

Determine the area between 0 and Z.025 which is -.5-.025 = .4750. We search through the probability part of the table looking for .4750. we see that the Z value is 1.96. Thus, Z .025 =1.96 with means that P (Z>1.96) = .025

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Finding Z.05

Find the values of a standard normal random variable such that the probability that the random variable is greater than it is .05SOLUTION We wish to determine Z.05. If .05 is the area in the tail, then the area between 0 and Z.45. To find Z.05, we search the table looking for the probability .4500. We don’t find this probability, but we find two values that are equally close : .4495 and .4505. The Z-values associated with these probabilities are 1.64 and 1.65, respectively. The average is taken as Z.05. Thus, Z.05 = 1.645.

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Example 8.4

Finding -Z.05

Find the values of a standard normal random variable such that the probability that the random variable is less than it is .05SOLUTIONBecause the standard normal curve is symmetric about 0, we wish to find –Z.05. In the example above, we found Z.05= 1.645. Thus, -Z.05= -1.645

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EXPONENTIAL DISTRIBUTION

Another important continuous distribution.Normal distribution is a two-parameter distribution, it is completely satisfied once the values of and are known.

In exponential distribution, is a one parameter distribution. The distribution is completely specified once the value of the parameter is known.

The exponential density function is easier to work with than the normal.The probability statements,If x is an exponential random variable

P (x>x)= e-x

P (x<x)= 1-e-x

P (x1 < x< x2)= P(x<x2) - P(x<x

1)= e-x1 - e-x2

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Example 8.6

The lifetime of an alkaline battery (measured in hours) is exponentially distributed with

= .05

•What are the mean and standard deviation of the battery’s lifetime?

•Find the probability that a battery will last between 10 and 15 hours.

• What is the probability that a battery will last for more than 20 hours?

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SOLUTION

•The mean and standard deviation are equal to I/. Thus,

•Let X denote the lifetime of a battery. The required probability is

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• P(X>20) =e-.05(20)

=e -1

= .3679

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Thank you!