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Introduction to Continuous Probability Distributions

In this lesson, we will discuss two of these: the uniform distribution and the normal

distribution. The normal distribution is particularly important because many of the

methods used in statistics are based on this distribution

Uniform Distribution

We learned that a continuous random variable has a set of possible values that is an

interval on the number line. It is not possible to assin a probability to each point in

the interval. Instead, the probability distribution of a continuous random variable ! is

specified by a mathematical function f"#$ called the probability density function or 

 %ust density function. The raph of a density function is a smooth curve. & probability

density function "pdf$ must satisfy two conditions: "'$ f"#$ ( ) for all real values of #

and "*$ the total area under the density curve is e+ual to '. The raphs of three density

functions are shown in iure ''.'.

The probability that ! lies in any particular interval is shown by the area under the

density curve and above the interval. The followin three events are fre+uently

encountered: "'$ ! - a, the event that the random variable ! assumes a value less than

a "*$ a - ! - b, the event that the random variable ! assumes a value between a and

 b and "/$ ! 0 b, the event that the random variable ! is reater than b. We say that

we are interested in the lower tail probability for "'$ and the upper tail probability

when usin "/$. The areas associated with each of these are shown in iure ''.*.

 1otice that the probability that a - ! - b may be computed usin tail probabilities:

P"a - ! - b$ 2 P"! - b$ 3 P"! - a$.

If the random variable ! is e+ually li4ely to assume any value in an interval "a, b$,

then ! is a uniform random variable. The pdf is flat and is above the #5a#is between a

and b, and it is ) outside of the interval. The heiht of the curve must be such that the

area under the density and above the #5a#is is '. 6ecause this reion is a rectanle, the

area is the heiht times the width of the interval, which is b 3 a. Thus, the heiht must

 be that is, the pdf of a uniform random variable has the form

2 ), otherwise.& raph of the pdf is shown in iure ''./.

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7#ample

& roup of volcanoloists "people who study volcanoes$ has been monitorin a

volcano8s seismicity, or the fre+uency and distribution of underlyin earth+ua4es.

6ased on these readins, they believe that the volcano will erupt within the ne#t *9

hours, but the eruption is e+ually li4ely to occur any time within that period. What isthe probability that it will erupt within the ne#t eiht hours

;olution

Define ! 2 the time until the eruption of the volcano. ! has positive probability over 

the interval "),*9$ because the volcano will erupt durin that time interval. 6ecause

the lenth of the interval is *9 3 ) 2 *9, the heiht of the density curve must be for 

the area under the density and above the #5a#is to be one. That is, the pdf is

2 ), otherwise.

The probability that the volcano will erupt within the ne#t eiht hours is e+ual to the

area under the curve and above the interval "),<$ as shown in iure ''.9. This area is

.

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In the previous e#ample, notice that the area is the same whether we have P") - ! -

<$ or P") = ! - <$ or P") - ! = <$ or P") = ! = <$. Unli4e discrete random variables,

whether the ine+uality is strict or not, the probability is the same for the continuous

random variables. This also correctly implies that, for continuous random variables,

the probability that the random variable e+uals a specific value is ).

Normal Distribution

Normal Probability Distributions

 1ormal probability distributions are continuous probability distributions that are bell

shaped and symmetric. They are also 4nown as >aussian distributions or bell5shaped

curves.

The normal distribution is perhaps the most widely used probability distribution,

larely because it provides a reasonable appro#imation to the distribution of many

random variables. It also plays a central role in many of the statistical methods that

will be discussed in later lessons. 1ormal probability distributions are continuous

 probability distributions that are bell shaped and symmetric as displayed in iure

''.?. The distribution is also called the >aussian distribution or the bell5shaped curve.

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The normal distribution has two parameters: the mean @ and the standard deviation A.The notation ! B 1"@ ,A$ means that ! is normally distributed with a mean of @ and

a standard deviation of A. The distribution is symmetric about the mean. The mean,

median, and mode are all e+ual. The mean is often referred to as the location

 parameter because it determines where the distribution is centered. The standard

deviation determines the spread of the distribution. The effect of the mean and

standard deviation on the normal distribution is displayed in iure ''..

or any normal distribution, about <E of the observations are within one standard

deviation of the mean. &bout F?E and FF.GE of the observations are, respectively,

within two and three standard deviations of the mean.

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It is important to remember that, althouh the location and spread may chane, the

area under the curve and above the #5a#is is always '. Unfortunately, the probabilities

associated with intervals cannot be computed easily as with the uniform distribution.

To overcome this difficulty, we rely on a table of areas for a reference of normal

distribution called the standard normal distribution. The standard normal distribution

is the normal distribution with @ 2 ) and A 2 '. It is customary to use the letter H torepresent a standard normal random variable.

We will first learn to compute probabilities for a standard normal random variable and

then learn how to find them for any random variable. We will also want to be able to

determine e#treme values of H, such as the value that only ?E of the population

e#ceeds or the value that 'E of the population is less than. To find either probabilities

or e#treme values, we need a table of standard normal curve areas, or we need a

calculator or computer that can be used to find these values. ere, we will restrict

ourselves to the use of tables. The standard normal table used here in Table ''.'

tabulates the probability of observin a value less than or e+ual to H "see iure ''.G$.

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>raphs are e#tremely useful tools to help us understand what values we are searchin

for. We will do this for each problem we wor4.

 

7#amples of Continuous Probability Distributions

6elow are nine e#amples of continuous probability distributions problems and

solutions.

7#ample '

ind P"H - './*$.

;olution '

Usin the standard normal table, we find the row with './ in the H column and move

alon that row to the ).)* column to find ).F). Thus, P"H - './*$ 2 ).F). iure

''.< shows the raphic imae of this.

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7#ample *

ind P"H 0 './*$.

;olution *

rom the table, we find P"H - './*$ as we did in the previous e#ample. Usin some of 

the ideas of probability we learned earlier, we have P"H 0 )./*$ 2 ' 3P"H = './*$ 2 ' 3 

).F) 2 ).)F/9. ;ee iure ''.F.

7#ample /

ind P"H - 3).?$.

;olution /

There are no neative H5values in the table, so we cannot loo4 this up directly. Instead,

we use the symmetry of the normal distribution to find the probability "see iure

''.')$. That is,

P"H - 3).?$ 2P"H 0 ).?$

  2 ' 3 P"H - ).?$

  2 ' 3 ).F'?

  2 )./)<?

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7#ample 9

ind P"3'.9? - H - ).G$.

;olution 9

iure ''.'' shows the solution.

irst, we notice that P"3'.9? - H - ).G$ 2 P"H - ).G$ 3 P"H - 3'.9?$. 1ow P"H -).G$ can be found directly from the table to be ).GG9. Usin the symmetry of the

normal distribution aain, P"H - 3'.9?$ 2 P"H 0 '.9?$ 2 ' 3 P"H = '.9?$ 2 ' 3 ).F*? 2

).)G/?. inally, P"3'.9? - H - ).G$ 2 P"H - ).G$ 3 P"H - 3'.9?$ 2 ).GG9 3 ).)G/? 2

).G)*F.

7#ample ?

ind the value HJ such that P"H - HJ$ 2 ).G?.

;olution ?

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This is different from the other problems we have considered. Instead of findin a

 probability, we are loo4in for a H5value. owever, the same table will allow us to

solve the problem. The difference is that we will loo4 in the table for a probability and

then find the H5value associated with the probability. Koo4in in the body of the table,

we find the values ).G9< and ).G?'G, which are the closest to the ).G? of interest. 6y

loo4in at the correspondin row and column headins, we find that P"H - ).G$ 2).G9< and P"H - 3).<$ 2 ).G?'G. 6ecause ).G9< is closer to ).G? than ).G?'G, we

ta4e HJ 2 ).G. "1ote: We could interpolate to find a more precise value of HJ, but we

will not o throuh this process here.$ ;ee iure ''.'*.

7#ample

ind the value HJ such that P"H 0 HJ$ 2 ).)?.

;olution

We need to have the probabilities in the form P"H - HJ$ to use the table. owever, P"H

0 HJ$ 2 ' 3 P"H = HJ$. We can rewrite this as P"H = HJ$ 2 ' 3 P"H 0 HJ$ 2 ' 3 ).)? 2

).F?. That is, if ?E of the population values are reater than HJ, then F?E of the

 population values must be less than or e+ual to HJ. Thus, we loo4 for ).F? in the body

of the table and find ).F9F? and ).F?)? correspondin to H 2 '.9 and H 2 '.?,

respectively. 6ecause ).F? is e#actly halfway between ).F9F? and ).F?)?, we have HJ2 '.9?. "This is the only time we don8t %ust round to the closest value.$ ;ee iure

''.'/.

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7#ample G

ind the value HJ such that P"H - HJ$ 2 ).)'.

;olution G

6ecause the standard normal is symmetric about its mean ), we 4now P"H - )$ 2 ).?,

we 4now that HJ must be less than ). &lso, because we have only positive values of H

in the table, we cannot loo4 for ).)' directly in the table. owever, aain because of 

symmetry, we 4now that, if P"H - HJ$ 2 ).)', then P"H 0 3HJ$ 2 ).)'. To use the table,

we must find P"H= 3HJ$ 2' 3 P"H 0 3HJ$ 2 ' 3 ).)' 2 ).FF. Koo4in in the body of the

table, we find ).F<F< and ).FF)', correspondin to H 2 *./* and H 2 *.//,

respectively, to be the closest to ).FF. 6ecause ).FF)' is the closer of the two to *.//,

we find HJ 2 3*.//. ;ee iure ''.'9.

ew normal random variables actually have a standard normal distribution. owever,

any normal random variable can be transformed to a standard normal, and anystandard normal random variable can be transformed to a normal random variable

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with any mean @ and standard deviation A. ;pecifically, if ! B 1"@,A$$,

. urther, if H B 1"),'$, then ! 2 @ L AH B 1"@,A$. Usin these

relationships, we can find probabilities and e#treme values for any normal random

variable usin the H5table. When doin this, it is important to do all calculations

carefully.

7#ample <

Ket ! B 1"'),?$. ind P"! - '?$.

;olution <

P"! - '?$

2 P"H - '$

2 ).<9'/.

 1otice that inside the parentheses, we had to transform both the ! and the '? to avoid

chanin the ine+uality. When wor4in with !, we used symbols, and we used

numbers when wor4in with '?. owever, we used the numbers that were associated

with each symbol. Mnce we have the e#pression in terms of H, then the problem is

e+uivalent to the earlier ones we wor4ed. ;ee iure ''.'?.

7#ample F

Ket !, 1"?,*$. ind !J such that P"! 0 !J$ 2 ).)?.

;olution F

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irst, we find HJ such that P"H 0 HJ$ 2 ).)?. rom our earlier wor4, we 4now that HJ 2

'.9?. Then !J 2 @ ' AHJ 2 L *"'.9?$ 2 <.*F.

Continuous Probabiity Distributions In ;hort

We have discussed two continuous distributions: the uniform and the normal. When

every value in an interval is e+ually li4ely to occur, then we have a uniform

distribution. The normal distribution is the most commonly used continuous

distribution. Probabilities associated with a normal random variable must be found by

usin tables, calculators, or computers. When usin tables, it is possible to use only

one table. 6y tradition, the probabilities of a standard normal distribution are

tabulated. Probabilities for other normal random variables are found by transformin

the problem to one on the standard normal.

Notes:

Continuous Probability Distributions

a random variable that can assume all possible random values "ie city

temperature$

Probability Density Function: a function that describes how li4ely

this random variable will occur at a iven point.

Height formula: heiht 2 'N"b5a$ where b is the top rane, and a is the

 bottom rane iven.

 The Normal Distribution

used to solve continuous probabilities

symmetry about the mean

total area under the curve is '

standard deviation is the distance from the mean to the point of 

inflection

&ny normal distribution can be described as by the mean and the

variance: so we often write 1"mean, variance$ to describe a distribution

The distribution chart shows area under the raph from the ! value tothe left end

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Z-Scores can be calculated usin 1ormal distributions

O 2 # 3 mean N standard deviation

;ometimes, you will have to subtract the mean to e+ualiHe. This

ma4es it so the mean is on the center.

 

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