continuous probability distributions( 连续型概率分布 ) the uniform distribution( 均匀...

Continuous Probability Continuous Probability

Distributions(Distributions(连续型概率分布连续型概率分布 ))

The Uniform Distribution(The Uniform Distribution( 均匀均匀分布分布 ))

The Normal Distribution(The Normal Distribution( 正态分正态分布布 ))

Chapter 5 Continuous Random Variables(连续型随机变量 )

Section 5.1 Continuous Probability DistributionsRecall: A continuous random variable may assume any numerical value in one or more intervals

Use a continuous probability distribution to assign probabilities to intervals of values

The curve f(x) is the continuous probability distribution of the continuous random variable x if the probability that x will be in a specified interval of numbers is the area under the curve f(x) corresponding to the interval

•Other names for a continuous probability distribution:•probability curve(概率曲线 ), or•probability density function(概率密度函数 )

Properties of ContinuousProbability Distributions

Properties of f(Properties of f(xx):): f( f(xx) is a continuous function such ) is a continuous function such thatthat

1.1. f(f(xx) ) 0 for all 0 for all xx2.2. The total area under the curve of f(The total area under the curve of f(xx) is equal to 1) is equal to 1

Essential point: An area under a continuous Essential point: An area under a continuous probability distribution is a probabilityprobability distribution is a probability

Area and Probability The blue-colored area under the probability curve The blue-colored area under the probability curve f(x)f(x)

from the value from the value xx = = aa to to xx = = bb is the probability that is the probability that xx could take any value in the range could take any value in the range aa to to bb Symbolized as Symbolized as PP((aa xx bb))

Or as Or as PP((aa < < xx < < bb), because each of the interval ), because each of the interval endpoints has a probability of endpoints has a probability of 00

Continuous Probability Distributions

For a continuous random variable its probability For a continuous random variable its probability density function gives a complete information about the density function gives a complete information about the variable.variable.

The most useful density functions are The most useful density functions are Normal Normal distribution (distribution ( 正态分布正态分布 ), uniform distribution (), uniform distribution ( 均匀均匀分布分布 ) ) andand Exponetial distribution ( Exponetial distribution ( 指数分布指数分布 ))

Distribution Shapes Symmetrical and rectangularSymmetrical and rectangular

The uniform distributionThe uniform distribution Section 5.2Section 5.2

Symmetrical and bell-shapedSymmetrical and bell-shaped The normal distributionThe normal distribution

Section 5.3Section 5.3 SkewedSkewed

Skewed either left or rightSkewed either left or right

For Example, the right-skewed exponential distribution

Section 5.2 The Uniform Distribution

dx c cd=x

otherwise0

for1

f

If If cc and and dd are numbers on the real line ( are numbers on the real line (cc < < dd), the ), the probability curve describing the probability curve describing the uniform distributionuniform distribution is is

The probability that The probability that xx is any value between the given is any value between the given values values aa and and bb ( (aa < < bb) is) is

cd

abbxaP

Note: The number ordering is Note: The number ordering is c c < < aa < < bb < < dd

f(x)

xc d

The Uniform Probability Curve

The Uniform Distribution

12

2

cd

dc

X

X

The mean X and standard deviation X of a uniform random variable x are

The uniform distribution is symmetrical Symmetrical about its center

is the median

The uniform distribution is rectangular For endpoints c and d (c < d and c ≠ d) the width of

the distribution is d – c and the height is1/(d – c)

The area under the entire uniform distribution is 1 Because width height = (d – c) [1/(d – c)] =

1 So P(c x d) = 1

Notes on the Uniform Distribution

XX

Uniform Waiting Time #1 The amount of time, The amount of time, xx, that a randomly selected hotel , that a randomly selected hotel

patron spends waiting for the elevator at dinnertimepatron spends waiting for the elevator at dinnertime.. Record suggests Record suggests xx is uniformly distributed between zero is uniformly distributed between zero

and four minutesand four minutes So So cc = 0 and = 0 and dd = 4, = 4,

x =x

otherwise0

40for4

1

04

1

f

404

ababbxaP

Example 5.1Example 5.1

Uniform Waiting Time #2 What is the probability that a randomly selected hotel patron What is the probability that a randomly selected hotel patron

waits at least 2.5 minutes for the elevator?waits at least 2.5 minutes for the elevator?

The probability is the area under the uniform distribution The probability is the area under the uniform distribution in the interval [2.5, 4] minutesin the interval [2.5, 4] minutes

The probability is the area of a rectangle with The probability is the area of a rectangle with height ¼ and base 4 – 2.5 = 1.5height ¼ and base 4 – 2.5 = 1.5

PP((xx ≥≥ 2.5) = 2.5) = PP(2.5 ≤ (2.5 ≤ xx ≤ 4) = ¼ ≤ 4) = ¼ 1.5 = 0.375 1.5 = 0.375

What is the probability that a randomly selected hotel patron waits less than one minutes for the elevator? P(x ≤ 1) = P(0 ≤ x ≤ 1) = ¼ 1 = 0.25

Uniform Waiting Time #3 Expect to wait the mean timeExpect to wait the mean time X

with a standard deviationwith a standard deviation X ofof

minutes 22

40

X

minutes 1.154712

04

X

Section5.3 The Normal DistributionThe The normal probability distributionnormal probability distribution is defined by is defined by the equationthe equation

for all values x on the real number line, where is the mean and is the standard deviation, = 3.14159,e = 2.71828 is the base of natural logarithms

2

1=)f(

2

2

1

eπσ

x

x

xx

ff((xx))

The normal curve is symmetrical and bell-shaped• The normal is symmetrical

about its meanμ• The mean is in the middle

under the curve• So μ is also the median

• The normal is tallest over its mean μ

• The area under the entire normal curve is 1• The area under either half

of the curve is 0.5

Properties of the Normal Distribution There is an infinite number of possible normal curves

The particular shape of any individual normal depends on its specific mean μ and standard deviation

The tails of the normal extend to infinity in both directions The tails get closer to the horizontal axis but never

touch it mean = median = mode

The left and right halves of the curve are mirror images of each other

The Position and Shape of the Normal Curve

(a) The mean positions the peak of the normal curve over the real axis

(b) The variance 2 measures the width or spread of the normal curve

The Position and Shape of the Normal Curve

Normal ProbabilitiesSuppose x is a normally distributed random variable with mean and standard deviation

The probability that x could take any value in the range between two given values a and b (a < b) is P(a ≤ x ≤ b)

PP((aa ≤ ≤ xx ≤≤ bb) is ) is the the area colored in bluearea colored in blue under the normal under the normal curve and between curve and between the valuesthe valuesx x == a a and and x x == b b

Three Important Areas under the Normal Curve1. P ( – ≤ x ≤ + ) = 0.6826

So 68.26% of all possible observed values of x are within (plus or minus) one standard deviation of

2. P ( – 2 ≤ x ≤ + 2) = 0.9544 So 95.44% of all possible observed values of x

are within (plus or minus) two standard deviations of

3. P ( – 3 ≤ x ≤ + 3) = 0.9973 So 99.73% of all possible observed values of x

are within (plus or minus) three standard deviations of

Three Important Areasunder the Normal Curve (Visually)

The Empirical Rule for Normal Populations

The Standard Normal Distribution

If x is normally distributed with mean and standard deviation , then the random variable z

is normally distributed with mean 0 and standard is normally distributed with mean 0 and standard deviation 1; this normal is called the deviation 1; this normal is called the standard standard normal distribution.normal distribution.

x

z

z measures the number of standard deviations that x is from the mean

• The algebraic sign on z indicates on which side of is x• z is positive if x > (x is to the right of on the number line)• z is negative if x < (x is to the left of on the number line)

The Standard Normal Distribution

24

The Standard Normal Table Page 860-861 The standard normal table is a table that lists the The standard normal table is a table that lists the

cumulative areas under the standard normal curve. cumulative areas under the standard normal curve.

This table is very important.This table is very important.

Always look at the Always look at the accompanying figureaccompanying figure for for guidance on how to use the tableguidance on how to use the table

25

The Standard Normal Table

The values of The values of zz (accurate to the nearest (accurate to the nearest tenth) in the table range from -3.99 to 3.99 tenth) in the table range from -3.99 to 3.99 in increments of 0.01in increments of 0.01

The areas under the normal curve to the left The areas under the normal curve to the left of any value of of any value of zz are given in the body of are given in the body of the tablethe table

26

The Standardized Normal Table The Standardized The Standardized Normal table in the Normal table in the textbook gives the textbook gives the probability that zprobability that zwill be less than or will be less than or equal to 2.00 equal to 2.00

P(Z <= 2.00) = 0.9772, P(Z<=1.96)=0.9750

.9772

The column The column shows the value shows the value of z to the first of z to the first decimal pointdecimal point

The row gives the value of z to the second decimal point

2.0

.

.1.9.

Z 0.00 0.01 0.02 …

0.0

0.1

0.06

0.9750

Z0 2.00

0.9772

Cumulative area under the standard normal curve

A Standard Normal Table

29

Find Find P P ((zz ≤ 2) ≤ 2) Find the area listed in the table corresponding to a Find the area listed in the table corresponding to a zz

value of 2.00value of 2.00 Starting from the top of the far left column, go down Starting from the top of the far left column, go down

to “2.0”to “2.0” Read across the row Read across the row zz = 2.0 until under the column = 2.0 until under the column

headed by “.00”headed by “.00” The area is in the cell that is the intersection of this The area is in the cell that is the intersection of this

row with this columnrow with this column As listed in the table, the area is 0.9772, so As listed in the table, the area is 0.9772, so

P P ((zz ≤ 2) = 0.9772 ≤ 2) = 0.9772

30

Calculating P (-2.53 ≤ z ≤ 2.53) First, find P(z ≤ 2.53)

Go to the table of areas under the standard normal curve

Go down left-most column for z = 2.5 Go across the row 2.5 to the column headed by .03 The cumulative area to the value of z = 2.53 is the

value contained in the cell that is the intersection of the 2.5 row and the .03 column

The table value for the area is 0.9943

31

From last slide, From last slide, P P ( ( zz ≤ 2.53)=0.9943 ≤ 2.53)=0.9943 By symmetry of the normal curve, By symmetry of the normal curve, P(z ≤ -2.53)=1-0.9943=0.0057P(z ≤ -2.53)=1-0.9943=0.0057 Then Then P P (-2.53 ≤ (-2.53 ≤ zz ≤ 2.53) ≤ 2.53) = = P P ( ( zz ≤ 2.53)- ≤ 2.53)-PP(z ≤ -2.53)(z ≤ -2.53) = 0.9943 -0.0057 = 0.9886= 0.9943 -0.0057 = 0.9886

Alternative:Alternative:P P (0 ≤ (0 ≤ zz ≤ 2.53)=0.9943-0.5=0.4943 ≤ 2.53)=0.9943-0.5=0.4943P P (-2.53 ≤ (-2.53 ≤ zz ≤ 2.53)=0.4943+0.4943=0.9886 ≤ 2.53)=0.4943+0.4943=0.9886

Calculating P (-2.53 ≤ z ≤ 2.53)

32

Calculating P(z 1)

An example of finding tail An example of finding tail

areasareas• the right-hand tail area the right-hand tail area

for for zz ≥≥ 1.00 1.00

(1)(1) (2) )1(1)1( ZPZP)1()1( ZPZP

Exercises: Calculating P (z -1), P(-1<z<2)

Finding Normal Probabilities

General procedure:

1. Formulate the problem in terms of 1. Formulate the problem in terms of xx values values

2. Calculate the corresponding 2. Calculate the corresponding zz values, and restate the values, and restate the problem in terms of these problem in terms of these zz values values

3. Find the required areas under the standard normal curve 3. Find the required areas under the standard normal curve by using the tableby using the table

Note: It is always useful to draw a picture showing the Note: It is always useful to draw a picture showing the required areas before using the normal tablerequired areas before using the normal table

The Car Mileage Case

Want the probability that the mileage of a randomly selected midsize car will be between 32 and 35 mpg

• Let x be the random variable of mileage of midsize cars, in mpg

• Want P (32 ≤ x ≤ 35 mpg)

Given:x is normally distributed

= 33 mpg

= 0.7 mpg


The Car Mileage Case #2

Procedure (from previous slide):Procedure (from previous slide):

1. Formulate the problem in terms of 1. Formulate the problem in terms of xx (as above) (as above)

2. Restate in terms of corresponding 2. Restate in terms of corresponding zz values (see values (see next slide)next slide)

3. Find the indicated area under the standard normal 3. Find the indicated area under the standard normal curve using the normal table (see the slide after that)curve using the normal table (see the slide after that)

For x = 32 mpg, the corresponding z value is

(so the mileage of 32 mpg is 1.43 standard deviations below (to the left of) the mean = 32 mpg)

For x = 35 mpg, the corresponding z value is

(so the mileage of 35 mpg is 2.86 standard deviations above (to the right of) the mean = 32 mpg)

Then P(32 ≤ x ≤ 35 mpg) = P(-1.43 ≤ z ≤ 2.86)


43107

1

70

3332.

..

xz

86207

2

70

3335.

..

xz


Want: the area under the normal curve between Want: the area under the normal curve between 3232 and 35 mpg and 35 mpg

Will find: the area under the standard normal Will find: the area under the standard normal curve between -1.43 and 2.86curve between -1.43 and 2.86

The Car Mileage Case #5 Break this into two pieces, around the mean

The area to the left of between -1.43 and 0 By symmetry, this is the same as the area between 0

and 1.43 From the standard normal table, this area is 0.4236

The area to the right ofThe area to the right of between 0 and 2.86 From the standard normal table, this area is 0.4979

The total area of both pieces is 0.4236 + 0.4979 = 0.9215 Then, P(-1.43 ≤ z ≤ 2.86) = 0.9215

Returning to x, P(32 ≤ x ≤ 35 mpg) = 0.9215

Exercise

The daily water usage per person in Providence, Rhode Island is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons. What percent of the population use between 18 and 26 gallons per day?

Finding z Points ona Standard Normal Curve

Stocking out of inventoryStocking out of inventory

A large discount store sells blank VHS tapes want to A large discount store sells blank VHS tapes want to know how many blank VHS tapes to stock (at the know how many blank VHS tapes to stock (at the beginning of the week) so that there is only a 5 percent beginning of the week) so that there is only a 5 percent chance of stocking out during the weekchance of stocking out during the week

• Let Let xx be the random variable of weekly demand be the random variable of weekly demand

• Let Let ss be the number of tapes so that there is only a 5% be the number of tapes so that there is only a 5% probability that weekly demand will exceed probability that weekly demand will exceed ss

• Want the value of Want the value of ss so that so that PP((xx ≥ ≥ ss) = 0.05) = 0.05

Given:Given:xx is normally distributed is normally distributed

= 100 tapes

= 10 tapes= 10 tapes


Example 5.3 #1

• Refer to Figure 5.20 in the textbook

• In panel (a), s is located on the horizontal axis under the right-tail of the normal curve having mean = 100 and standard deviation = 10

• The z value corresponding to s is

• In panel (b), the right-tail area is 0.05, so the area In panel (b), the right-tail area is 0.05, so the area under the standard normal curve to the right of the under the standard normal curve to the right of the mean mean zz = 0 is 0.5 – 0.05 = 0.45 = 0 is 0.5 – 0.05 = 0.45

10

100

ssz

Example 5.3 #2 Use the standard normal table to find the Use the standard normal table to find the zz value associated with a table value associated with a table

entry of 0.45entry of 0.45 But do not find 0.45; instead find values that bracket 0.45But do not find 0.45; instead find values that bracket 0.45 For a table entry of 0.4495, For a table entry of 0.4495, zz = 1.64 = 1.64 For a table entry of 0.4505, For a table entry of 0.4505, zz = 1.65 = 1.65 For an area of 0.45, use the For an area of 0.45, use the zz value midway between these value midway between these So So zz = 1.645 = 1.645

645.110

100

s

Solving for s gives s = 100 + (1.645 10) = 116.45Rounding up, 117 tapes should be stocked so that the probability of running out will not be more than 5 percent

Finding Z Points on a Normal Curve

Example: scores of IQ test Most of the IQ tests are standardized with a

mean score of 100 and a standard deviation of 15. If scores on an IQ test are normally distributed, what is the probability that a person who takes the test will score between 90 and 110?

Solution: We want to know the probability that the test score falls between 90 and 110. Since the mean is 100 and the sd is 15, we need to find the probability between z>-0.67 and z<0.67 from table. The answer is 2X0.2486 = 0.4972.

Example: scores of IQ test 2

With the same assumption as last example, what is the probability that a person with IQ score above 130?

Solution:We have z= (130-100)/15 = 2. From the table, P(z>2) = 0.5-0.4772 = 2.28%

Normal Approximation to the Binomial(二项分布的正态近似 )

• The figure below shows several binomial distributions

• Can see that as n gets larger and as p gets closer to 0.5, the graph of the binomial distribution tends to have the symmetrical, bell-shaped, form of the normal curve

Normal Approximation to the Binomial

• Generalize observation from last slide for large p

• Suppose x is a binomial random variable, where n is the number of trials, each having a probability of success p

• Then the probability of failure is 1 – p

• If n and p are such that np 5 and n(1 – p) 5, then x is approximately normal with

pnpnp 1 and

Normal Approximation to the Binomial

• Suppose there is a binomial variable Suppose there is a binomial variable xx with with nn = 50 = 50 and and pp = 0.5 = 0.5

• Want the probability of Want the probability of xx = 23 successes in the = 23 successes in the nn = = 50 trials50 trials

• Want Want PP((xx = 23) = 23)

• Approximating by using the normal curve withApproximating by using the normal curve with

5355.350.050.0501

2550.050

pnp

np

Normal Approximation to a Binomial #1



• With continuity correction, find the normal probability P(22.5 ≤ x ≤ 23.5)

• See figure below

For x = 22.5, the corresponding z value is

For x = 23.5, the corresponding z value is

Then P(22.5 ≤ x ≤ 23.5) = P(-0.71 ≤ z ≤ -0.42) = 0.2611 – 0.1628 = 0.0983

Therefore the estimate of the binomial probability P(x = 23) is 0.0983

71053553

25522.

.

.xz

42053553

25523.

.

.xz


continuous probability distributions( 连续型概率分布 ) the uniform distribution( 均匀...

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