1530 6 probability distributions

7/27/2019 1530 6 Probability Distributions

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Lesson 6: Probability Distributions

1 Distributions 2

2 Binomial Random Variables 6

3 Normal Random Variables 8

http://sites.google.com/site/mattjones1204/ 1


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1 Distributions

A random variable is a quantitative variable whose value depends on chance.

Well use them to model the behavior of populations.

A discrete random variable is a random variable whose possible values form

a finite or countably infinite set of numbers.

A continuous random variable is a random variable whose possible values

form a continuum or interval of numbers.

Use capital letters for random variables and lower case letters for their

possible values.

The distribution of a random variable is TWO things:

1. a list of the possible values it can take on;

2. a list of the probabilities of those possible values.



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Example 1 Let X = # of siblings for a person chosen at random.

# of Siblings x 0 1 2 3 4

P(X = x) 0.2 0.425 0.275 0.075 0.025

X is the random variable representing the number of siblings.

x represents a possible value for the # of siblings and can be 0, 1, 2, 3, or 4.

P(X = x) is the probability that the number of children equals x.

For any discrete random variable X,x

P(X = x) = 1

because the events {X = x} are disjoint and the probability of getting someoutcome in the sample space is 1.



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Mean of a Random Variable

The mean or expected value of X is

= E[X] =x

xP(X = x)

The mean is a weighted average of the possible values of x, where the

weights are the probabilities.

If the distribution of X describes a population, the sample means of large

samples tend to be close to the mean .

Example 2 Find the expected number of siblings for a person chosen at

random.


P(X = x) 0.2 0.425 0.275 0.075 0.025



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Standard Deviation of a Discrete Random Variable

If there are N equally likely possible outcomes for a random variable X (so

each occurs with probability 1/N), the standard deviation of X is given by

=

x(x )2

N=

x

(x )2 1/N

If the outcomes are not equally likely, this generalizes to

=

x

(x )2P(X = x)

Example 3 Find the standard deviation of the number of siblings.


P(X = x) 0.2 0.425 0.275 0.075 0.025



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2 Binomial Random Variables

X is a binomial random variable with parameters n and p if

P(X = x) =

n

x

px(1 p)nx

where x is any number in

{0, 1, 2, . . . , n

}. X denotes the number of successes

that occur in n independent trials where each trial has success probability p.

Example 4 There are 49 senators, each of whom shows for the senate

meeting with probability 0.95. What is the probability that no one shows?

What is the probability that everyone shows? What is the probability that

exactly 46 show? What is the probability that at least 47 show?



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Mean and Standard Deviation of a Binomial Random Variable

If X is a binomial random variable with parameters n and p, its mean is

given by

= np

and its standard deviation is given by

=np(1 p)Example 5 Of the 49 senators, each showing up with probability 0.95

independently of the others, what is the mean number who show? What is

the standard deviation of the number who show?



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3 Normal Random Variables

A random variable has the normal distribution with mean and standard

deviation if its density function is

f(x) =1

2e(x)2

22

where is the mean and is the standard deviation. The graph of thedensity function looks like this:



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More about Normal Random Variables

Normal random variables arise in the study of natural and industrial systems.

Many statistical tests are based on normal distributions.

If X has a normal distribution with mean and standard deviation , then

P(a X b) is the area of the region under the normal curve between thepoints a and b above the x-axis:

If X is distributed normal with mean and standard deviation , use the

calculator to find probabilities by doing

P(a

X

b) = normalcdf(a,b,,)



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Example 6The amount of coffee X your statistics professor drinks eachday has a normal distribution with mean 55 oz and standard deviation 6 oz.

Shade the appropriate regions beneath the density function and find the

following:

P(49

X

55) =

P(55 X 61) =P(X 55) =P(X > 65) =

P(X < 45 or X > 65) =

P(|X 55| 12) =



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Empirical Rule

The empirical rule states if X is distributed normal with mean and

standard deviation , then

P(|X | ) 68.27%P(|X | 2) 95.45%P(|X | 3) 99.73%

Another way to think about this: for a normally distributed population with

mean and standard deviation , approximately

68.27% of all observations are within one

standard deviation of the mean,

95.45% of all observations are within two

standard deviations of the mean,

99.73% of all observations are within three

standard deviations of the mean.



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Example 7 The heights of men in the United States are distributed

approximately normal with mean 69 inches and standard deviation 2.8

inches. Then the empirical rule tells us

Example 8 Ball bearings have masses distributed normal with mean 1.2

grams and standard deviation 0.03 grams. Then the empirical rule tells us



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Standard Normal Random Variables

A standard normal random variable Z has a normal distribution with

mean 0 and standard deviation 1.

Setting = 0 and = 1 in the density function on Slide 2 gives the density

function for a standard normal random variable:

f(z) =1

2ez

2/2

The graph of the density function looks like this:

If Z is distributed normal with mean 0 and standard deviation 1, use your

calculator to find probabilities as

P(a Z b) = normalcdf(a,b,0,1) or simply normalcdf(a,b)



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Relationship Between Standard Normals and Other Normals

If X is distributed normal with mean and standard deviation , thenX

has a normal distribution with mean 0 and standard deviation 1. That is, for

any numbers a and b,

P(a X b) = Pa

Z b

This observation allows us to use normal tables to obtain probabilities for

any normal random variable.



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Example 9 Suppose X normal(5, 2). Calculate the following with yourcalculator or using normal tables:

P(X > 6) =

P(2.5

X

6.5) =

P(|X 5| < 2.2) =

P(X < 4.5) =

P(X < 3.1 or X > 5.5) =



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z Notation

z is the number on the real line such that the probability of a standard

normal random variable taking on a greater value is :

P(Z > z) =

We find these values using normal tables in reverse, or using a calculator:

z = invNorm() = invNorm(1 - )



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Example 10 Find the following with (1) your calculator, (2) using a normal

table. Also mark locations of the z-scores beneath density functions:

z0.025 =

z0.05 =

z0.5 =

z0.975 =

z0.62 =

z0.12 =



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Example 11 Evidence suggests IQs are distributed approximatelynormal(100, 16).

1. Find the 90th percentile.

2. Find the 84th percentile.

3. Find the IQ such that only 5% of the population has higher IQs.


1530 6 probability distributions

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