random variable and distribution

Chapter 4: Random Variables and Distribution

Statistics

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Where We’re Going

Develop the notion of a random variable

Numerical data and discrete random variables

Discrete random variables and their probabilities

4.1: Two Types of Random Variables

A random variable is a variable hat assumes numerical values associated with the random outcome of an experiment, where one (and only one) numerical value is assigned to each sample point.

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4.1: Two Types of Random Variables

A discrete random variable can assume a countable number of values. Number of steps to the top of the Eiffel Tower*

A continuous random variable can assume any value along a given interval of a number line. The time a tourist stays at the top

once s/he gets there

*Believe it or not, the answer ranges from 1,652 to 1,789. See Great Buildings

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http://www.greatbuildings.com/cgi-bin/glk?http://www.endex.com/gf/buildings/eiffel/eiffel.html

4.1: Two Types of Random Variables Discrete random variables

Number of sales Number of calls Shares of stock People in line Mistakes per page

Continuous random variables Length Depth Volume Time Weight

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4.2: Probability Distributions for Discrete Random Variables

The probability distribution of a discrete random variable is a graph, table or formula that specifies the probability associated with each possible outcome the random variable can assume. p(x) ≥ 0 for all values of x p(x) = 1

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4.2: Probability Distributions for Discrete Random Variables

Say a random variable x follows this pattern: p(x) = (.3)(.7)x-1

for x > 0. This table gives the

probabilities (rounded to two digits) for x between 1 and 10.

x P(x)

1 .30

2 .21

3 .15

4 .11

5 .07

6 .05

7 .04

8 .02

9 .02

10 .01

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4.3: Expected Values of Discrete Random Variables

The mean, or expected value, of a discrete random variable is

( ) ( ).E x xp x

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The variance of a discrete random variable x is

The standard deviation of a discrete random variable x is

2 2 2[( ) ] ( ) ( ).E x x p x

2 2 2[( ) ] ( ) ( ).E x x p x

9

)33(

)22(

)(

xP

xP

xP

Chebyshev’s Rule Empirical Rule

≥ 0 .68

≥ .75 .95

≥ .89 1.00

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In a roulette wheel in a U.S. casino, a $1 bet on “even” wins $1 if the ball falls on an even number (same for “odd,” or “red,” or “black”).

The odds of winning this bet are 47.37%

9986.0526.5263.1$4737.1$

5263.)1$(4737.)1$(

losePwinP

On average, bettors lose about a nickel for each dollar they put down on a bet like this.(These are the best bets for patrons.)

Binomial Distribution

Tree Diagram

4 Properties of Binomial Distribution

1. Fixed number of Trials (n)

Tree Diagram



2. Two outcomes in a trial, SUCCESS or FAILURE

Tree Diagram




3. Trials are independent

Tree Diagram




3. Trials are independent

4. Probability of success (p) remains constant

Tree Diagram

Tree Diagram

Throwing a die

Tree Diagram

X ~ B(n,p)

X – number of successes in a trial

X ~ B(3, 1/6)

Is there a formula for calculating Binomial Probabilities rather than draw a tree diagram?

There are five things you need to do to work a binomial story problem.

1. Define Success first. Success must be for a single

trial. Success = "Rolling a 6 on a single die"

2. Define the probability of success (p): p = 1/6

3. Find the probability of failure (q): q = 5/6

4. Define the number of trials: n = 3

5. Define the number of successes out of those trials (r)

The General Binomial Probability Formula

r – number of successes out of those trialsn – number of trialsp – probability of successq – probability of failure

Where: q = 1 - p

The General Binomial Probability Formula

In the old days, there was a probability of 0.8 of success in any attempt to make a telephone call. Calculate the

probability of having 7 successes in 10 attempts.

Mean and Variance

4.5: The Poisson Distribution

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The Poisson distribution is a discrete probability distribution for the counts of events that occur randomly in a given interval of time (or space).

The Poisson distribution can be used to calculate the probabilities of various numbers of "successes" based on the mean number of successes. In order to apply the Poisson distribution, the various events must be independent.


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Many experimental situations occur in which we observe the counts of eventswithin a set unit of time, area, volume, length etc. For example, The number of cases of a disease in different towns The number of mutations in set sized regions of a chromosome The number of dolphin pod sightings along a flight path through

a region The number of particles emitted by a radioactive source in a

given time The number of births per hour during a given day


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FORMULAThe formula for the Poisson probability mass function is

where•e is the base of natural logarithms (2.7183)•μ is the mean number of "successes"•x is the number of "successes" in question


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EXAMPLE The average number of homes sold by the Acme Realty company is 2

homes per day. What is the probability that exactly 3 homes will be sold tomorrow?

Solution: This is a Poisson experiment in which we know the following: μ = 2; x = 3; e = 2.71828; since e is a constant equal to approximately 2.71828. We plug these values into the Poisson formula as follows: P(x; μ) = (e-μ) (μx) / x!

P(3; 2) = (2.71828-2) (23) / 3! P(3; 2) = (0.13534) (8) / 6 P(3; 2) = 0.180

Thus, the probability of selling 3 homes tomorrow is 0.180 .


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EXAMPLE

Suppose you knew that the mean number of calls to a fire station on a weekday is 8. What is the probability that on a given weekday there would be 11 calls? μ = 8; x = 11; e = 2.71828; since e is a constant equal to approximately

2.71828.


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Changing the size of the intervalSuppose we know that births in a hospital occur randomly at an average rate of1.8 births per hour.What is the probability that we observe 5 births in a given 2 hour interval? Well, if births occur randomly at a rate of 1.8 births per 1 hour intervalThen births occur randomly at a rate of 3.6 births per 2 hour interval Let Y = No. of births in a 2 hour period

P(Y=5) = (e-3.6)(3.65) / (5!)

= 0.13768


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Sum of two Poisson variablesNow suppose we know that in hospital A births occur randomly at an average rateof 2.3 births per hour and in hospital B births occur randomly at an average rateof 3.1 births per hour.What is the probability that we observe 7 births in total from the two hospitalsin a given 1 hour period? So if we let X = No. of births in a given hour at hospital Aand Y = No. of births in a given hour at hospital B

P (X + Y = 7) = (e-5.4)(5.47) / (7!)

= 0.11999


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Cumulative Poisson Probability

A cumulative Poisson probability refers to the probability that the Poisson random variable is greater than some specified lower limit and less than some specified upper limit.


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Example 1een on a 1-day safari is 5. What is the probability that tourists will see fewer than four lions on the next 1-day safari?

Solution: This is a Poisson experiment inSuppose the average number of lions s which we know the following: μ = 5; x = 0, 1, 2, or 3; e = 2.71828; since e is a constant equal to approximately

2.71828.

To solve this problem, we need to find the probability that tourists will see 0, 1, 2, or 3 lions. Thus, we need to calculate the sum of four probabilities: P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5). To compute this sum, we use the Poisson formula:

P(x < 3, 5) = P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5)P(x < 3, 5) = [ (e-5)(50) / 0! ] + [ (e-5)(51) / 1! ] + [ (e-5)(52) / 2! ] + [ (e-5)(53) / 3! ] P(x < 3, 5) = [ (0.006738)(1) / 1 ] + [ (0.006738)(5) / 1 ] + [ (0.006738)(25) / 2 ] + [ (0.006738)(125) / 6 ] P(x < 3, 5) = [ 0.0067 ] + [ 0.03369 ] + [ 0.084224 ] + [ 0.140375 ] P(x < 3, 5) = 0.2650

Thus, the probability of seeing at no more than 3 lions is 0.2650.

4.6: The Hypergeometric Distribution

In the binomial situation, each trial was independent. Drawing cards from a deck and replacing

the drawn card each time If the card is not replaced, each trial

depends on the previous trial(s). The hypergeometric distribution can be

used in this case.

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Randomly draw n elements from a set of N elements, without replacement. Assume there are r successes and N-r failures in the N elements.

The hypergeometric random variable is the number of successes, x, drawn from the r available in the n selections.


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nN

xnrN

xr

xP )(

where N = the total number of elementsr = number of successes in the N elementsn = number of elements drawnX = the number of successes in the n elements


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nN

xnrN

xr

xP )(

)1()()(

22

NNnNnrNr

Nnr


44.22.2)2()2()2or2(

22.45

)1)(10(

210

22510

25

)2()2(

FPMPFMP

FPMP

Suppose a customer at a pet store wants to buy two hamsters for his daughter, but he wants two males or two females (i.e., he wants only two hamsters in a few months)

If there are ten hamsters, five male and five female, what is the probability of drawing two of the same sex? (With hamsters, it’s virtually a random selection.)

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Continuous Random Variable

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The normal distribution refers to a family of continuous probability distributions described by the normal equation.

• Normal Distribution

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• Normal Distributionz = (X - μ) / σ

where X is a normal random variable, μ is the mean of X, and σ is the standard deviation of X


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• Normal DistributionExample

An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed, what is the probability that an Acme light bulb will last at most 365 days?

Solution: The value of the normal random variable is 365 days.• The mean is equal to 300 days.• The standard deviation is equal to 50 days.

z = (X - μ) / σ = (365-300)/50z= 1.3

The answer is: P( X < 365) = 0.90. Hence, there is a 90% chance that a light bulb will burn out within 365 days.


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• Standard Normal DistributionThe standard normal distribution is a special case of the normal distribution. It is the distribution that occurs when a normal random variable has a mean of zero and a standard deviation of one.Standard Score (aka, z Score)The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation:

z = (X - μ) / σwhere X is a normal random variable, μ is the mean of X, and σ is the standard deviation of X.

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• Standard Normal DistributionThe standard normal distribution is a special case of the normal distribution. It is the distribution that occurs when a normal random variable has a mean of zero and a standard deviation of one.Standard Score (aka, z Score)The normal random variable of a standard normal distribution is called a standard score or a z-score. Every normal random variable X can be transformed into a z score via the following equation:

z = (X - μ) / σwhere X is a normal random variable, μ is the mean of X, and σ is the standard deviation of X.

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• Standard Normal DistributionExampleMolly earned a score of 940 on a national achievement test. The mean test score was 850 with a standard deviation of 100. What proportion of students had a higher score than Molly? (Assume that test scores are normally distributed.)

(A) 0.10 (B) 0.18 (C) 0.50 (D) 0.82 (E) 0.90


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• Standard Normal Distribution• First, we transform Molly's test score into a z-score, using the z-

score transformation equation.

z = (X - μ) / σ = (940 - 850) / 100 = 0.90

• Then, using the standard normal distribution table, we find the cumulative probability associated with the z-score. In this case, we find P(Z < 0.90) = 0.8159.

• Therefore, the P(Z > 0.90) = 1 - P(Z < 0.90) = 1 - 0.8159 = 0.1841.

Thus, we estimate that 18.41 percent of the students tested had a higher score than Molly.

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