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Probability DistributionsProbability DistributionsChapter 4.1Chapter 4.1
Objectives• Distinguish between discrete random
variables and continuous random variables• Learn to construct a discrete probability
distribution and its graph• Learn to find the mean, variance, and
standard deviation of a discrete probability distribution
• Learn to find the expected value of a discrete probability distribution
Discrete & Continuous• Discrete data is countable
– Number of students– Number of sales calls
• Continuous data has an infinite number of possible outcomes – Amount of time– Temperature
Random Variables• A random variable x represents a numerical
value associated with each outcome of a probability experiment.
• The random variable is discrete if it has a finite or countable number of possible outcomes that can be listed.
• The variable is continuous if it has an uncountable number of possible outcomes, represented by an interval on the number line.
Discrete or Continuous?• The number of stocks in the Dow
Jones Industrial Average that have a share price increase on a given day.
Discrete or Continuous?• The length of time it takes to
complete a test.
Discrete or Continuous?• The number of home runs hit during a
baseball game
Discrete Probability Distributions
• A discrete probability distribution lists each possible value the random variable can assume, together with its probability.
• The Rules:– The probability of each value is between
0 and 1.– The sum of all the probabilities is 1.
Example – Roll 2 dice:Total Probability
2 0.02783 0.05564 0.08335 0.11116 0.13897 0.16678 0.13899 0.111110 0.083311 0.055612 0.0278
1.0000
Constructing a Discrete Probability Distribution
• Let x be a discrete random variable with the possible outcomes x1, x2, . . . xn
2. Make a frequency distribution for the possible outcomes.
3. Find the sum of the frequencies.4. Find the probability of each possible outcome by
dividing its frequency by the sum of the frequencies.
5. Check that each probability is between 0 and 1 and that the sum is 1.
Example• A sociologist surveyed 200 households
in a small town and asked how many dependents they had.
• The results were:Dependent Children Number of Households
0 14
1 40
2 76
3 44
4 26
Example• Find the sum of the frequencies:• 200• Find the probability of each possible
outcome:Dependent Children
Number of Households
Percentage
0 14 7%
1 40 20%
2 76 38%
3 44 22%
4 26 13%
Example• Check that each probability is
between 0 and 1 and that the sum is 1Dependent Children
Number of Households
Percentage
0 14 7%
1 40 20%
2 76 38%
3 44 22%
4 26 13%
Mean• How would you find the average?
Dependent Children
Number of Households
Percentage
0 14 7%
1 40 20%
2 76 38%
3 44 22%
4 26 13%
Mean• How would you find the average?
Dependent Children
Number of Households
Percentage
x(P(x)
0 14 7% 0*.07 0
1 40 20% 1*.20 .20
2 76 38% 2*.38 .76
3 44 22% 3*.22 .66
4 26 13% 4*.13 .52
Total 200 100% 2.14
Mean• The mean of a discrete random variable is
given by:• μ = ∑xP(x)• Each value of x is multiplied by its
corresponding probability and the products are added.
How would you find the variance & standard deviation?Dependent Children
Number of Households
Percentage
0 14 7%
1 40 20%
2 76 38%
3 44 22%
4 26 13%
How would you find the variance & standard deviation?
• standard deviation is √10.098 = 3.177
Dependent Children
Number of Households
Percentage x - μ (x – μ)2
0 14 7% -2.14 4.5796
1 40 20% -1.14 1.2996
2 76 38% -.14 .0196
3 44 22% .86 .7396
4 26 13% 1.86 3.4596
Total 200 100% 10.098
Mean, Variance, & Standard Deviation
• The variance of a discrete random variable is given by:
• σ2 = ∑(x-μ) 2 P(x)• A shortcut of this formula is:• σ2 = ∑x2 P(x) - μ2
• The standard deviation is:• σ = √σ2
Expected Value• The expected value of a discrete
random variable is equal to the mean of the random variable.
Expected Value Example• At a raffle, 1500 tickets are sold at $2 each for
4 prizes of $500, $250, $150, and $75. You buy 1 ticket. What is the expected value of your gain?
• What are your possible gains?• $500-$2 = $498• $248, $148, $73, and $-2• What is the possibility of each of these?• 1/1500 for the first 4, 1496/1500 for the -2
Expected Value Example• E(x) = $498 * 1/1500 + $248*1/1500 +
$148*1/1500 + $73* 1/1500 + -2*1496/1500
• = -$1.35• Because the expected value is negative,
you can expect to lose an average of $1.35 for each ticket you purchase.
Homework• P. 179 Do 1-10 together• P. 179 12-28, 34 evens