Download AP Stat 6.1

CHAPTER 6
Random Variables
6.1: Discrete and Continuous
Random Variables
HW due Friday: p.359 (2, 3, 6, 8, 9, 11, 14,
15, 19, 21, 23, 26, 30)
Tutorial next week Tuesday for QM review
The Practice of Statistics, 5th Edition
Starnes, Tabor, Yates, Moore
Bedford Freeman Worth Publishers
Random Variables and Probability Distributions
A probability model describes the possible outcomes of a chance
process and the likelihood that those outcomes will occur.
Consider tossing a fair coin 3 times.
Define X = the number of heads obtained
X = 0: TTT
X = 1: HTT THT TTH
X = 2: HHT HTH THH
X = 3: HHH
Value
0
1
2
3
Probability
1/8
3/8
3/8
1/8
A random variable takes numerical values that describe the
outcomes of some chance process. The probability distribution of
a random variable gives its possible values and their probabilities.
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Discrete Random Variables
There are two main types of random variables: discrete and
continuous. If we can find a way to list all possible outcomes for a
random variable and assign probabilities to each one, we have a
discrete random variable.
Discrete Random Variables And Their Probability Distributions
A discrete random variable X takes a fixed set of possible values with
gaps between. The probability distribution of a discrete random variable
X lists the values xi and their probabilities pi:
Value:
x1 x2 x3
Probability: p1 p2 p3 …
…
The probabilities pi must satisfy two requirements:
1.Every probability pi is a number between 0 and 1.
2.The sum of the probabilities is 1.
To find the probability of any event, add the probabilities pi of the
particular values xi that make up the event.
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Mean (Expected Value) of a Discrete Random Variable
When analyzing discrete random variables, we follow the same
strategy we used with quantitative data – describe the shape, center,
and spread, and identify any outliers.
The mean of any discrete random variable is an average of the
possible outcomes, with each outcome weighted by its probability.
Suppose that X is a discrete random variable whose probability
distribution is
Value:
x1 x2 x3 …
Probability:
p1 p2 p3 …
To find the mean (expected value) of X, multiply each possible
value by its probability, then add all the products:
 x  E ( X )  x1 p1  x2 p2  x3 p3  ...
  xi pi
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Mean (Expected Value) of a Discrete Random Variable
A baby’s Apgar score is the sum of the ratings on each of five scales, which
gives a whole-number value from 0 to 10.
Let X = Apgar score of a randomly selected newborn
Compute the mean of the random variable X. Interpret this value in context.
We see that 1 in every 1000 babies would have an Apgar score of 0; 6 in
every 1000 babies would have an Apgar score of 1; and so on. So the
mean (expected value) of X is:
The mean Apgar score of a randomly selected newborn is 8.128. This is
the average Apgar score of many, many randomly chosen babies.
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Standard Deviation of a Discrete Random Variable
Since we use the mean as the measure of center for a discrete random
variable, we use the standard deviation as our measure of spread. The
definition of the variance of a random variable is similar to the
definition of the variance for a set of quantitative data.
Suppose that X is a discrete random variable whose probability
distribution is
Value:
x1 x2 x3 …
Probability:
p1 p2 p3 …
and that µX is the mean of X. The variance of X is
Var(X) = s X2 = (x1 - m X ) 2 p1 + (x 2 - m X ) 2 p2 + (x 3 - m X ) 2 p3 + ...
= å (x i - m X ) 2 pi
To get the standard deviation of a random variable, take the square
root of the variance.
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Standard Deviation of a Discrete Random Variable
A baby’s Apgar score is the sum of the ratings on each of five scales, which
gives a whole-number value from 0 to 10.
Let X = Apgar score of a randomly selected newborn
Compute and interpret the standard deviation of the random variable X.
The formula for the variance of X is
The standard deviation of X is σX = √(2.066) = 1.437.
A randomly selected baby’s Apgar score will typically differ from the mean
(8.128) by about 1.4 units.
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Analyzing Random Variables On Graphing Calculator
• Enter the values of the random variable in L1/list 1 and the corresponding
probabilities in L2/list 2.
1. TI-83/84: Press Stat and select enter for lists.
2. TI-89: Go to data/matrix editor and create a name for list.
• To graph the histogram:
1. Set up Stat Plot:
TI-83/84: Press 2nd Stat Plot: change to histogram, xlist: L1, freq: L2
TI-89: Press F2 (Plots): change to histogram, xlist: L1, freq: L2
2. Adjust window: leave some extra space for min and max
3. Press Graph. Press Trace button & scroll left/right to move along histogr
• Calculate the mean and standard deviation: use one variable statistics with
the values in L1 and L2.
1. TI-83/84: Press Stat, arrow over to Calc, select 1-Var Stats type L1, L2,
Press Enter.
2. TI-89: Go back to your data in data/matrix editor, Press F4 (Calc), select
1-Var Stats, inputs: c1, freq: c2, Press Enter.
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Continuous Random Variables
Discrete random variables commonly arise from situations that involve
counting something. Situations that involve measuring something often
result in a continuous random variable.
A continuous random variable X takes on all values in an interval
of numbers. The probability distribution of X is described by a density
curve. The probability of any event is the area under the density
curve and above the values of X that make up the event.
The probability model of a discrete random variable X assigns a
probability between 0 and 1 to each possible value of X.
A continuous random variable Y has infinitely many possible values. All
continuous probability models assign probability 0 to every individual
outcome. Only intervals of values have positive probability.
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Example: Normal probability distributions
The heights of young women closely follow the Normal distribution with
mean µ = 64 inches and standard deviation σ = 2.7 inches.
Now choose one young woman at random. Call her height Y.
If we repeat the random choice very many times, the distribution of values
of Y is the same Normal distribution that describes the heights of all young
women.
Problem: What’s the
probability that the chosen
woman is between 68 and 70
inches tall?
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Example: Normal probability distributions
Step 1: State the distribution and the values of interest.
The height Y of a randomly chosen young woman has the N(64, 2.7) distribution.
We want to find P(68 ≤ Y ≤ 70).
Step 2: Perform calculations—show
your work! Remember: P(68 ≤ Y ≤ 70)
= area under the curve.
1.) Find z-score
2.) Use table A and look up area
under curve with corresponding z-score.
The standardized scores (z-scores) for
the two boundary values are
z
x

  mean
  std. dev.
(Area under normal distribution
curve is from ch 2)
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Discrete and Continuous Random Variables
Section Summary
In this section, we learned how to…
 COMPUTE probabilities using the probability distribution of a discrete
random variable.
 CALCULATE and INTERPRET the mean (expected value) of a
discrete random variable.
 CALCULATE and INTERPRET the standard deviation of a discrete
random variable.
 COMPUTE probabilities using the probability distribution of certain
continuous random variables.
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