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155S5.1­2_3 Review and Preview; Random Variables
MAT 155
Dr. Claude Moore
Cape Fear Community College
Chapter 5 Probability Distributions
5­1 Review and Preview
5­2 Random Variables
5­3 Binomial Probability Distributions
5­4 Mean, Variance and Standard Deviation for the Binomial Distribution
5­5 Poisson Probability Distributions
Preview
In order to fully understand probability distributions, we must first understand the concept of a random variable, and be able to distinguish between discrete and continuous random variables. September 20, 2010
Review and Preview
This chapter combines the methods of descriptive statistics presented in Chapters 2 and 3 and those of probability presented in Chapter 4 to describe and analyze probability distributions.
Probability Distributions describe what will probably happen instead of what actually did happen, and they are often given in the format of a graph, table, or formula.
Combining Descriptive Methods and Probabilities
In this chapter we will construct probability distributions by presenting possible outcomes along with the relative frequencies we expect.
In this chapter we focus on discrete probability distributions. In particular, we discuss binomial and Poisson probability distributions. 1
155S5.1­2_3 Review and Preview; Random Variables
Key Concept
This section introduces the important concept of a probability distribution, which gives the probability for each value of a variable that is determined by chance.
Give consideration to distinguishing between outcomes that are likely to occur by chance and outcomes that are “unusual” in the sense they are not likely to occur by chance.
Random Variable
Probability Distribution
• Random variable ­ a variable (typically represented by x) that has a single numerical value, determined by chance, for each outcome of a procedure.
• Probability distribution ­ a description that gives the probability for each value of the random variable; often expressed in the format of a graph, table, or formula.
September 20, 2010
Key Concept
1. The concept of random variables and how they relate to probability distributions.
2. Distinguish between discrete random variables and continuous random variables.
3. Develop formulas for finding the mean, variance, and standard deviation for a probability distribution.
4. Determine whether outcomes are likely to occur by chance or they are unusual (in the sense that they are not likely to occur by chance).
Discrete and Continuous Random Variables
• Discrete random variable ­ either a finite number of values or countable number of values, where “countable” refers to the fact that there might be infinitely many values, but they result from a counting process.
• Continuous random variable ­ infinitely many values, and those values can be associated with measurements on a continuous scale without gaps or interruptions
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155S5.1­2_3 Review and Preview; Random Variables
Graphs The probability histogram is very similar to a relative frequency histogram, but the vertical scale shows probabilities.
September 20, 2010
Requirements for Probability Distribution
0 ≤ P(x) ≤ 1 for every individual value of x.
Mean, Variance and Standard Deviation of a Probability Distribution
µ = Σ [x • P(x)] Mean
σ 2 = Σ [(x – µ)2 • P(x)] Variance
σ 2 = Σ [x2 • P(x)] – µ 2 Variance (shortcut)
σ = Σ [x 2 • P(x)] – µ 2 Roundoff Rule for µ, s, and s2
Round results by carrying one more decimal place than the number of decimal places used for the random variable x.
If the values of x are integers, round µ, s, and s2 to one decimal place.
Standard Deviation
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155S5.1­2_3 Review and Preview; Random Variables
Identifying Unusual Results
Range Rule of Thumb
According to the range rule of thumb, most values should lie within 2 standard deviations of the mean.
We can therefore identify “unusual” values by determining if they lie outside these limits:
Maximum usual value = μ + 2σ
Minimum usual value = μ – 2σ
September 20, 2010
Rare Event Rule for Inferential Statistics
If, under a given assumption (such as the assumption that a coin is fair), the probability of a particular observed event (such as 992 heads in 1000 tosses of a coin) is extremely small, we conclude that the assumption is probably not correct.
Identifying Unusual Results Probabilities
• Unusually high: x successes among n trials is an unusually high number of successes if P(x or more) ≤ 0.05.
• Unusually low: x successes among n trials is an unusually low number of successes if P(x or fewer) ≤ 0.05.
Expected Value
The expected value of a discrete random variable is denoted by E, and it represents the mean value of the outcomes. It is obtained by finding the value of Σ [x • P(x)].
E = Σ [x • P(x)]
Recap
In this section we have discussed:
• Combining methods of descriptive statistics with probability.
• Random variables and probability distributions.
• Probability histograms.
• Requirements for a probability distribution.
• Mean, variance, and standard deviation of a probability distribution.
• Identifying unusual results.
• Expected value.
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155S5.1­2_3 Review and Preview; Random Variables
In Exercises 7–12, determine whether or not a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied.
221/7. Genetic Disorder Three males with an X­linked genetic disorder have one child each. The random variable x is the number of children among the three who inherit the X­linked genetic disorder.
Pea Hybridization Experiment. In Exercises 13–16, refer to the accompanying table, which describes results from eight offspring peas. The random variable x represents the number of offspring peas with green pods.
222/14. Range Rule of Thumb for Unusual Events Use the range rule of thumb to identify a range of values containing the usual number of peas with green pods. Based on the result, is it unusual to get only one pea with a green pod? Explain.
September 20, 2010
In Exercises 7–12, determine whether or not a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied.
222/11. American Televisions In the accompanying table, the random variable x represents the number of televisions in a house­
hold in the United States (based on data from Frank N. Magid Associates).
Pea Hybridization Experiment. In Exercises 13–16, refer to the accompanying table, which describes results from eight offspring peas. The random variable x represents the number of offspring peas with green pods.
222/16. Using Probabilities for Unusual Events a. Find the probability of getting exactly 3 peas with green pods. b. Find the probability of getting 3 or fewer peas with green pods. c. Which probability is relevant for determining whether 3 is an unusually low number of peas with green pods: the result from part (a) or part (b)? d. Is 3 an unusually low number of peas with green pods? Why or why not?
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155S5.1­2_3 Review and Preview; Random Variables
223/21. Finding Mean and Standard Deviation Let the random variable x represent the number of girls in a family of three children. Construct a table describing the probability distribution, then find the mean and standard deviation. (Hint: List the different possible outcomes.) Is it unusual for a family of three children to consist of three girls?
September 20, 2010
223/22. Finding Mean and Standard Deviation Let the random variable x represent the number of girls in a family of four children. Construct a table describing the probability distribution, then find the mean and standard deviation. (Hint: List the different possible out­comes.) Is it unusual for a family of four children to consist of four girls?
Solution to 223/21.
There are eight equally likely possible outcomes: GGG GGB GBG BGG GBB BGB BBG BBB.
mean ­ 2 standard deviations = 2 ­ 2(1.0) = 0
The mean is 1.5 girls with a standard deviation of 0.9 girls ﴾rounded to one decimal place﴿ for a family of 3 children.
mean + 2 standard deviations = 2 + 2(1.0) = 4
No, because P﴾x = 3﴿ = 0.125 > 0.05, it is not unusual for a family of 3 children to have all girls.
It is not unusual for a family of 4 children to have 4 girls because P( x = 4) = 0.0625 > 0.05.
223/26. Expected Value in New Jersey’s Pick 4 Game In New Jersey’s Pick 4 223/29. Expected Value for Life Insurance There is a 0.9986 probability that a lottery game, you pay to select a sequence of four digits, such as 1332. If you select the same sequence of four digits that are drawn, you win and collect $ 2788. a. How many different selections are possible? (10)(10)(10)(10) = 104 = 10,000 possible selections
randomly selected 30­year­old male lives through the year (based on data from the U. S. Department of Health and Human Services). A Fidelity life insurance company charges $161 for insuring that the male will live through the year. If the male does not survive the year, the policy pays out $ 100,000 as a death benefit. a. From the perspective of the 30­year­old male, what are the values corresponding to the two events of surviving the year and not surviving? b. If a 30­year­old male purchases the policy, what is his expected value? c. Can the insurance company expect to make a profit from many such policies? Why?
b. What is the probability of winning? P(winning) = 1 / 10,000 = 0.0001
c. If you win, what is your net profit? Net profit = $2788.00 ­ $0.50 = $2787.50
Solution to 223/29.
d. Find the expected value. E = (­0.50)(0.9999) + (2787.50)(0.0001) = ­$0.221 or loss 22.1 cents
a. From the 30­year­old male’s perspective, the two possible outcome values are
­$161, if he lives; $100,000 ­ $161 = $99,839, if he dies.
e. If you bet in Illinois’ Pick 4 game, the expected value is . Which bet is better: A bet in the Illinois Pick 4 game or a bet in New Jersey’s Pick 4 game? Explain.
Since ­22.1 > ­25, New Jersey's Pick 4 lottery game is the better bet.
b. The following table describes the situation.
Expected value is E = ­21.0000, rounded to ­$21.0
c. Yes; the insurance company can expect to make an average of $21.0 per policy.
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