Probability and statistics

1. Mathematics Generalist EC-6 Standard IV Probability and statistics

2. Probability of a Simple Event • The probability of a simple event is a ratio of the number of favorable outcomes for the event to the total number of possible outcomes of the event. • The probability of an event a can be expressed as:

3. Finding the Outcomes of Simple Events • For Simple Events – count the outcomes • Examples: • One Die- 6 outcomes • One coin- 2 outcomes • One deck of cards- 52 outcomes

4. Finding Outcomes of More Than One Event • The total outcomes of each event are found by using a tree diagram or by using the fundamental counting principle. • Example: At football games, a student concession stand sells sandwiches on either wheat or rye bread. The sandwiches come with salami, turkey, or ham, and either chips, a brownie, or fruit. Use a tree diagram to determine the number of possible sandwich combinations.

5. Tree Diagram

6. Answer • Using the fundamental counting principle bread x meat x side 2 x 3 x 3 = 18 outcomes

7. Fundamental Counting Principle • Sometimes the number of outcomes changes after each event depending upon the situation • Example: There are 8 students in the Algebra Club at Central High School. The students want to stand in a line for their yearbook picture. How many different ways could the 8 students stand for their picture?

8. Counting Principle cont. • The number of ways to arrange the students can be found by multiplying the number of choices for each position. • There are eight people from which to choose for the first position. • After choosing a person for the first position, there are seven people left from which to choose for the second position.

9. Counting Principle • There are now six choices for the third position. • This process continues until there is only one choice left for the last position. Let n represent the number of arrangements. Answer: There are 40,320 different ways they could stand.

10. Probability of Compound Events • A compound event consists of two or more simple events. • Examples: • rolling a die and tossing a penny • spinning a spinner and drawing a card • tossing two dice • tossing two coins

11. Compound Events • When the outcome of one event doesnotaffect the outcome of a second event, theseare calledindependentevents. • The probability of two independent events isfound bymultiplyingthe probability of the first event by the probability of the second event.

12. Compound Event Notations

13. Compound Probability • P(roll even #, spin odd) =

14. Probability of Compound Events P(jack, tails)

15. Compound Events • Events that cannot occur at the same time are called mutually exclusive. • Suppose you want to find the probability of rolling a 2 or a 4 on a die. P(2 or 4) • Since a die cannot show both a 2 and a 4 at the same time, the events are mutually exclusive.

16. Compound Mutually Exclusive

17. Mutually Exclusive • Example: Alfred is going to the Lakeshore Animal Shelter to pick a new pet. Today, the shelter has 8 dogs, 7 cats, and 5 rabbits available for adoption. If Alfred randomly picks an animal to adopt, what is the probability that the animal would be a cat or a dog?

18. Add the probability of getting a dog plus the probability of getting a cat. 8/20 + 7/20 Mutually Exclusive cont. • Since a pet cannot be both a dog and a cat, the events are mutually exclusive. The probability of randomly picking a cat or a dog is ¾ or 75%

19. Mutually Exclusive cont. • Example Two: The French Club has 16 seniors, 12 juniors, 15 sophomores, and 21 freshmen as members. What is the probability that a member chosen at random is a junior or a senior?

20. Compound ProbabilityMutually Inclusive The P(king or diamond) (there is a king of diamonds that can only be counted once) This is called mutually inclusive.

21. Normal Distribution • The normal distribution refers to a family of continuous probability distributions described by the normal equation. • The normal distribution is defined by the following equation: Normal equation. The value of the random variable Y is: Y = [ 1/σ * sqrt(2π) ] * e-(x - μ)2/2σ2 where X is a normal random variable, μ is the mean, σ is the standard deviation, π is approximately 3.14159, and e is approximately 2.71828.

22. The Normal Curve • The graph of the normal distribution depends on two factors - the mean and the standard deviation. The mean of the distribution determines the location of the center of the graph, and the standard deviation determines the height and width of the graph. When the standard deviation is large, the curve is short and wide; when the standard deviation is small, the curve is tall and narrow. All normal distributions look like a symmetric, bell-shaped curve, as shown below. • The curve on the left is shorter and wider than the curve on the right, because the curve on the left has a bigger standard deviation.

23. Probability and the Normal Curve • The normal distribution is a continuous probability distribution. This has several implications for probability. • The total area under the normal curve is equal to 1. • The probability that a normal random variable X equals any particular value is 0. • The probability that X is greater than a equals the area under the normal curve bounded by a and plus infinity (as indicated by the non-shaded area in the figure below). • The probability that X is less than a equals the area under the normal curve bounded by a and minus infinity (as indicated by the shaded area in the figure below).

24. Probability and the Normal Curve cont. • Additionally, every normal curve (regardless of its mean or standard deviation) conforms to the following "rule". • About 68% of the area under the curve falls within 1 standard deviation of the mean. • About 95% of the area under the curve falls within 2 standard deviations of the mean. • About 99.7% of the area under the curve falls within 3 standard deviations of the mean. • Collectively, these points are known as the empirical rule or the 68-95-99.7 rule. Clearly, given a normal distribution, most outcomes will be within 3 standard deviations of the mean.

25. Standard Normal Distribution • The 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. • 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) / σ where X is a normal random variable, μ is the mean of X, and σ is the standard deviation of X.

26. Standard Normal Distribution Table • A standard normal distribution table shows a cumulative probability associated with a particular z-score. Table rows show the whole number and tenths place of the z-score. Table columns show the hundredths place. The cumulative probability (often from minus infinity to the z-score) appears in the cell of the table. • For example, a section of the standard normal table is reproduced below. To find the cumulative probability of a z-score equal to -1.31, cross-reference the row of the table containing -1.3 with the column containing 0.01. The table shows that the probability that a standard normal random variable will be less than -1.31 is 0.0951; that is, P(Z < -1.31) = 0.0951.

27. Standard Normal Distribution Table cont. • Of course, you may not be interested in the probability that a standard normal random variable falls between minus infinity and a given value. You may want to know the probability that it lies between a given value and plus infinity. Or you may want to know the probability that a standard normal random variable lies between two given values. These probabilities are easy to compute from a normal distribution table. Here's how. • Find P(Z > a). The probability that a standard normal random variable (z) is greater than a given value (a) is easy to find. The table shows the P(Z < a). The P(Z > a) = 1 - P(Z < a). Suppose, for example, that we want to know the probability that a z-score will be greater than 3.00. From the table (see above), we find that P(Z < 3.00) = 0.9987. Therefore, P(Z > 3.00) = 1 - P(Z < 3.00) = 1 - 0.9987 = 0.0013. • Find P(a < Z < b). The probability that a standard normal random variables lies between two values is also easy to find. The P(a < Z < b) = P(Z < b) - P(Z < a). For example, suppose we want to know the probability that a z-score will be greater than -1.40 and less than -1.20. From the table (see above), we find that P(Z < -1.20) = 0.1151; and P(Z < -1.40) = 0.0808. Therefore, P(-1.40 < Z < -1.20) = P(Z < -1.20) - P(Z < -1.40) = 0.1151 - 0.0808 = 0.0343.

28. The Normal Distribution as a Model for Measurements • Often, phenomena in the real world follow a normal (or near-normal) distribution. This allows researchers to use the normal distribution as a model for assessing probabilities associated with real-world phenomena. Typically, the analysis involves two steps. • Transform raw data. Usually, the raw data are not in the form of z-scores. They need to be transformed into z-scores, using the transformation equation presented earlier: z = (X - μ) / σ. • Find probability. Once the data have been transformed into z-scores, you can use standard normal distribution tables, online calculators, or handheld graphing calculators to find probabilities associated with the z-scores.

29. Test Your Understanding • Molly 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

30. Correct!!! • You’re right. The correct answer is B. • 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, 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% of the students tested had a higher score than Molly. Continue

31. Incorrect. Here is the solution • The correct answer is B. As part of the solution to this problem, we assume that test scores are normally distributed. In this way, we use the normal distribution as a model for measurement. Given an assumption of normality, the solution involves three steps. • 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 an online calculator, a handheld graphing calculator, or 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% of the students who tested had a higher score than Molly.

32. Works Cited • Probability information from: • http://janellmcclure.typepad.com/files/probability-of-compound-events.ppt • Statistics information from: • http://stattrek.com/Lesson2/Normal.aspx?Tutorial=Stat • http://www.unt.edu/rss/class/mike/3610/Normal%20Distribution.ppt • http://php.spea.iupui.edu/jottensm/V506/class4.ppt • http://www.netmba.com/statistics/distribution/normal/ • http://www.childrens-mercy.org/stats/definitions/norm_dist.htm • http://www.statsoft.com/textbook/esc.html#Illustration%20of%20how%20the%20normal%20distribution%20is%20used%20in%20statistical%20reasoning%20(induction)