1. Chapter 7 Continuous Probability Distributions and the Normal Distribution

2. From Chapter 6 …  A discrete random variable can only assume a finite or countable number of values with a break between successive values of the variable. (X=0, X=1, X=2)  Consider the number of televisions in a household: X=0, X=1, X=2, X=3, X=4 ….  It’s impossible for a discrete variable to have the value 2.45.  It’s impossible for a household to have 2.45 television sets.  This is why we call them DISCRETE random variables vs …  Chapter 6: CONTINUOUS random variables

3. From Chapter 6 …  This is the probability histogram we created for the discrete random variable, number of correct answers, on a 5 question multiple choice test.  Each question has four choices.  What is the probability of getting 1 correct answer?  0.3955 = 39.55% 0 1 2 3 4 5 X number of correct answers Probability 0.3955 0.2637 0.2373 0.0879 0.0146 0.0010

4. Probability Histograms  What is the area of this rectangle?  1x0.3955 = 0.3955  What is the probability of getting 1 or 2 correct answers?  0.3955 + 0.2637 = 0.6592  What are the areas of these rectangles?  0.6592 0 1 2 3 4 5 X number of correct answers Probability 0.3955 0.2637 0.2373 0.0879 0.0146 0.0010

5. Probability Histograms  What is the area of the sum of all the rectangles?  100%  There is an important relationship between area and probability:  The probability of an event occurring is equal to the area of the rectangles that “cover” that event. 0 1 2 3 4 5 X number of correct answers Probability 0.3955 0.2637 0.2373 0.0879 0.0146 0.0010

6. Continuous Random Variables  Let continuous random variable X represents the amount of time a college student needs to travel between classes at NCC.  The amount of time is recorded to the nearest minute for up to a maximum time of twenty minutes.

7. Continuous Random Variables  There is a rectangle associated for each minute to get to class. The area of each rectangle represents the probability that the student will take X minutes to get to the next class.  What is the probability that it takes a student 10 minutes to get to class?

8. Continuous Random Variables  0.06 = 6%  The area of the rectangle for X = 10 is equal to the probability that it takes a student 10 minutes to get to class.  Area of Rectangle = Probability  What will the sum of the area of all the rectangles of this histogram equal?  100% - just like the probabilities of a probability distribution table.

9. Continuous Random Variables  The probability a student will require from 5 to 7 minutes to get to the next class is equal to the sum of the areas corresponding to the shaded rectangles associated with X = 5 minutes, X = 6 minutes, and X = 7 minutes.

10. Continuous Random Variables  But, wait! Time is a continuous random variable.  What if time to get to class were rounded to the nearest quarter of a minute instead of the nearest minute?  For example, 5.1 minutes, 5.2 minutes, etc …  More rectangles!  but the sum of all the areas of the rectangles = sum of the probabilities still is 1, or 100%.

11. Continuous Random Variables  Remember, probability = area.  Again, if we are interested in finding the probability that it takes a student between 4.5 and 7.5 minutes to get to class,  we add up the areas of the rectangles between 4.5 and 7.5.

12. Continuous Random Variables  Suppose that the amount of time a student needs to get to class is measured to a finer degree (such as one hundredth of a minute or one thousandth of a minute) …  The rectangles get more frequent and more narrower,  eventually, approaching a smooth curve This is the probability density curve.

13. Probability density curve of the continuous variable time  The area under the probability density curve between two possible values of the continuous variable represents the probability that the value of the continuous variable will fall between these two values.  The probability that the amount of time a student will take from 5 to 7 minutes to get to the next class is equal to the area under the probability density curve between 5 and 5. P(5 to 7)

14. Characteristics of a Continuous Probability Distribution  The probability that the continuous random variable X will assume a value between two possible values, X = a and X = b of the variable is equal to the area under the density curve between the values X = a and X = b.

15. Characteristics of a Continuous Probability Distribution  The total area (or probability) under the density curve is equal to 1.

16. Characteristics of a Continuous Probability Distribution  The probability that the continuous random variable X will assume a value between any two possible values X = a and X = b of the variable is between 0 and 1.

17. Normal Distribution  Most of the balls that fall through demonstrator settle in the center. Some balls form a tail on the right, and others form a tail on the left.  This is a demonstration of the normal distribution. http://www.youtube.com/watch?v=xDIyAOBa_yU http://www.youtube.com/watch?v=xDIyAOBa_yU

18. The Normal Distribution  A normal distribution is a distribution that represents the values of a continuous variable.  If a continuous variable is said to be approximately normal, then the normal distribution can be used to model the continuous variable.  Examples of distributions that can be modeled or approximated by the normal distribution include IQ scores of individuals, adult weights, blood pressure of men and women, tire wear, the size of red blood cells, and the time required to get to work.  The importance of a normal distribution is that it serves as a good model in approximating many distributions of real-world phenomena.  So if a variable is approximately normal, we can use mathematics to make predictions about the population regarding that variable.

19. Properties of a Normal Distribution  The normal curve is bell-shaped and has a single peak that is located at the center.

20. Properties of a Normal Distribution IQ Scores Probability Adult Weights Probability Blood Pressure Size of Red Blood Cells Probability

21. Properties of a Normal Distribution  The mean, median, and mode all have the same numerical value and are located at the center of the distribution.

22. Properties of a Normal Distribution  The normal curve is symmetric about the mean.

23. Properties of a Normal Distribution  In theory, the normal curve extends infinitely in both directions, always getting closer to the horizontal axis but never touching it. As the normal curve extends out away from the mean and gets closer to the horizontal axis, the area in the tails of the normal curve is getting closer to zero.

24. Properties of a Normal Distribution  The greatest concentration of scores are clustered around the center.  Relating this to probability, the likelihood of an event occurring is greatest if the value is close to the mean / median / mode.

25. Properties of a Normal Distribution  The total area under the normal curve is 1 which can also be interpreted as probability. Thus, the area under the normal curve represents 100% of all the data values.

26. NormalCDF on the TI 83 / 84  The calculator’s normalcdf function finds the area between two z-scores in a normal distribution.  To find the area between z=0 and z=2:  2 nd  DISTR  2: normalcdf ENTER (0,2)  The lower z-score is entered first (0), the higher z-score is entered second (2)  A z-score of -100000 can be used to model negative infinity on the calculator. (the textbook uses –E99)  A z-score of 100000 can be used to model positive infinity. (the textbook uses E99)  normalcdf(2, E99) will calculate the area “greater than z=2. i.e. from 2 to infinity.

27. Using the calculator to get the area to the left of a z-score  Example 7.1 - Find the proportion of area under the normal curve:  a) To the left of z = – 0.53 (or less than a z-score of – 0.53)

28. Example 7.1  Using your TI83/84 calculator:  2 nd DISTR 2: normalcdf ( -E99, -0.53) ENTER

29. Example 7.1  Find the proportion of area under the normal curve:  b) To the left of z = 2.56 (or Less than a z score of 2.56)  Using your TI83/84 calculator:  2 nd DISTR 2: normalcdf ( -E99, 2.56 ) ENTER  Remember, the lower z-score is entered first, the higher z-score second!

30. Example 7.2  Find the proportion of area under the normal curve:  a) To the right of z = – 1.37 (or greater than a z score of – 1.37)  Using your TI83/84 calculator:  2 nd DISTR 2: normalcdf ( -1.37, E99) ENTER  0.9147

31. Example 7.2  Find the proportion of area under the normal curve:  b) To the right of z = 0.67.  2 nd DISTR 2: normalcdf (0.67, E99) ENTER  0.2514

32. Example 7.3  Find the proportion of area under the normal curve:  a) Between z = 0 and z = 1.5  2 nd DISTR 2: normalcdf (smaller z score, larger z score) ENTER  2 nd DISTR 2: normalcdf (0, 1.5) ENTER  0.4332

33. Example 7.3  Find the proportion of area under the normal curve:  c) Between z = – 1.25 and z = 1.0  2 nd DISTR 2: normalcdf (-1.25, 1.0) ENTER  0.7357

34. Proof of the Empirical Rule  The Empirical Rule states that:  About 68% of the data values will lie within 1 standard deviation of the mean  About 95% of the data values will lie within 2 standard deviations of the mean.  About 99-100% of the data values will lie within 3 standard deviations of the mean.  We now know enough about the normal distribution to prove that this is true!

35. What if the problem gives area? Can we find the z-score?  So far, we have seen how to find the proportion of area under the normal curve if given z-scores.  But what if we are given the proportion of area? Can the calculator give us the corresponding z-score?  Yes! The calculator function is called invnorm and it takes as input an area and gives us a z-score. [2 nd ->DISTR->invnorm]  But there is a catch… invnorm will only accept as input area to the left and will return the z-score that gives that area.

36. Example 7.5  For a normal distribution, find the z score(s) that cut(s) off:  a) The lowest 20% of area. (or cuts off the bottom 20 % or 0.20 of the z scores)  invnorm (0.20) = z = – 0.84

37. Example 7.5  Recognize that the z-score that cuts off the bottom 20% is the same as the z-score that cuts off the top 80%.  invnorm (0.20) = z = – 0.84

38. Example 7.5  For a normal distribution, find the z score(s) that cut(s) off:  b) The highest of 10% of area (or cuts off the top 10 % or 0.10 of the z scores)  Realize that invnorm only takes in area to the left of a score. What is the area to the left if the area to the right is 10%?  90%  invnorm (0.90) =1.28

39. Example 7.5  For a normal distribution, find the z score(s) that cut(s) off:  c) The middle 90% of the area (or of the z scores).  In a normal distribution, z = 0 is located in the center so middle 90% represents 45% to the left of z = 0 and 45% to the right of z = 0.  What is the area to the left of z 1 ?  5%  z 1 = invnorm (0.05) = – 1.64  Since the normal curve is symmetric, the other value for z 2 is 1.64

40. Normalcdf vs invnorm  The normalcdf function on the calculator takes z-scores as inputs, and gives you an area as output.  The area it outputs is always between 0 and 1.  This area is also interpreted as the probability between the two input z-scores.  The invnorm function on the calculator reverses this process.  This function takes in an area (to the left) and outputs the z-score that corresponds to that area.  So invnorm(0.25) gives the z-score that cuts off the bottom 25% of all the data values.  What would invnorm(0.50) be? 00