1. Statistics for Business and Economics Module 1:Probability Theory and Statistical Inference Spring 2010 Lecture 3: Continuous probability distributions Priyantha Wijayatunga, Department of Statistics, Umeå University priyantha.wijayatunga@stat.umu.se These materials are altered ones from copyrighted lecture slides (© 2009 W.H. Freeman and Company) from the homepage of the book: The Practice of Business Statistics Using Data for Decisions :Second Edition by Moore, McCabe, Duckworth and Alwan.

2. Continuous probability distributions  Probability density  Uniform probability density  Normal distributions, standard normal distribution  Law of large numbers  Sampling distributions  The mean and standard deviation of sample mean  The central limit theorem

3. Recall: Discrete Probability Distributions Let X denote the # of days a student comes to class (in a week). Probability distibution is then 1)what is the probability that a student comes to the class more than 3 days? 2)what is the probability that a student comes to the class 2 or 3 days?

4. A continuous random variable X takes all values in an interval. Example: There is an infinity of numbers between 0 and 1 (e.g., 0.001, 0.4, 0.0063876). The probability distribution of a continuous random variable is described by a density curve ( also called density function or probability density). The probability of any event is the area under the density curve for the values of X that make up the event. Continuous Probability Distributions The probability that X falls between 0.3 and 0.7 is the area under the density curve for that interval: P(0.3 ≤ X ≤ 0.7) = (0.7 – 0.3)*1 = 0.4 Density function: f(x)= 1; for 0 ≤ x ≤ 1 f(x)= 0; for x 1 This is a uniform density curve for the variable X. X

5. P(X 0.8) = P(X 0.8) = 1 – P(0.5 < X < 0.8) = 0.7 The probability of a single event is zero: P(X=1) = (1 – 1)*1 = 0 Intervals All continuous probability distributions assign probability 0 to every individual outcome. Only intervals can have a positive probability, represented by the area under the density curve for that interval. Height = 1 X The probability of an interval is the same whether boundary values are included or excluded: P(0 ≤ X ≤ 0.5) = (0.5 – 0)*1 = 0.5 P(0 < X < 0.5) = (0.5 – 0)*1 = 0.5 P(0 ≤ X < 0.5) = (0.5 – 0)*1 = 0.5

6. Assigning Probabilities: intervals of outcomes  A sample space may contain all numbers within a range.  For continuous outcomes, the probability model is a density curve.  Area under the entire density curve is equal to 1.  Probability model assigns probabilities as areas under the density curve.

7. Assigning Probabilities: intervals of outcomes If all possible outcomes are equally likely: for example, obtaining a value from 0 to 1 is equally likely. Uniform density curve (uniform probability distribution) on [0,1]. Probabilities are computed as areas P(0.3  X  0.7) = 0.4 Similarly, P(X 0.8) = 0.5 +0.2 = 0.7

8. General uniform probability distribution If the outcomes are equally likely for any value in between two numbers a and b (random variable X can take any value in between a and b) where a<b, then the probability density of X is Ex: The number of minutes that a student takes to solve a math problem is known to be any number in between 10 to 20 with equal chances. Find the probability that a student takes more than 6 but less than 12 minutes to solve a given math problem.

9. Because the probability of drawing one individual at random depends on the frequency of this type of individual in the population, the probability is also the shaded area under the curve. The shaded area under a density curve shows the proportion, or %, of individuals in a population with values of X between x 1 and x 2. % individuals with X such that x 1 < X < x 2 Continuous random variable and population distribution

10. Normal probability models  Normal probability models look like:  The scores of students on the ACT college entrance examination in a recent year had the normal distribution with mean  =18.6 and standard deviation  = 5.9.  What is the probability that a randomly chosen student scores 21 or higher?

11. The probability distribution of many random variables is a normal distribution. It shows what values the random variable can take and is used to assign probabilities to those values. Example: Probability distribution of women’s heights. Here since we chose a woman randomly, her height, X, is a random variable. To calculate probabilities with the normal distribution, we will standardize the random variable (z score) and use Table A. Normal probability distributions

12. Normal distributions e = 2.71828… The base of the natural logarithm π = pi = 3.14159… Normal – or Gaussian – distributions are a family of symmetrical, bell shaped density curves defined by a mean  (mu) and a standard deviation  (sigma) : N(  ). xx

13. A family of density curves Here means are different (  = 10, 15, and 20) while standard deviations are the same (  = 3) Here means are the same (  = 15) while standard deviations are different (  = 2, 4, and 6).

14. mean µ = 64.5 standard deviation  = 2.5 N(µ,  ) = N(64.5, 2.5) The 68-95-99.7 rule  About 68% of all observations are within 1 standard deviation (  of the mean (  ).  About 95% of all observations are within 2  of the mean .  Almost all (99.7%) observations are within 3  of the mean. Inflection point

15. Because all Normal distributions share the same properties, we can standardize our data to transform any Normal curve N(  ) into the standard Normal curve N(0,1). The standard Normal distribution For each x we calculate a new value, z (called a z-score). N(0,1) => N(64.5, 2.5) Standardized height (no units)

16. A z-score measures the number of standard deviations that a data value x is from the mean . Standardizing: calculating z-scores When x is larger than the mean, z is positive. When x is smaller than the mean, z is negative. When x is 1 standard deviation larger than the mean, then z = 1. When x is 2 standard deviations larger than the mean, then z = 2.

17. mean µ = 64.5" standard deviation  = 2.5" x (height) = 67" We calculate z, the standardized value of x: Because of the 68-95-99.7 rule, we can conclude that the percent of women shorter than 67” should be, approximately,.68 + half of (1 -.68) =.84 or 84%. Area= ??? N(µ,  ) = N(64.5, 2.5)  = 64.5” x = 67” z = 0z = 1 Ex. Women heights Women heights follow the N(64.5”,2.5”) distribution. What percent of women are shorter than 67 inches tall (that’s 5’6”)?

18. What is the probability, if we pick one woman at random, that her height will be some value X? For instance, between 68 and 70 inches P(68 < X < 70)? Because the woman is selected at random, X is a random variable. As before, we calculate the z- scores for 68 and 70. For x = 68", For x = 70", The area under the curve for the interval [68" to 70"] is 0.9861 − 0.9192 = 0.0669. Thus, the probability that a randomly chosen woman falls into this range is 6.69%. P(68 < X < 70) = 6.69% 0.9861 0.9192 N(µ,  ) = N(64.5, 2.5)

19. Using Table A (…) Table A gives the area under the standard Normal curve to the left of any z value..0082 is the area under N(0,1) left of z = - 2.40.0080 is the area under N(0,1) left of z = -2.41 0.0069 is the area under N(0,1) left of z = -2.46

20. Area ≈ 0.84 Area ≈ 0.16 N(µ,  ) = N(64.5”, 2.5”)  = 64.5” x = 67” z = 1 Conclusion: 84.13% of women are shorter than 67”. By subtraction, 1 - 0.8413, or 15.87% of women are taller than 67". For z = 1.00, the area under the standard Normal curve to the left of z is 0.8413. Percent of women shorter than 67”

21. Tips on using Table A Because the Normal distribution is symmetrical, there are 2 ways that you can calculate the area under the standard Normal curve to the right of a z value. area right of z = 1 - area left of z Area = 0.9901 Area = 0.0099 z = -2.33 area right of z = area left of -z

22. Tips on using Table A To calculate the area between 2 z-values, first get the area under N(0,1) to the left for each z-value from Table A. area between z 1 and z 2 = area left of z 1 – area left of z 2 A common mistake made by students is to subtract both z- values, but the Normal curve is not uniform. Then subtract the smaller area from the larger area.  The area under N(0,1) for a single value of z is zero (Try calculating the area to the left of z minus that same area!)

23. The National Collegiate Athletic Association (NCAA) requires Division I athletes to score at least 820 on the combined math and verbal SAT exam to compete in their first college year. The SAT scores of 2003 were approximately normal with mean 1026 and standard deviation 209. What proportion of all students would be NCAA qualifiers (SAT ≥ 820)? Note: The actual data may contain students who scored exactly 820 on the SAT. However, the proportion of scores exactly equal to 820 is 0 for a normal distribution is a consequence of the idealized smoothing of density curves. area right of 820= total area - area left of 820 =1 - 0.1611 ≈ 84%

24. The NCAA defines a “partial qualifier” eligible to practice and receive an athletic scholarship, but not to compete, as a combined SAT score is at least 720. What proportion of all students who take the SAT would be partial qualifiers? That is, what proportion have scores between 720 and 820? About 9% of all students who take the SAT have scores between 720 and 820. area between = area left of 820 - area left of 720 720 and 820=0.1611 - 0.0721 ≈ 9%

25. N(0,1) The cool thing about working with normally distributed data is that we can manipulate it and then find answers to questions that involve comparing seemingly non- comparable distributions. We do this by “standardizing” the data. All this involves is changing the scale so that the mean now = 0 and the standard deviation = 1. If you do this to different distributions it makes them comparable.

26. Finding a value when given a proportion Backward normal calculations: We may also want to find the observed range of values that correspond to a given proportion under the curve. For that, we use Table A backward:  we first find the desired area/proportion in the body of the table  we then read the corresponding z-value from the left column and top row For an area to the left of 1.25 % (0.0125), the z-value is -2.24

27. Backward Normal Calculations  Miles per gallon ratings of compact cars (2001 models) follow approximately the N(25.7, 5.88) distribution. How many miles per gallon must a vehicle get to place in the top 10% of all 2001 model compact cars? 1. z = 1.28 is the standardized value with area 0.9 to its left and 0.1 to its right. 2. Unstandardize Solving for x gives x = 33.2 miles per gallon.

28. Other Standard Normal probability tables

29. One way to assess if a distribution is indeed approximately normal is to plot the data on a normal quantile plot. The data points are ranked and the percentile ranks are converted to z- scores with Table A. The z-scores are then used for the x axis against which the data are plotted on the y axis of the normal quantile plot.  If the distribution is indeed normal the plot will show a straight line, indicating a good match between the data and a normal distribution.  Systematic deviations from a straight line indicate a nonnormal distribution. Outliers appear as points that are far away from the overall pattern of the plot. Assessing the Normality of data

30. Normal quantile plot of the earnings of 15 black female hourly workers at National Bank. This distribution is roughly Normal except for one low outlier. Normal quantile plot of the salaries of Cincinnati Reds players on opening day of the 2000 season. This distribution is skewed to the right. The Normal Distributions

31. Law of large numbers As the number of randomly drawn observations in a sample increases, the mean of the sample gets closer and closer to the population mean . This is the law of large numbers. It is valid for any population. Note: We often intuitively expect predictability over a few random observations, but it is wrong. The law of large numbers only applies to really large numbers.

32. Reminder: What is a sampling distribution? The sampling distribution of a statistic is the distribution of all possible values taken by the statistic when all possible samples of a fixed size n are taken from the population. It is a theoretical idea — we do not actually build it. The sampling distribution of a statistic is the probability distribution of that statistic.

33. Sampling distribution of sample mean We take many random samples of a given size n from a population with mean  and standard deviation  Some sample means will be above the population mean  and some will be below, making up the sampling distribution. Sampling distribution of “x bar” Histogram of some sample averages

34. Sampling distribution of x bar  √n√n For any population with mean  and standard deviation  :  The mean of the sampling distribution is equal to the population mean    The standard deviation of the sampling distribution is  /√n, where n is the sample size.

35. Mean and standard deviation of sample mean Mean of a sampling distribution of There is no tendency for a sample mean to fall systematically above or below  even if the distribution of the raw data is skewed. Thus, the mean of the sampling distribution is an unbiased estimate of the population mean  — it will be “correct on average” in many samples. Standard deviation of a sampling distribution of The standard deviation of the sampling distribution is smaller than the standard deviation of the population by a factor of √n.  Averages are less variable than individual observations. Also, the results of large samples are less variable than the results of small samples.

36. For normally distributed populations When a variable in a population is normally distributed, the sampling distribution of the sample mean for all possible samples of size n is also normally distributed. If the population is N(  ) then the sample means distribution is N(  /√n). Population Sampling distribution

37. The central limit theorem Central Limit Theorem: When randomly sampling from any population with mean  and standard deviation , when n is large enough, the sampling distribution of x bar is approximately normal: ~ N(  /√n). Population with strongly skewed distribution Sampling distribution of for n = 2 observations Sampling distribution of for n = 10 observations Sampling distribution of for n = 25 observations

38. The central limit theorem From a highly skewed distribution (mean=3.5, sd=1.024695) get random samples with n=50 and get their sample means Relative frequency distribution is pproximately normal (bell –shaped) mean=3.50164 and sd=0.1471508

39. IQ scores: population vs. sample In a large population of adults, the mean IQ is 112 with standard deviation 20. Suppose 200 adults are randomly selected for a market research campaign.  The distribution of the sample mean IQ is: A) Exactly normal, mean 112, standard deviation 20 B) Approximately normal, mean 112, standard deviation 20 C) Approximately normal, mean 112, standard deviation 1.414 D) Approximately normal, mean 112, standard deviation 0.1 C) Approximately normal, mean 112, standard deviation 1.414

40. Application Hypokalemia is diagnosed when blood potassium levels are low, below 3.5mEq/dl. Let’s assume that we know a patient whose measured potassium levels vary daily according to a normal distribution N(  = 3.8,  = 0.2). If only one measurement is made, what is the probability that this patient will be misdiagnosed hypokalemic? z = −1.5, P(z < −1.5) = 0.0668 ≈ 7% If instead measurements are taken on 4 separate days, what is the probability of such a misdiagnosis? z = −3, P(z < −1.5) = 0.0013 ≈ 0.1% Note: Make sure to standardize (z) using the standard deviation for the sampling distribution.

41. Income distribution Let’s consider the very large database of individual incomes from the Bureau of Labor Statistics as our population. It is strongly right skewed.  We take 1000 SRSs of 100 incomes, calculate the sample mean for each, and make a histogram of these 1000 means.  We also take 1000 SRSs of 25 incomes, calculate the sample mean for each, and make a histogram of these 1000 means. Which histogram corresponds to the samples of size 100? 25?

42. In many cases, n = 25 isn’t a huge sample. Thus, even for strange population distributions we can assume a normal sampling distribution of the mean and work with it to solve problems. How large a sample size? It depends on the population distribution. More observations are required if the population distribution is far from normal.  A sample size of 25 is generally enough to obtain a normal sampling distribution from a strong skewness or even mild outliers.  A sample size of 40 will typically be good enough to overcome extreme skewness and outliers.