Understanding Random Variables and Probability Distributions

1. Chapter 7 Random Variables and Probability Distributions

2. Random Variable - A grocery store manager might be interested in the number of broken eggs in each carton (dozen of eggs). OR An environmental scientist might be interested in the amount of ozone in an air sample. • A numerical variable whose value depends on the outcome of a chance experiment • Associates a numerical value with each outcome of a chance experiment • Two types of random variables • Discrete • Continuous Since these values change and are subject to some uncertainty, these are examples of random variables.

3. Two Types of Random Variables: • Discrete– its set of possible values is a collection of isolated points along a number line • Continuous - its set of possible values includes an entire interval on a number line In this chapter, we will look at different distributions of discrete and continuous random variables. This is typically a “count” of something This is typically a “measure” of something

4. Identify the following variables as discrete or continuous • The number of broken eggs in each carton • The amount of ozone in samples of air • The weight of a pineapple • The amount of time a customer spends in a store • The number of gas pumps in use Discrete Continuous Continuous Continuous Discrete

5. Probability Distributions for Discrete Random Variables Probability distribution is a model that describes the long-run behavior of a variable.

6. Probability Number of Pets In a Wolf City (a fictional place), regulations prohibit no more than five dogs or cats per household. Let x = the number of dogs and cats in a randomly selected household in Wolf City x 0 1 2 3 4 5 P(x) .26 .31 .21 .13 .06 .03 This is called a discrete probability distribution. It can also be displayed in a histogram with the probability on the vertical axis. What do you notice about the sum of these probabilities? Is this variable discrete or continuous? What are the possible values for x? The Department of Animal Control has collected data over the course of several years. They have estimated the long-run probabilities for the values of x.

7. Discrete Probability Distribution • Gives the probabilities associated with each possible x value • Each probability is the long-run relative frequency of occurrence of the corresponding x-value when the chance experiment is performed a very large number of times • Usually displayed in a table, but can be displayed with a histogram or formula

8. Properties of Discrete Probability Distributions 1) For every possible x value, 0 <P(x) < 1. 2) For all values of x, SP(x) = 1.

9. Dogs and Cats Revisited . . . Let x = the number of dogs or cats per household in Wolf City x 0 1 2 3 4 5 P(x) .26 .31 .21 .13 .06 .03 What is the probability that a randomly selected household in Wolf City has at most 2 pets? Just add the probabilities for 0, 1, and 2 What does this mean? P(x< 2) = .26 + .31 + .21 = .78

10. Dogs and Cats Revisited . . . Let x = the number of dogs or cats per household in Wolf City x 0 1 2 3 4 5 P(x) .26 .31 .21 .13 .06 .03 What is the probability that a randomly selected household in Wolf City has less than 2 pets? Notice that this probability does NOT include 2! What does this mean? P(x < 2) = .26 + .31 = .57

11. Dogs and Cats Revisited . . . Let x = the number of dogs or cats per household in Wolf City x 0 1 2 3 4 5 P(x) .26 .31 .21 .13 .06 .03 What is the probability that a randomly selected household in Wolf City has more than 1 but no more than 4 pets? When calculating probabilities for discrete random variables, you MUST pay close attention to whether certain values are included (< or >) or not included (< or >)in the calculation. What does this mean? P(1 < x< 4) = .21 + .13 + .06 = .40

12. Probability Distributions for Continuous Random Variables

13. Consider the random variable: x = the weight (in pounds) of a full-term newborn child Suppose that weight is reported to the nearest pound. The following probability histogram displays the distribution of weights. Now suppose that weight is reported to the nearest 0.1 pound. This would be the probability histogram. What type of variable is this? If weight is measured with greater and greater accuracy, the histogram approaches a smooth curve. What is the sum of the areas of all the rectangles? The area of the rectangle centered over 7 pounds represents the probability 6.5 <x< 7.5 Notice that the rectangles are narrower and the histogram begins to have a smoother appearance. The shaded area represents the probability 6 <x< 8. This is an example of a density curve.

14. Probability Distributions for Continuous Variables • Is specified by a curve called a density curve. • The function that describes this curve is denoted by f(x) and is called the density function. • The probability of observing a value in a particular interval is the area under the curve and above the given interval.

15. Properties of continuous probability distributions 1. f(x) > 0 (the curve cannot dip below the horizontal axis) 2. The total area under the density curve equals one.

16. Let x denote the amount of gravel sold (in tons) during a randomly selected week at a particular sales facility. Suppose that the density curve has a height f(x) above the value x, where The density curve is shown in the figure:

17. This area can be found by use the formula for the area of a trapezoid: OR, more easily, by finding the area of the triangle, and subtracting that area from 1. Gravel problem continued . . . What is the probability that at most ½ ton of gravel is sold during a randomly selected week? P(x< ½) = 1 – ½(0.5)(1) = .75 The probability would be the shaded area under the curve and above the interval from 0 to 0.5.

18. Gravel problem continued . . . What is the probability that exactly ½ ton of gravel is sold during a randomly selected week? P(x = ½) = 0 How do we find the area of a line segment? The probability would be the area under the curve and above 0.5. Since a line segment has NO area, then the probability that exactly ½ ton is sold equals 0.

19. Gravel problem continued . . . What is the probability that less than ½ ton of gravel is sold during a randomly selected week? P(x < ½) = P(x< ½) = 1 – ½(0.5)(1) = .75 Hmmm . . . This is different than discrete probability distributions where it does change the probability whether a value is included or not! Does the probability change whether the ½ is included or not?

20. Density Time (in minutes) Suppose x is a continuous random variable defined as the amount of time (in minutes) taken by a clerk to process a certain type of application form. Suppose x has a probability distribution with density function: The following is the graph of f(x), the density curve:

21. Density Time (in minutes) Application Problem Continued . . . What is the probability that it takes more than 5.5 minutes to process the application form? P(x > 5.5) = .5(.5) = .25 When the density is constant over an interval (resulting in a horizontal density curve), the probability distribution is called a uniform distribution. Find the probability by calculating the area of the shaded region (base × height).

22. Other Density Curves Some density curves resemble the one below. Integral calculus is used to find the area under the these curves. Don’t worry – we will use tables (with the values already calculated). We can also use calculators or statistical software to find the area.

23. The probability that a continuous random variable x lies between a lower limit a and an upper limit b is P(a < x < b) = (cumulative area to the left of b) – (cumulative area to the left of a) P(a < x < b) = P(x < b) – P(x < a) This will be useful later in this chapter!

24. Means and Standard Deviations of Probability Distributions • The mean value of a random variable x, denoted by mx, describes where the probability distribution of x is centered. • The standard deviation of a random variable x, denoted by sx, describes variability in the probability distribution

25. Mean and Variance for Discrete Probability Distributions • Mean is sometimes referred to as the expected value (denoted E(x)). • Variance is calculated using • Standard deviation is the square root of the variance.

26. Dogs and Cats Revisited . . . Let x = the number of dogs and cats in a randomly selected household in Wolf City x 0 1 2 3 4 5 P(x) .26 .31 .21 .13 .06 .03 What is the mean number of pets per household in Wolf City? xP(x) 0 .31 .42 .39 .24 .15 xP(x) 0 + .31 + .42 + .39 + .24 + .15 First multiply each x-value times its corresponding probability. Next find the sum of these values. mx = 1.51 pets

27. Dogs and Cats Revisited . . . Let x = the number of dogs or cats per household in Wolf City x 0 1 2 3 4 5 P(x) .26 .31 .21 .13 .06 .03 What is the standard deviation of the number of pets per household in Wolf City? This is the variance – take the square root of this value. Next multiply by the corresponding probability. Then add these values. First find the deviation of each x-value from the mean. Then square these deviations. sx2 = (0-1.51)2(.26) + (1-1.51)2(.31)+(2-1.51)2(.21) +(3-1.51)2(.13) +(4-1.51)2(.06) +(5-1.51)2(.03) = 1.7499 sx = 1.323 pets

28. Mean and Variance for Continuous Random Variables For continuous probability distributions, mx and sx can be defined and computed using methods from calculus. • The mean value mx locates the center of the continuous distribution. • The standard deviation, sx, measures the extent to which the continuous distribution spreads out around mx.

29. 4300 4500 4700 4900 The first supplier is preferred to the second both in terms of mean value and variability. A company receives concrete of a certain type from two different suppliers. Let x = compression strength of a randomly selected batch from Supplier 1 y = compression strength of a randomly selected batch from Supplier 2 Suppose that mx = 4650 pounds/inch2sx = 200 pounds/inch2 my = 4500 pounds/inch2sy = 275 pounds/inch2 my mx

30. What would happen to the mean and standard deviation if we had to deduct $100 from everyone’s salary because of business being bad? Suppose Wolf City Grocery had a total of 14 employees. The following are the monthly salaries of all the employees. The mean and standard deviation of the monthly salaries are mx = $1700 and sx = $603.56 Suppose business is really good, so the manager gives everyone a $100 raise per month. The new mean and standard deviation would be mx = $1800 and sx = $603.56 Let’s graph boxplots of these monthly salaries to see what happens to the distributions . . . What happened to the standard deviations? What happened to the means? We see that the distribution just shifts to the right 100 units but the spread is the same.

31. Wolf City Grocery Continued . . . mx = $1700 and sx = $603.56 Suppose the manager gives everyone a 20% raise - the new mean and standard deviation would be mx = $2040 and sx = $724.27 Let’s graph boxplots of these monthly salaries to see what happens to the distributions . . . Notice that multiplying by a constant stretches the distribution, thus, changing the standard deviation. Notice that both the mean and standard deviation increased by 1.2.

32. If x is a random variable with mean, mx, and standard deviation, sx, and a and b are numerical constants, and the random variable y is defined by and Mean and Standard Deviation of Linear functions

33. Consider the chance experiment in which a customer of a propane gas company is randomly selected. Let x be the number of gallons required to fill a propane tank. Suppose that the mean and standard deviation is 318 gallons and 42 gallons, respectively. The company is considering the pricing model of a service charge of $50 plus $1.80 per gallon. Let y be the random variable of the amount billed. What is the equation for y? What are the mean and standard deviation for the amount billed? y = 50 + 1.8x my = 50 + 1.8(318) = $622.40 sy = 1.8(42) = $75.60

34. Move 1 s ? 1 2 4 3 Spinner A Spinner B 2 1 3 6 4 5 Suppose we are going to play a game called Stat Land! Players spin the two spinners below and move the sum of the two numbers. mA = 2.5 mB = 3.5 sA = 1.118 sB = 1.708 List all the possible sums (A + B). 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 Find the mean and standard deviation for these sums. Not sure – let’s think about it and return in just a few minutes! Here are the mean and standard deviation for each spinner. Notice that the mean of the sums is the sum of the means! How are the standard deviations related? mA+B= 6 sA+B =2.041

35. Move 1 s ? 1 2 4 3 Spinner A Spinner B 2 1 3 6 4 5 Stat Land Continued . . . Suppose one variation of the game had players move the difference of the spinners mA = 2.5 mB = 3.5 sA = 1.118 sB = 1.708 List all the possible differences (B - A). 0 -1 -2 -3 1 0 -1 -2 2 1 0 -1 3 2 1 0 4 3 2 1 5 4 3 2 Find the mean and standard deviation for these differences. How do we find the standard deviation for the sums or differences? Notice that the mean of the differences is the difference of the means! WOW – this is the same value as the standard deviation of the sums! mB-A= 1 sB-A =2.041

36. Mean and Standard Deviations for Linear Combinations If x1, x2, …, xn are random variables with means m1, m2, …, mn and variances s12, s22, …, sn2, respectively, and y = a1x1 + a2x2 + … + anxn then This result is true ONLY if the x’s are independent. This result is true regardless of whether the x’s are independent.

37. A commuter airline flies small planes betweenSan Luis Obispo and San Francisco. For small planes the baggage weight is a concern. Suppose it is known that the variable x = weight (in pounds) of baggage checked by a randomly selected passenger has a mean and standard deviation of 42 and 16, respectively. Consider a flight on which 10 passengers, all traveling alone, are flying. The total weight of checked baggage, y, is y = x1 + x2 + … + x10

38. Airline Problem Continued . . . mx = 42 and sx = 16 The total weight of checked baggage, y, is y = x1 + x2 + … + x10 What is the mean total weight of the checked baggage? mx= m1 + m2 + … + m10 = 42 + 42 + … + 42 = 420 pounds

39. Airline Problem Continued . . . mx = 42 and sx = 16 The total weight of checked baggage, y, is y = x1 + x2 + … + x10 What is the standard deviation of the total weight of the checked baggage? Since the 10 passengers are all traveling alone, it is reasonable to think that the 10 baggage weights are unrelated and therefore independent. To find the standard deviation, take the square root of this value. sx2= sx12 + sx22 + … + sx102 = 162 + 162 + … + 162 = 2560 pounds s = 50.596 pounds

40. Special Distributions Two Discrete Distributions: Binomial and Geometric One Continuous Distribution: Normal Distributions

41. Suppose we decide to record the gender of the next 25 newborns at a particular hospital. What is the chance that at least 15 are female? What is the chance that between 10 and 15 are female? Out of the 25 newborns, how many can we expect to be female? These questions can be answered using a binomial distribution.

42. Properties of a Binomial Experiment • There are a fixed number of trials • Each trial results in one of two mutually exclusive outcomes. (success/failure) • Outcomes of different trials are independent • The probability that a trial results in success is the same for all trials The binomial random variable x is defined as x = the number of successes observed when a binomial experiment is performed We use n to denote the fixed number of trials.

43. Are these binomial distributions? • Toss a coin 10 times and count the number of heads Yes • Deal 10 cards from a shuffled deck and count the number of red cards No, probability does not remain constant • The number of tickets sold to children under 12 at a movie theater in a one hour period No, no fixed number

44. Binomial Probability Formula: Let n = number of independent trials in a binomial experiment p = constant probability that any trial results in a success Where: Appendix Table 9 can be used to find binomial probabilities. Technology, such as calculators and statistical software, will also perform this calculation.

45. Instead of recording the gender of the next 25 newborns at a particular hospital, let’s record the gender of the next 5 newborns at this hospital. Is this a binomial experiment? Yes, if the births were not multiple births (twins, etc). Define the random variable of interest. x = the number of females born out of the next 5 births What are the possible values of x? x 0 1 2 3 4 5 What is the probability of “success”? What will the largest value of the binomial random value be? Will a binomial random variable always include the value of 0?

46. Newborns Continued . . . What is the probability that exactly 2 girls will be born out of the next 5 births? What is the probability that less than 2 girls will be born out of the next 5 births?

47. Since this is a discrete distribution, we could use: Newborns Continued . . . Let’s construct the discrete probability distribution table for this binomial random variable: What is the mean number of girls born in the next five births? Notice that this is the same as multiplying n × p mx = 0(.03125) + 1(.15625) + 2(.3125) + 3(.3125) + 4(.15625) + 5(.03125) =2.5

48. Formulas for mean and standard deviation of a binomial distribution

49. Newborns Continued . . . How many girls would you expect in the next five births at a particular hospital? What is the standard deviation of the number of girls born in the next five births?

50. Remember, in binomial distributions, trials should be independent. However, when we sample, we typically sample without replacement, which would mean that the trials are not independent. . . In this case, the number of success observed would not be a binomial distribution but rather hypergeometric distribution. But when the sample size, n, is small and the population size, N, is large, probabilities calculated using binomial distributions and hypergeometric distributions are VERY close! When sampling without replacement if n is at most 5% of N, then the binomial distribution gives a good approximation to the probability distribution of x. The calculation for probabilities in a hypergeometric distribution are even more tedious than the binomial formula!