Chapter 7

1. Chapter 7 Please pick up an assignment sheet and notes packet

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

12. Finish the dog and cat probability problems on the second page of your notes

13. 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

14. 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

15. Suppose that each of four random selected customers purchasing a hot tub at a certain store chooses either an electric (E) or a gas (G) model. Assume that these customers makes their choices independently of one another and that 40% of all customers select an electric model. This implies that for any particular one of the four customers P(E) = 0.40 and P(G) = 0.60. One possible experimental outcome is EFFE, where the first and fourth customers select electric models and the other two choose gas models. Because the customers make their choices independently the multiplication rule for independent events implies that P(EGGE) = P(1st chooses E AND 2nd chooses G AND 3rd chooses G AND 4th chooses E) = = P(E)P(G)P(G)P(E) = (0.4)(0.6)(0.6)(0.4) = 0.0576

16. Suppose that each of four random selected customers purchasing a hot tub at a certain store chooses either an electric (E) or a gas (G) model. Assume that these customers makes their choices independently of one another and that 40% of all customers select an electric model. This implies that for any particular one of the four customers P(E) = 0.40 and P(G) = 0.60. One possible experimental outcome is EFFE, where the first and fourth customers select electric models and the other two choose gas models. Because the customers make their choices independently the multiplication rule for independent events implies that P(EGGE) = P(1st chooses E AND 2nd chooses G AND 3rd chooses G AND 4th chooses E) = = P(E)P(G)P(G)P(E) = (0.4)(0.6)(0.6)(0.4) = 0.0576

17. P(x = 0) = 0.1296 • p(x = 1) =0.3456 • p(x = 2) = 0.3456 • p(x = 3) = 0.1536 • p(x = 4) = 0.0256 • P(2 ≤ x ≤ 4) =0.3456 + 0.1536 + 0.0256 = 0.5248 • P(x ≤ 3) = 0.1296 + 0.3456 + 0.3456 + 0.1536 = 0.9744

18. Probability Distributions for Continuous Random Variables

19. 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.

20. 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.

21. 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.

22. 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:

23. 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.

24. 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.

25. 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 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?

26. 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:

27. 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).

28. 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.

29. 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!

32. 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

33. 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.

34. 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

35. 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

36. 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.

37. 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

38. 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.

39. 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.

40. 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

41. 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

42. 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

43. 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

44. 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.

46. 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

47. 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

48. 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

49. The Attila Barbell Company makes bars for weight lifting. The weights of the bars are independent and are normally distributed with a mean of 720 ounces (45 pounds) and a standard deviation of 4 ounces. The bars are shipped 10 in a box to the retailers. The weights of the empty boxes are normally distributed with a mean of 320 ounces and a standard deviation of 8 ounces. The weights of the boxes filled with 10 bars are expected to be normally distributed with a mean of 7520 ounces and a standard deviation of:

50. The Attila Barbell Company makes bars for weight lifting. The weights of the bars are independent and are normally distributed with a mean of 720 ounces (45 pounds) and a standard deviation of 4 ounces. The bars are shipped 10 in a box to the retailers. The weights of the empty boxes are normally distributed with a mean of 320 ounces and a standard deviation of 8 ounces. The weights of the boxes filled with 10 bars are expected to be normally distributed with a mean of 7520 ounces and a standard deviation of: