Honors Statistics. 3. Review OTL C6#3. 4. Normal Curve Quiz. Chapter 6 Section 2 Day s Notes.notebook. May 02, 2016.

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1 Honors Statistics Aug 23-8:26 PM 3. Review OTL C6#3 4. Normal Curve Quiz Aug 23-8:31 PM 1

2 May 1-9:09 PM Apr 28-10:29 AM 2

3 27, 28, 29, 30 Nov 21-8:16 PM Working out Choose a person aged 19 to 25 years at random and ask, In the past seven days, how many times did you go to an exercise or fitness center or work out? Call the responsey for short. Based on a large sample survey, here is a probability model for the answer you will get: 8 (a) Show that this is a legitimate probability distribution = 1 (b) Make a histogram of the probability distribution. Describe what you see. frequency (c) Describe the event Y < 7 in words. What is P(Y < 7)? Number of workout days What is the probability that a randomly selected persons aged 19 to 25 went to the gym less than seven days this week? P(Y < 7) = 1 - P(Y = 7) = = 0.98 (d) Express the event worked out at least once in terms of Y. What is the probability of this event? P(Y 1) = 1 - P(Y = 0) = = 0.32 Nov 29-10:57 AM 3

4 Working out Refer to Exercise 6. Consider the events A = works out at least once and B = works out less than 5 times per week. (a) What outcomes make up the event A? What is P(A)? outcomes = 1,2,3,4,5,6,7 P(Y 1) = 1 - P(Y = 0) = = 0.32 (b) What outcomes make up the event B? What is P(B)? outcomes = 0,1,2,3,4 P(Y < 5) = = 0.93 (c) What outcomes make up the event A and B? What isp(a and B)? Why is this probability not equal to P(A) P(B)? P(A and B) = = 0.25 The events working out at least once and working out less than 5 times per week are not INDEPENDENT events. So Multiplication cannot be used to determine the probability of P(A and B) Nov 29-11:00 AM Keno Keno is a favorite game in casinos, and similar games are popular with the states that operate lotteries. Balls numbered 1 to 80 are tumbled in a machine as the bets are placed, then 20 of the balls are chosen at random. Players select numbers by marking a card. The simplest of the many wagers available is Mark 1 Number. Your payoff is $3 on a $1 bet if the number you select is one of those chosen. Because 20 of 80 numbers are chosen, your probability of winning is 20/80, or Let X= the net amount you gain on a single play of the game. OR Based on what you "get" back OR µ x = -1(0.75) + 2(0.25) = Nov 29-11:02 AM 4

5 Size of American households In government data, a household consists of all occupants of a dwelling unit, while a family consists of two or more persons who live together and are related by blood or marriage. So all families form households, but some households are not families. Here are the distributions of household size and family size in the United States: Legitimate Distributions... Let X = the number of people in a randomly selected U.S. household and Y = the number of people in a randomly chosen U.S. family. (a) Make histograms suitable for comparing the probability distributions of X and Y. Describe any differences that you observe. Dec 5-5:48 PM ` > (b) Find the mean for each random variable. Explain why this difference makes sense. µ x = 1(0.25) + 2(0.32) + 3(0.17) + 4(0.15) + 5(0.07) + 6(0.03) + 7(0.01) = 2.6 The expected number of people in a randomly selected household is 2.6 people. If many, many random households were selected, they would on average have 2.6 people. µ y = 1(0.0) + 2(0.42) + 3(0.23) + 4(0.21) + 5(0.09) + 6(0.03) + 7(0.02) = 3.14 The expected number of people in a randomly selected family is 3.14 people. If many, many random families were selected, they would on average have 3.14 people. > (c) Find and interpret the standard deviations of both Xand Y. Nov 21-10:40 PM 5

6 > (c) Find and interpret the standard deviations of both X and Y. σ 2 x = (1-2.6) 2 (0.25) + (2-2.6) 2 (0.32) + (3-2.6) 2 (0.17) + (4-2.6) 2 (0.15) + (5-2.6) 2 (0.07) + (6-2.6) 2 (0.03) (7-2.6) 2 (0.01) = 2.02 σ x = 2.02 = The standard deviation of X is σ x = A the number of people in a randomly selected household will typically differ from the mean (2.6) by about people. σ 2 y = (1-3.14) 2 (0.0) + (2-3.14) 2 (0.42) + (3-3.14) 2 (0.23) + (4-3.14) 2 (0.21) + (5-3.14) 2 (0.09) + (6-3.14) 2 (0.03) (7-3.14) 2 (0.02) = σ x = = The standard deviation of X is σ y = A the number of people in a randomly selected family will typically differ from the mean (3.14) by about people. Dec 4-7:00 AM THIS IS ON THE CLASS WORKSHEET! Random numbers Let Y be a number between 0 and 1 produced by a random number generator. Assuming that the random variable Y has a uniform distribution, find the following probabilities: HOMEWORK PROBLEM #22 Random numbers a) P(Y 0.4) = (0.4)(1) = 0.40 b) P(Y < 0.4) = (0.4)(1) = 0.40 c) P(0.1 < Y 0.15 or 0.77 Y < 0.88) = (0.05)(1) + (0.11)(1) = 0.16 d) What important fact about continuous random variables does comparing your answer to a and b illustrate? The area of a line segment (Y = 0.4) is equal to zero Nov 29-11:16 AM 6

7 the University of Illinois found that their times for the mile run This interprets (in the context of this problem)... The probability of randomly choosing a random student who can run the mile in less than 6 minutes is approximately 6.68% or 6.7 out of 100. Nov 29-11:19 AM Professional tennis player Rafael Nadal hits the ball extremely hard. His This interprets (in the context of this problem)... The probability of randomly choosing one of Nadal's first serves that is faster than 120 mph is approximately 20.33% or 20 out of 100. Nov 29-11:21 AM 7

8 This interprets (in the context of this problem)... The probability of randomly choosing one of Nadal's first serves that is slower than mph is approximately 15% or 15 out of 100. Dec 1-9:51 PM follows a Normal distribution with mean 266 days and standard This interprets (in the context of this problem)... The probability of randomly choosing one pregnant lady that carries her baby at least 240 days is approximately 94.84% or 95 out of 100 pregnant women. Dec 4-7:12 AM 8

9 This interprets (in the context of this problem)... The probability of randomly choosing one pregnant lady that carries her baby at least days is approximately 20% or 20 out of 100 pregnant women. Dec 4-7:12 AM trucks) they own. Here is the probability model if we ignore the few households that own = 0(0.09) + 1(0.36) + 2(0.35) + 3(0.13) + 4(0.05) + 5(0.02) = 1.75 Dec 4-7:12 AM 9

10 random, which of the following would be the best interpretation of the value Dec 4-7:12 AM About what percentage of households have a number of cars within 2 Dec 4-11:45 AM 10

11 ace. Otherwise, you lose $1. If you make this wager very many times, what will be the Dec 4-11:45 AM Nov 27-10:28 PM 11

12 Nov 27-9:53 PM Nov 30-7:23 PM 12

13 Nov 30-7:23 PM Nov 30-7:23 PM 13

14 Nov 30-7:23 PM Nov 30-7:24 PM 14

15 May 11-9:19 AM Nov 30-7:22 PM 15

16 Nov 30-7:37 PM May 11-9:20 AM 16

17 Can you see why this is called a linear transformation? The equation describing the sequence of transformations has the form Y = a + bx, which you should recognize as a linear equation. Nov 30-7:39 PM Nov 28-11:07 PM 17

18 Dec 5-6:53 PM May 2-8:18 AM 18

19 new σ = old σ * b σ σ new σ = old σ σ σ σ σ σ 2 a+bx =b 2 * σ2 Dec 5-7:04 PM Dec 5-7:16 PM 19

20 Dec 5-7:16 PM 20 4 x i 4 2x i - 10 x i x i Dec 5-7:16 PM 20

21 Dec 5-7:16 PM Dec 5-7:16 PM 21

22 A large auto dealership keeps track of sales made during each hour of the day. Let X = the number of cars sold during the first hour of business on a randomly selected Friday. Based on previous records, the probability distribution of X is as follows: The random variable X has mean µ X = 1.1 and standard deviation σ X = Suppose the dealership s manager receives a $500 bonus from the company for each car sold. Let Y = the bonus received from car sales during the first hour on a randomly selected Friday. Find the mean and standard deviation of Y. 2. To encourage customers to buy cars on Friday mornings, the manager spends $75 to provide coffee and doughnuts. The manager s net profit T on a randomly selected Friday is the bonus earned minus this $75. Find the mean and standard deviation of T. Nov 30-7:28 PM Normal Curve (Continuous Random Variable) practice Scaling a Test In a large introductory statistics class, the distribution of raw scores on a test X follows a Normal Distribution with a mean of 17.2 and a standard deviation of 3.8. The professor decides to scale the scores by multiplying the raw scores by 4 and adding 10. a) Define the variable Y to be the scaled score of a randomly selected student from this class. Find the mean and the standard deviation of Y. b) What is the probability that a randomly selected student has a scaled test score of at least 90? Nov 30-8:11 PM 22

23 Dec 1-2:08 PM Sep 26-6:57 PM 23

24 Sep 26-6:58 PM Honors Statistics POP Quiz Chapter 6 Section 2 Nov 1-1:41 PM 24

25 A-Skip 35, 39, 40 Nov 21-8:16 PM Crickets The length in inches of a cricket chosen at random from a field is a random variable X with mean 1.2 inches and standard deviation 0.25 inches. Find the mean and standard deviation of the length Y of a randomly chosen cricket from the field in centimeters. There are 2.54 centimeters in an inch. Formula Centimeters = 2.54(inches) Given the following... µ x = 1.2 and σ x = 0.25 Find µ y and σ y If Y = 2.54X µ y = 2.54(1.2) = cm σ y = 2.54(0.25) = cm Nov 30-7:46 PM 25

26 Men s heights A report of the National Center for Health Statistics says that the height of a 20-year-old man chosen at random is a random variable H with mean 5.8 feet and standard deviation 0.24 feet. Find the mean and standard deviation of the height J of a randomly selected 20-year-old man in inches. There are 12 inches in a foot. Formula Inches = 12(feet) µ H = 5.8 and σ H = 0.24 Find µ J and σ J µ J = 12(5.8) = 69.6 inches σ J = 12(0.24) = 2.88 inches Dec 4-10:55 PM Get on the boat! A small ferry runs every half hour from one side of a large river to the other. The number of cars X on a randomly chosen ferry trip has the probability distribution shown below. You can check that µ X = 3.87 and σ X = > (a) The cost for the ferry trip is $5. Make a graph of the probability distribution for the random variable M = money collected on a randomly selected ferry trip. Describe its shape. > M = 5(X) 0.5 Probability > (b) Find and interpret µm. Dollars This distribution of money is NOT symmetric. It is skewed to the left. The data should be used to analyzed with the 5 number summary. µ M = 5(3.87) = $19.35 The expected money collected on a randomly selected ferry trip is $17. If many, many ferry trips are randomly selected, this is the average amount of money collected. (In the long run!!) > (c) Find and interpret σm. σ M = 5(1.29) = $6.45 The standard deviation of the mean is $6.45. The amounts collected on the ferry trips will typically differ from the mean of $19.35 by $6.45. Dec 4-10:58 PM 26

27 Exercises 39 and 40 refer to the following setting. Ms. Hall gave her class a 10- question multiple-choice quiz. Let X = the number of questions that a randomly selected student in the class answered correctly. The computer output below gives information about the probability distribution of X. To determine each student s grade on the quiz (out of 100), Ms. Hall will multiply his or her number of correct answers by 5 and then add 50. Let G = the grade of a randomly chosen student in the class. G = 5X + 50 Easy quiz (a) Find the mean of G. Show your method. µ G = 5(7.6) + 50 = 88 points (b) Find the standard deviation of G. Show your method. σ G = 5(1.32) = 6.6 points (c) How do the variance of X and the variance of G compare? Justify your answer. The variance is the square of the standard deviation. σ 2 X = (1.32) 2 = σ 2 G = 5 2 (1.32) 2 = The variance of G is 25 times the variance of X. Dec 4-11:04 PM Easy quiz G = 5X + 50 (a) Find the median of G. Show your method. Median G = 5(8.5) + 50 = 92.5 points (b) Find the IQR of G. Show your method. IQR G = 5(9-8) = 5 points (c) What shape would the probability distribution of G have? Justify your answer. The shape of G will be the same as the shape of X, because multiplying by a constant and adding a constant do not change the shape of a distribution. Dec 4-11:06 PM 27

28 Get on the boat! Refer to Exercise 37. The ferry company s expenses are $20 per trip. Define the random variable Y to be the amount of profit (money collected minus expenses) made by the ferry company on a randomly selected trip. That is, Y = M 20 Remember... µ M = $19.35 and (a) Find and interpret the mean of Y. µ Y = $ = $-0.65 (b) Find and interpret the standard deviation of Y. σ Y = $6.45 σ M = $6.45 The expected profit on a randomly selected ferry trip is $ (NOT GOOD BUSINESS) If many, many ferry trips are randomly selected, this is the average amount of profit loss. (In the long run!!) The standard deviation of the mean is $6.45. The profit on the ferry trips will typically differ from the mean of $-0.65 by $6.45. Dec 4-11:07 PM Get on the boat! Based on the analysis in Exercise 41, the ferry company decides to increase the cost of a trip to $6. We can calculate the company s profit Y on a randomly selected trip from the number of cars X. Find the mean and standard deviation of Y. Show your work. Y = 6X - 20 remember µ X = 3.87 and σ X = µ Y = 6(3.87) - 20 = $3.22 The expected money collected on a randomly selected ferry trip is $3.22. If many, many ferry trips are randomly selected, this is the average amount of money collected. (In the long run!!) σ Y = 6(1.29) = $7.74 The standard deviation of the mean is $7.74. The amounts collected on the ferry trips will typically differ from the mean of $3.22 by $7.74. Dec 4-11:08 PM 28

29 Making a profit Rotter Partners is planning a major investment. From experience, the amount of profit X (in millions of dollars) on a randomly selected investment of this type is uncertain, but an estimate gives the following probability distribution: Based on this estimate, µ x = 3 and σ X = Rotter Partners owes its lender a fee of $200,000 plus 10% of the profits X. So the firm actually retains Y = 0.9X 0.2 from the investment. Find the mean and standard deviation of Y. Show your work. µ Y = 0.9(3) = $2.5 million The expected net profit collected on a randomly selected investment $2.5 million. If many, many investments are randomly selected, this is the average amount of net profit. (In the long run!!) σ Y = 0.9(2.52) = $2.268 million The standard deviation of the mean is $2.27 million. The amounts of investment net profit will typically differ from the mean of $2.5 million by $2.27 million. Dec 4-11:08 PM Too cool at the cabin? During the winter months, the temperatures at the Starneses Colorado cabin can stay well below freezing (32 F or 0 C) for weeks at a time. To prevent the pipes from freezing, Mrs. Starnes sets the thermostat at 50 F. She also buys a digital thermometer that records the indoor temperature each night at midnight. Unfortunately, the thermometer is programmed to measure the temperature in degrees Celsius. Based on several years worth of data, the temperature T in the cabin at midnight on a randomly selected night follows a Normal distribution with mean 8.5 C and standard deviation 2.25 C. (a) Let Y = the temperature in the cabin at midnight on a randomly selected night in degrees Fahrenheit (recall that F = (9/5)C + 32). Find the mean and standard deviation of Y. 9 µ Y = ( )(8.5) + 32 = F 5 9 σ Y = ( )(2.25) = F 5 (b) Find the probability that the midnight temperature in the cabin is below 40 F. Show your work. X: y = 40 z = = P( Y < 40 ) = P(z < -1.80) This interprets (in the context of this problem)... The probability of randomly choosing one night in the cabin where the midnight temperature is less than 40 0 F is approximately 3.59% Dec 4-11:12 PM 29

30 Cereal A company s single-serving cereal boxes advertise 9.63 ounces of cereal. In fact, the amount of cereal X in a randomly selected box follows a Normal distribution with a mean of 9.70 ounces and a standard deviation of 0.03 ounces. (a) Let Y = the excess amount of cereal beyond what s advertised in a randomly selected box, measured in grams (1 ounce = grams). Find the mean and standard deviation of Y ounces = 9.63(28.35) = grams Y = 28.35X µ Y = 28.35(9.7) = grams σ Y = 28.35(0.03) = grams (b) Find the probability of getting at least 3 grams more cereal than advertised. Show your work. X: z = = y = 3 P( Y > 3 ) = P(z > 1.19) This interprets (in the context of this problem)... The probability of randomly choosing one single serving cereal box and getting at least 3 grams more than advertised is approximately 11.7% Dec 4-11:13 PM 30