Honors Statistics. Daily Agenda

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1 Honors Statistics Aug 23-8:26 PM Daily Agenda 1. Review OTL C6#4 Chapter 6.2 rules for means and variances Aug 23-8:31 PM 1

Nov 21-8:16 PM Working out Choose a person aged 19 to 25 years at random and ask, In the past

(a) Show that this is a legitimate probability distribution. 0.68 + 0.05 + 0.07 + 0.08 + 0.

Describe what you see. frequency 0.5 0.4 0.3 0.2 0.

5 6 7 Number of workout days What is the probability that a randomly selected persons aged 19

2 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 2

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?

05 + 0.07 + 0.08 + 0.05 = 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) = 0.05 + 0.07 + 0.08 + 0.05 = 0.25 The events working out at least once and working out less than 5 times per week are not INDEPENDENT events.

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.

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 0.25.

3 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 3

Running a mile the University of Illinois found that their times for the mile run were approximately Normal with mean 7.

This interprets (in the context of this problem).

7 out of 100. Nov 29-11:19 AM Professional tennis player Rafael Nadal hits the ball extremely hard.

Choose one of Nadal s first serves at random. Let Y = its speed, measured in miles per hour.

4 Running a mile the University of Illinois found that their times for the mile run were approximately Normal with mean 7.11 minutes and standard deviation 0.74 minute. Choose a student at random and interpret the result. 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 first-serve speeds follow a Normal distribution with mean 115 miles per hour (mph) and standard deviation 6 mph. Choose one of Nadal s first serves at random. Let Y = its speed, measured in miles per hour. (a) Find P(Y > 120) and interpret the result 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. (b) What is P(Y 120)? Explain. The answer is equal to P(Y > 120) Because the P(Y = 120) is zero. Nov 29-11:21 AM 4

(c) Find the value of c such that P(Y c) = 0.15. Show your work. This interprets (in the context of this problem).

Dec 1-9:51 PM The length of human pregnancies from conception to birth follows a Normal distribution with mean 266 days and standard deviation 16 days. Choose a pregnant woman at random.

5 (c) Find the value of c such that P(Y c) = Show your work. 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 The length of human pregnancies from conception to birth follows a Normal distribution with mean 266 days and standard deviation 16 days. Choose a pregnant woman at random. Let = the length of her N(266, 16) 240) and interpret the result. > 240)? Explain 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. The answer is equal to P(X > 240) Because the P(X = 240) is zero. Dec 4-7:12 AM 5

Find the value of ) = 0.20. Show your work. N(266, 16) c = 0.84(16) + 266 P( X 279.

.. The probability of randomly choosing one pregnant lady that carries her baby at least

6 Find the value of ) = Show your work. N(266, 16) c = 0.84(16) P( X ) = P(z 0.84) 0.20 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 NORMAL CURVE QUIZ Apr 28-10:26 AM 6

1. Multiple choice: Select the best answer for Exercises 27 to 30. Exercises 27 to 29 refer to the following setting.

Here is the probability model if we ignore the few households that own more than 5 cars: What s the expected number of cars in a randomly selected American household? (a) 1.

If many households were selected at random, which of the following would be the best interpretation of the value 1.08? (a) The mean number of cars would be about 1.08. (b) The number of cars would typically be about 1.

7 1. Multiple choice: Select the best answer for Exercises 27 to 30. Exercises 27 to 29 refer to the following setting. Choose an American household at random and let the random variable X be the number of cars (including SUVs and light trucks) they own. Here is the probability model if we ignore the few households that own more than 5 cars: What s the expected number of cars in a randomly selected American household? (a) 1.00 (b) 1.75 (c) 1.84 (d) 2.00 (e) 2.50 µ x = 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 2. The standard deviation of X is σ X = If many households were selected at random, which of the following would be the best interpretation of the value 1.08? (a) The mean number of cars would be about (b) The number of cars would typically be about 1.08 from the mean. (c) The number of cars would be at most 1.08 from the mean. (d) The number of cars would be within 1.08 from the mean about 68% of the time. (e) The mean number of cars would be about 1.08 from the expected value. Dec 4-7:12 AM 7

3. About what percentage of households have a number of cars within 2 standard deviations of the mean? (a) 68% (b) 71% (c) 93% (d) 95% (e) 98% C we know the following... µ x = 1.

08) = -0.41 between 0 and 3.91 cars (must use 3) 0.09 + 0.36 + 0.35 + 0.13 = 0.93 What is the probability that they have more than 5 cars????? Dec 4-11:45 AM 4.

Otherwise, you lose $1. If you make this wager very many times, what will be the mean amount you win? (a) About $1, because you will lose most of the time.

8 3. About what percentage of households have a number of cars within 2 standard deviations of the mean? (a) 68% (b) 71% (c) 93% (d) 95% (e) 98% C we know the following... µ x = 1.75 σ x = 1.08 but it is not stated that the distribution is approximately NORMAL so use the table above... mean + 2 St. Dev. = (1.08) = 3.91 mean - 2 St. Dev. = (1.08) = between 0 and 3.91 cars (must use 3) = 0.93 What is the probability that they have more than 5 cars????? Dec 4-11:45 AM 4. A deck of cards contains 52 cards, of which 4 are aces. You are offered the following wager: Draw one card at random from the deck. You win $10 if the card drawn is an ace. Otherwise, you lose $1. If you make this wager very many times, what will be the mean amount you win? (a) About $1, because you will lose most of the time. (b) About $9, because you win $10 but lose only $1. (c) About $0.15; that is, on average you lose about 15 cents. (d) About $0.77; that is, on average you win about 77 cents. (e) About $0, because the random draw gives you a fair bet. Dec 4-11:45 AM 8

choose a 3 digit number. The lottery commission announces the The "box" pays $83.

9 5. Question 5 In which setting does is P(X < 6) = P(X 6)? (a) Binomial (b) Probability (c) Symmetric (d) Continuous (e) Discrete Nov 29-11:21 AM You choose a 3 digit number. The lottery commission announces the The "box" pays $83.33 if the number you choose has the same digits as on the box. (Assume a number with three different digits is chosen) let X = Dec 5-1:10 PM 9

10 Nov 30-7:23 PM Nov 30-7:23 PM 10

11 Nov 30-7:23 PM Nov 30-7:24 PM 11

12 Nov 30-7:37 PM May 11-9:20 AM 12

13 May 11-9:19 AM May 1-8:44 AM 13

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.

14 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 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 µx = 1.1 and standard deviation σx = 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 14

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.

15 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 Dec 1-2:08 PM 15

16 Sep 26-6:57 PM Sep 26-6:58 PM 16

17 A-Skip 35, 39, 40 Nov 21-8:16 PM 17

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

Honors Statistics. 3. Review OTL C6#3. 4. Normal Curve Quiz. Chapter 6 Section 2 Day s Notes.notebook. May 02, 2016. Honors Statistics Aug 23-8:26 PM 3. Review OTL C6#3 4. Normal Curve Quiz Aug 23-8:31 PM 1 May 1-9:09 PM Apr 28-10:29 AM 2 27, 28, 29, 30 Nov 21-8:16 PM Working out Choose a person aged 19 to 25 years at