Honors Statistics. Daily Agenda

Transcription

1 Honors Statistics Aug 23-8:26 PM Daily Agenda 1. Review OTL C6#7 emphasis Normal Distributions Aug 23-8:31 PM 1

2 1. Multiple choice: Select the best answer for Exercises 65 and 66, which refer to the following setting. The number of calories in a 1-ounce serving of a certain breakfast cereal is a random variable with mean 110 and standard deviation 10. The number of calories in a cup of whole milk is a random variable with mean 140 and standard deviation 12. For breakfast, you eat 1 ounce of the cereal with 1/2 cup of whole milk. Let T be the random variable that represents the total number of calories in this breakfast. The mean of T is (a) 110 (b) 140 (c) 180 (d) 195 (e) 250 Dec 7-3:48 PM 2. Multiple choice: Select the best answer for Exercises 65 and 66, which refer to the following setting. The number of calories in a 1-ounce serving of a certain breakfast cereal is a random variable with mean 110 and standard deviation 10. The number of calories in a cup of whole milk is a random variable with mean 140 and standard deviation 12. For breakfast, you eat 1 ounce of the cereal with 1/2 cup of whole milk. Let T be the random variable that represents the total number of calories in this breakfast. The standard deviation of T is (a) 22. (b) 16. (c) (d) (e) 4. Dec 7-3:48 PM 2

3 3. T6.5. A certain vending machine offers 20-ounce bottles of soda for $1.50. The number of bottles X bought from the machine on any day is a random variable with mean 50 and standard deviation 15. Let the random variable Y equal the total revenue from this machine on a given day. Assume that the machine works properly and that no sodas are stolen from the machine. What are the mean and standard deviation of Y? > (a)µ Y = $1.50, σ Y = $22.50 > (b)µ Y = $1.50, σ Y = $33.75 > (c)µ Y = $75, σ Y = $18.37 > (d)µ Y = $75, σ Y = $22.50 > (e)µ Y = $75, σ Y = $33.75 Dec 9-9:49 PM 4. T6.4. Suppose a student is randomly selected from your school. Which of the following pairs of random variables are most likely independent? > (a)x = student s height; Y = student s weight > (b)x = student s IQ; Y = student s GPA > (c)x = student s PSAT Math score; Y = student s PSAT Verbal score > (d)x = average amount of homework the student does per night; Y = student s GPA > (e)x = average amount of homework the student does per night; Y = student s height Dec 9-9:48 PM 3

4 Nov 21-8:16 PM Rainy days Imagine that we randomly select a day from the past 10 years. Let X be the recorded rainfall on this date at the airport in Orlando, Florida, and Y be the recorded rainfall on this date at Disney World just outside Orlando. Suppose that you know the means µ X and µ Y and the variances and of both variables. (a) Is it reasonable to take the mean of the total rainfall X+ Y to be µ X + µ Y? Explain your answer. It is reasonable to take the mean of the total rainfall because the rainfall two cities that are close in location does not depend on independence. You can always combine means of data sets. (b) Is it reasonable to take the variance of the total rainfall to be Explain your answer. It is NOT reasonable to take the variance of the total rainfall because the rainfall of two closely located cities will NOT be independent. Dec 6-11:13 PM 4

5 Get on the boat! Refer to Exercise 41. Find the expected value and standard deviation of the total amount of profit made on two randomly selected days. Show your work. Big assumption... two randomly selected days are INDEPENDENT so... go ahead and calculate! Remember... µ Y = $ = $-0.65 σ Y = $6.45 Use the formula T = Y 1 + Y 2 µ T = $ $-0.65 = $-1.30 Remember you can never add STANDARD DEVIATIONS. σ Y = $6.45 so σ 2 Y = (6.45) 2 = σ 2 T = σ 2 Y1 + σ 2 Y2 σ 2 T = = σ T = = $9.12 Dec 6-11:13 PM Essay errors Typographical and spelling errors can be either nonword errors or word errors. A nonword error is not a real word, as when the is typed as teh. A word error is a real word, but not the right word, as when lose is typed as loose. When students are asked to write a 250-word essay (without spell-checking), The number of nonword errors X in a randomly selected essay has the following probability distribution: The number of word errors Y has this probability distribution: Assume that X and Y are independent. An English professor deducts 3 points from a student s essay score for each nonword error and 2 points for each word error. Find the mean and standard deviation of the total score deductions for a randomly selected essay. Show your work. Use Formula D = 3(X) + 2(Y) D = deductions, X = "non-word error" and Y = word error µ D = 3(2.1) + 2(1.0) = 8.3 points deducted The expected number of points deducted on two randomly selected essays is 8.3 points. If many, many essays are randomly selected, this is the average amount of point deductions on two essays. (In the long run!!) Remember you can never add STANDARD DEVIATIONS. σ X = so σ 2 X = (1.136) 2 = σ Y = 1.0 so σ 2 Y = (1.0) 2 = 1.0 σ 2 D = (3) 2 σ 2 X + (2) 2 σ 2 Y = 9(1.136) 2 + 4(1.0) 2 = σ D = = points The standard deviation of the mean is 3.95 points. The deductions on the 2 essays will typically differ from the mean of 8.3 by 3.95 points. Dec 6-11:14 PM 5

6 = = -1.1 The expected difference in number of points (for word vs. non-word errors) deducted on a randomly selected essay is -1.1 points. If many, many essays are randomly selected, -1.1 points is the average amount of difference between word vs non-word errors point (In the long run!!)... Fewer word errors than non-word errors σ 2 D = σ 2 Y + σ 2 X σ 2 D = (1.136) 2 + (1) 2 = σ D = = 1.51 If one essay is randomly selected the difference in error types (word vs. non-word errors) will typically differ from the mean of -1.1 points by 1.51 points The results in the yellow triangle make up the event a randomly selected student makes more word errors than non-word errors. P( (Y-X) > 0) = = 0.15 Dec 6-11:14 PM The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures academic motivation and study habits. The distribution of SSHA scores among the women at a college has mean 120 and standard deviation 28, and the distribution of scores among male students has mean 105 and standard deviation 35. You select a single male student and a single female student at random and give them the SSHA test. Find the mean and standard deviation of the difference (female minus male) between their scores. Interpret each value in context. Use Formula D = X - Y = = 15 points The expected number of points a randomly selected female outscores a randomly selected male on the SSHA test is 15 points. If many, many essays are randomly selected, 15 points is the average amount of points a female outscores a male in the SSHA (In the long run!!) 2009 = The standard deviation of the mean is points. The amount of points in which the female outscores the males will typically differ from the mean of 15 by points. From the information given, can you find the probability that the woman chosen scores higher than the man? If so, find this probability. If not, explain why you cannot. If we new the shape of the distribution (specifically if it was normally distributed) then we could answer the question. Currently we do not have enough information. Dec 6-11:14 PM 6

7 Essay scores Refer to Exercise 51. Find the mean and standard deviation of the difference in score deductions (nonword word) for a randomly selected essay. Show your work. Use Formula D = 3(X) - 2(Y) D = deductions, X = "non-word error" and Y = word error µ D = 3(2.1) - 2(1.0) = 4.3 points deducted If many, many essays are randomly selected, the expected difference in score deduction will average 4.3 points. Remember you can never Remember you can never subtract VARIANCES. = so = (1.136) = = 1.0 so = (1.0) = 9(1.136) = = = points The standard deviation of the mean is 3.95 points. The deductions on a randomly selected essay will typically differfrom the mean of 8.3 by 3.95 points. Jan 27-1:35 PM Exercises 57 and 58 refer to the following setting. In Exercises 14 and 18 of Section 6.1,we examined the probability distribution of the random variable X = the amount a life insurance company earns on a randomly chosen 5-year term life policy. Calculations reveal that µ X = $ and σ X = $ Life insurance The risk of insuring one person s life is reduced if we insure many people. Suppose that we insure two 21-year-old males, and that their ages at death are independent. If X1 and X2 are the insurer s income from the two insurance policies, the insurer s average income W on the two policies is Find the mean and standard deviation of W. (You see that the mean income is the same as for a single policy but the standard deviation is less.) µ W = $ $ = $ Remember you can never add STANDARD DEVIATIONS. σ X = $ so σ 2 x = ( ) 2 = σ 2 W = σ 2 X1 + σ 2 X2 σ 2 W = ( ) 2 + ( ) σ 2 W = σ W = = $ Nov 30-7:46 PM 7

8 Life insurance If four 21-year-old men are insured, the insurer s average income is where X i is the income from insuring one man. Assuming that the amount of income earned on individual policies is independent, find the mean and standard deviation of V. (If you compare with the results of Exercise 57, you should see that averaging over more insured individuals reduces risk.) µ V = $ $ = $ Remember you can never add STANDARD DEVIATIONS. σ X = $ so σ 2 x = ( ) 2 = σ 2 V = σ 2 X1 + σ 2 X2 + σ 2 X3 + σ 2 X4 4 2 σ 2 V = ( ) 2 + ( ) 2 + ( ) 2 + ( ) 2 σ 2 V = σ V = = $ Dec 6-11:13 PM = 7 pounds = so σ = pounds = 0.12 = 21.6 pounds Jan 22-7:07 PM 8

9 or σ 2 B = σ M = (0.15)2 (50) 2 = σ B = ~ 7.5 dollars Jan 22-7:07 PM Nov 30-7:23 PM 9

10 Nov 30-7:23 PM Dec 1-2:08 PM 10

11 Sep 26-6:57 PM Sep 26-6:58 PM 11

12 Dec 9-9:54 AM Dec 11-7:07 PM 12

13 Dec 11-7:05 PM May 4-9:19 AM 13

14 (done in class) Nov 21-8:16 PM 14