Chapter 6 Section Review day s.notebook. May 11, Honors Statistics. Aug 23-8:26 PM. 3. Review team test.

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1 Honors Statistics Aug 23-8:26 PM 3. Review team test Aug 23-8:31 PM 1

2 Nov 27-10:28 PM Nov 27-9:53 PM 2

3 May 8-7:44 PM May 1-9:09 PM 3

4 Dec 1-2:08 PM Sep 26-6:57 PM 4

5 Sep 26-6:58 PM Nov 30-7:23 PM 5

6 Nov 30-7:23 PM 6 If many, many emergency rooms are randomly selected during a given hour the number of patients seeking treatment for the flu will average 1.87 persons 1.09 is the standard deviation of the mean. If an emergency room is randomly selected, the number of people seeking treatment for the flu any given hour will typically differ from the average value 1.87 by 1.09 persons. Dec 7-11:53 PM 6

7 µ X = 5 σ X = 2 X + 3 µ X+3 = = 8 σ X+3 = 2 3X µ 3X = 3(5) = 15 σ 3X = 3(2)= 6 3X - 4 µ 3X-4 = 3(5) - 4 = 11 σ 3X-4 = 3(2) = 6 May 8-8:08 PM 3 Dec 6-2:56 PM 7

8 Dec 8-11:07 AM 5. X = the money value of the change found in the car µ X = 1(0.60) + 5(0.30) + 20(0.10) = 4.10 If the boy randomly scrounges for many, many years the amount of cash found will average $4.10 per year. This is the expected value of the cash recovered. σ 2 x = (1-4.1) 2 (0.60) + (5-4.1) 2 (0.30) + (20-4.1) 2 (0.10) = σ x = = $5.59 The standard deviation of X is σ x = $5.59 The amount of cash in a randomly selected year will typically differ from the mean ($4.10) by about $5.59. T = Y 1 + Y 2 + Y 3 µ T = = $12.30 σ 2 T = = σ T = = $9.69 Dec 10-5:25 PM 8

9 µ X = cm σ x = 7.2 cm I = X 2.54 µ I = = in σ I = 7.2 = 2.83 in Dec 6-2:59 PM May 8-8:21 PM 9

10 May 8-8:21 PM 8 Dec 7-11:55 PM 10

11 May 8-8:22 PM May 8-8:22 PM 11

12 Nov 21-8:16 PM Section I: Multiple Choice Select the best answer for each question. Questions T6.1 to T6.3 refer to the following setting. A psychologist studied the number of puzzles that subjects were able to solve in a five-minute period while listening to soothing music. Let X be the number of puzzles completed successfully by a randomly chosen subject. The psychologist found that X had the following probability distribution: T6.1. What is the probability that a randomly chosen subject completes more than the expected number of puzzles in the five-minute period while listening to soothing music? > (a) 0.1 > (b) 0.4 > (c) 0.8 > (d) 1 µ X = 1(0.2) + 2(0.4) + 3(0.3) + 4(0.1) = 2.3 > (e) Cannot be determined P(X = 3 or X = 4) = 0.4 Dec 9-9:43 PM 12

13 T6.2. The standard deviation of X is 0.9. Which of the following is the best interpretation of this value? D > (a) About 90% of subjects solved 3 or fewer puzzles. > (b) About 68% of subjects solved between 0.9 puzzles less and 0.9 puzzles more than the mean. > (c) The typical subject solved an average of 0.9 puzzles. > (d) The number of puzzles solved by subjects typically differed from the mean by about 0.9 puzzles. > (e) The number of puzzles solved by subjects typically differed from one another by about 0.9 puzzles. Dec 9-9:47 PM T6.3. Let D be the difference in the number of puzzles solved by two randomly selected subjects in a five-minute period. What is the standard deviation of D? D > (a) 0 > (b) 0.81 > (c) 0.9 > (d) 1.27 µ X = 1(0.2) + 2(0.4) + 3(0.3) + 4(0.1) = 2.3 > (e) 1.8 σ 2 x = (1-2.3) 2 (0.2) + (2-2.3) 2 (0.4) + (3-2.3) 2 (0.3) + (4-2.3) 2 (0.1) = 0.81 σ x = 0.81 = 0.9 D = X - X σ 2 D = = 1.62 σ D = 1.62 = Dec 9-9:46 PM 13

14 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 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 Y = 1.50(X) µ Y = 1.50(50) = $75 σ Y = 1.5(15) = $22.5 Dec 9-9:49 PM 14

15 T6.6. The weight of tomatoes chosen at random from a bin at the farmer s market follows a Normal distribution with mean µ = 10 ounces and standard deviation σ = 1 ounce. Suppose we pick four tomatoes at random from the bin and find their total weight T. The random variable T is (a) Normal, with mean 10 ounces and standard deviation 1 ounce. (b) Normal, with mean 40 ounces and standard deviation 2 ounces. (c) Normal, with mean 40 ounces and standard deviation 4 ounces. (d) binomial, with mean 40 ounces and standard deviation 2 ounces. (e) binomial, with mean 40 ounces and standard deviation 4 ounces. µ T = = 40 ounces σ 2 T = = 4 σ T = 4.0 = 2 ounces B Dec 9-9:50 PM T6.11 Let Y denote the number of broken eggs in a randomly selected carton of one dozen store brand eggs at a local supermarket. Suppose that the probability distribution of Y is as follows. > (a) What is the probability that at least 10 eggs in a randomly selected carton are unbroken? P(Y 2) = P(Y=0) + P(Y=1) + P(Y=2) = = 0.96 > (b) Calculate and interpret µy µy = 0(0.78) + 1(0.11) + 2(0.07) + 3(0.03) = 0.38 If many, many egg cartons were randomly selected the number of broken eggs in the carton would average 0.38 broken eggs. > (c) Calculate and interpret σy. Show your work. σ 2 Y = (0.38-0) 2 (0.78) + (0.38-1) 2 (0.11) + (0.38-2) 2 (0.07) + (0.38-3) 2 (0.03) + (0.38-4) 2 (0.01) = σ Y = = If a carton of eggs is randomly selected, the number of broken eggs in the carton will typically differ from the average 0.38 by 0.82 eggs. > (d) A quality control inspector at the store keeps looking at randomly selected cartons of eggs until he finds one with at least 2 broken eggs. Find the probability that this happens in one of the first three cartons he inspects. THIS QUESTION IS A PREVIEW OF SECTION 6.3 P( at least two broken eggs) =P(Y=2) + P(Y=3) + P(Y=4) = = 0.11 P(less than 2 broken eggs) = = 0.89 Let X be the number of cartons selected until you find one with at least 2 broken eggs. P(X = 1) = 0.11 P(X = 2) = (0.89)(0.11) = P(X = 3) = (0.89)(0.89)(0.11) = P(X 3) = = Dec 9-9:50 PM 15