Thus, P(F or L) = P(F) + P(L) - P(F & L) = = 0.553

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1 Test 2 Review: Solutions 1) The following outcomes have at least one Head: HHH, HHT, HTH, HTT, THH, THT, TTH Thus, P(at least one head) = 7/8 2) The following outcomes have a sum of 9: (6,3), (5,4), (4,5), (3,6) Thus, P(sum of 9) = 4/36 = 1/9 3) The following outcomes include C and D: ACD, BCD, DCE Thus, P(C and D are included) = 3/10 4) = ) = ) Since a student cannot be under 21 and over 35 at the same time, outcomes for event A&B is zero. 7) P(F) = P(a student is a female) = P(L) = P(a student is left-handed) = P(F & L) = P(a student is female and left-handed) = Thus, P(F or L) = P(F) + P(L) - P(F & L) = = 0.553

2 8) Total number of people surveyed = = 148 Let M = event the person surveyed is male Let Y = event the person surveyed answered "Yes" Let N = event the person surveyed answered "No" Thus, P(M) = P(person surveyed is male = (9+15)/148; P(Y) = P(person surveyed answered "Yes") = 57/148; P(Y & M) = P(person answered "Yes" is a male) = 9/148 P(Y or M) = P(Y) + P(M) - P(Y & M) = ) See answer key 10) 1-1 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = 0 X P(X) 0 6/ /36 2 8/36 3 6/36 4 4/36 5 2/36

3 11) Total = = 617 X P(X) 0 37/ / / / /617 12) Possible outcomes: HHHH HHHT HHTH HHTT HTHH HTHT HTTH HTTT THHH THHT THTH THTT TTHH TTHT TTTH TTTT X P(X) (# of Heads) 0 1/16 1 4/16 2 6/16 3 4/16 4 1/16

4 13) Total = = 50 X P(X) 0 4/ / /50 3 5/50 4 6/50 14) = ) At least 30 minutes means 30 minutes or more. P(At least 30 minutes) = = ) = ) Mean = xp( x) = 0(0.24) + 1(0.01) (0.21) = ) Mean = xp( x) = 3(0.14) + 6(0.33) + 9(0.36) + 12(0.07) + 15(0.10) = ) Mean = xp( x) = 0(0.4096) + 1( ) + 2(0.1536) +3(0.0256) + 4(0.0016) = 0.80

5 20) A student in the survey either works full-time or does not work full-time. Thus, this is a binomial experiment with 16 trials and probability of success (or a student randomly surveyed works full time) is Mean of a binomial random variable = np = 16(0.22) = 3.52 Explanation: Suppose we repeat this experiment many, many times -- say then: For first experiment, we might have 2 out of 16 students work full time For second experiment, we might have 3 out of 16 students work full time For third experiment, we might have 4 out of 16 students work full time For 4th experiment, we might have 2 out of 16 students work full time For 5th experiment, we might have 2 out of 16 students work full time... For 1000th experiment, we might have 0 out of 16 students work full time Now calculate the average: ( )/1000 will be approximately 3.52.

6 21) When a die is rolled, the number on the upper face is either a 2 or not a 2. Thus, this is a binomial experiment with 6 trials. The probability of getting a 2 when a die is rolled is 1/6 since the die has 6 faces. Thus, p = P(getting 2) = 1/6 and n 6 ==> Mean = np = 6(1/6) = 1 Explanation: Suppose we repeat this experiment many, many times -- say then: For first experiment, we rolled the die 6 times and got no 2. For second experiment, we rolled the die 6 times and got three 2's. For third experiment, we rolled the die 6 times and got one 2. For 4th experiment, we rolled the die 6 times and got four 2's. For 5th experiment, we rolled the die 6 times and got two 2's.... For 1000th experiment, we rolled the die 6 times and got no 2. Now calculate the average: ( )/1000 will be approximately 1.

7 22) When an answer choice is selected, it is either correct or incorrect. Thus, this is a binomial experiment with 14 trials (14 questions; each question is a trial) P(of getting correct answer) = 1/4 Mean = np = (14)(0.25) = 3.5 Explanation: On average, if you guess the answer for all 14 questions on the test, you will get 3.5 questions correct. Note that you either 3 right or 4 right; no such thing as 3.5 right. Think of repeating this experiment times and then calculate the average. 23) np p 1 16(0.22)(1 0.22) ) np p 1 16(0.7)(1 0.7) ) np p 1 14(0.03)(1 0.03) 0.638

8 26) Using Table V, Area to the left of 0 is 0.5 Area = Area to the left of 3.01 is Area = Thus, area between 0 and 3.01 = = Area =

9 27) Using Table V, Area to the left of 1.13 is Area =

10 28) Using Table V, Area to the left of is Area = Area to the left of is Area = Thus, area between and = = Area =

11 29) Using Table V, Area to the left of 0.92 is Area = Area to the left of is Area = Thus, area between and 0.92 = = Area =

12 30) Using Table V, Area to the left of 1.08 is Area = Area to the left of is Area = Thus, area between and 1.08 = = Area =

13 31) Using Table V, Area to the left of 1.17 is Area = Area to the left of is Area = Thus, area between and 1.17 = = Area =

14 32) Look inside Table V and search for a value that is as close to 0.96 as possible and then find the corresponding z-score. z-score = 1.75 Area = ) Look inside Table V and search for a value that is as close to 0.40 as possible and then find the corresponding z-score. z-score = Area =

15 34) Look inside Table V and search for a value that is as close to 0.09 as possible and then find the corresponding z-score. z-score = Area = ) z-score for x = 53 is x z P(X < 53) = P(z < -1.75) = area to the left of = (Using Table V) Area = Interpretation: For a variable with population mean of 60 and population standard deviation of 4, about = 4.01% of the data values in this population are less than 53.

16 36) z-score for x = 16.1 is x z We want to find P(X > 16.1) = P(z > 1) = area to the right of 1. But Table only gives area of the left of a z-score of 1. Thus, we will find this left area and subtract it from the total area of 1. From Table V, we note that area to the left of z-score of 1 is Area = Thus, P(X > 16.1) = P(z > 1) = area to the right of 1 = = Interpretation: For a variable with population mean of 15.2 and population standard deviation of 0.9, about = 15.87% of the data values in this population are greater than 16.1.

17 37) z-score for x = 19.7 is x z Area to the left of z-score of is Area = z-score for x = 25.3 is x z Area to the left of z-score of is Area = P( 19.7 < X < 25.3) = P( < z < 1.375) = (area to the left of 1.375) - (area to the left of ) = = 0.74 Interpretation: For a variable with population mean of 22 and population standard deviation of 2.4, about 0.74 = 74% of the data values in this population are between 19.7 and 25.3.

18 38) Q1 25th percentile For this population, the mean is 293 and standard deviation is 12. We are looking for the 25th percentile. In other words, we are looking a value where 25% of the population values are less than this value. z-score corresponding to area of 0.25 to the left is (look inside Table V and find a value as close 0.25 as possible and then find correspond z-score.) Area = Now using the z-score formula: x z-score x = 12 x 293 Multiply both sides by 12: = x x x Solve for x, we get x = Hence, for this population, the data value of has a z-score of and area to the left of a z- score of is 25%. Thus, 25% of light bulbs have lifetimes less than hours.

19 39) Q3 75th percentile For this population, the mean is 2.6 and standard deviation is 0.6. We are looking for the 75th percentile. In other words, we are looking a data value in this population where 75% of the population values are less than this value. z-score corresponding to area of 0.75 to the left is 0.67 (look inside Table V and find a value as close 0.75 as possible and then find correspond z-score.) Area = Using z-score formula, x z-score x = 0.6 x 2.6 Multiply both sides by = = x 2.6 Solve for x, we get x = = x = x Hence, for this population, the data value of has a z-score of 0.67 and area to the left of a z-score of 0.67 is 75%. Thus, 75% of all GPA's are less than

20 40) Q1 25th percentile For this population, the mean is 65 and standard deviation is 11. We are looking for the 25th percentile. In other words, we are looking a data value where 25% of the population data values are less than this value. z-score corresponding to area of 0.25 to the left is (look inside Table V and find a value as close 0.25 as possible and then find correspond z-score. Area = Using z-score formula, x z-score x = 11 Multiply both sides by 11: = = x = x = x x Solve for x, we get x = Hence, for this population, the data value of has a z-score of and area to the left of a z-score of is 25%. Thus, 25% of Jen's monthly bills are less than $57.63.

21 41) 2nd Decile = 20th percentile For this population, the mean is 35.2 and standard deviation is 3.3. We are looking for the 20th percentile. In other words, we are looking a value where 20% of the population values are less than this value. z-score corresponding to area of 0.20 to the left is (look inside Table V and find a value as close 0.20 as possible and then find correspond z-score.) Area = Using z-score formula, x z-score x = 3.3 x 35.2 Multiply both sides by 3.3, we have = x x Solve for x, we get x = Hence, for this population, the data value of has a z-score of and area to the left of a z- score of is 20%. Thus, 20% of the annual precipitations are less than inches.

22 42) Third Quartile = 75th percentile For this population, the mean is 63.6 and standard deviation is 2.5. We are looking for the 75th percentile. In other words, we are looking a value where 75% of the population values are less than this value. z-score corresponding to area of 0.75 to the left is 0.67 (look inside Table V and find a value as close 0.75 as possible and then find correspond z-score.) Area = Using z-score formula, x z-score x = 2.5 Multiply both sides by 2.5, we have x = = x 63.6 x Solve for x, we get x = 65.3 Thus, 75% of all women have heights less than 65.3 inches.

23 43) n = 21; p = 0.83; q = 1 - p = = 0.17 np = (21)(0.83) = nq = (21)(0.17) = 3.57 Thus, we cannot use normal distribution since nq = (21)(0.17) = 3.57 < 5 44) ) n =39; p = 0.76; q = 1 - p = = 0.24 np = (39)(0.76) = nq = (39)(0.24) = 9.36 Thus, we can use normal distribution since np > 5 and nq > 5. We will use normal distribution with μ = np = and npq np(1 p) 39(.76)(.24) ) n = 50; p = 60% = 0.60; q = 1-p = 40% = 0.40 μ = np = (50)(0.60) = 30 and npq np(1 p) 50(.60)(.40) 3.46 We want to find P(x < 35). From page 366, we will rewrite P(X < 35) as PX ( 34). The approximate probability using normal distribution is P( X ) P( X 34.5). x The z-score for X = 34.5 is z Using Table V, z-score of 1.30 has corresponding area of to the left. Area = Hence, P( X 34.5) P( z 1.30) Thus, the probability that in a given winter month fewer than 35 cars will be rented is

24 46) n = 48; p = probability of guessing correctly = 1/4 = 0.25; q = probability of guessing incorrectly =1-p = 0.75 μ = np = (48)(0.25) = 12 and npq np(1 p) 48(0.25)(0.75) 3 We want to find probability that the student gets exactly 15 correct answers = P(X = 15) From page 366, The approximate probability using normal distribution is P( X ) P(14.5 X 15.5). x The z-score for X = 15.5 is z From Table V, z-score of has corresponding area of Area = x The z-score for X = 14.5 is z From Table V, z-score of has corresponding area of Area = Hence, P(14.5 X 15.5) = Thus, probability that the student gets exactly 15 correct answers is about 0.08.

Advantages of Sampling. Cheaper Often the only possible way Better quality control over measurement Investigation can be destructive

Advantages of Sampling. Cheaper Often the only possible way Better quality control over measurement Investigation can be destructive Soc 60 SAMPLING Advantages of Sampling Cheaper Often the only possible way Better quality control over measurement Investigation can be destructive Sample Planning Step 1. Define the Sample components