CMTH 380, SAMPLE TEST 1. The example of writing the test.

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Feb 20, 2024

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1 CMTH 380, SAMPLE TEST Duration: 3 hours Last Name ____________________________________First Name: ___________________________ Student ID Number _________________________ Total mark: 50 1. (3 marks) The birth weights of 25 full - term newborn babies born at a hospital are recorded to the nearest tenth of pound. 7.2, 8.0, 8.2, 5.8, 5.9, 8.5, 7.8, 8.2, 7.7, 6.8, 7.9, 6.8, 5.6, 7.5, 6.8, 9.4, 7.7 5.7, 8.6, 7.2, 9.0, 6.7, 7.1, 7.5, 7.7 Construct stem and leaf plot for this data set by using each number in the ones place twice. Indicate the leaf unit. 5 5 8 9 6 7 5 6 7 8 9 6 6 6 8 8 8 7 6 7 8 8 8 7 2 2 1 7 1 2 2 7 8 7 9 5 7 5 7 7 5 5 7 7 7 8 9 8 0 2 2 8 0 2 2 8 5 6 8 5 6 9 4 0 9 0 4 9 Leaf unit = 0.1

2 2. (3 marks) The ages in (months) at which 25 children were first enrolled in a preschool are listed below. 31, 32, 32, 33, 34, 34 , 35, 35, 36, 36, 36, 37, 38, 39, 40, 41, 41, 42, 44 , 46, 47, 49, 50 , 56, 58 Construct frequency and relative frequency distribution for this data set with the class width 5. Insert required values into the statements below the table. Class interval Frequency Relative frequency 30 and under 35 6 0.24 35 and under 40 8 0.32 40 and under 45 5 0.20 45 and under 50 3 0.12 50 and under 55 1 0.04 55 and under 60 2 0.08 25 1.00 Lower boundary of the first class = LBFC = _30_____ Upper boundary of the last class = UBLC = __60____ The total of all frequencies is ___25_____ The total of all relative frequencies is __1______

3 3. ( 6 marks) Fifteen randomly selected university students were asked to state the number of hours they slept the previous night (rounded the nearest hour). The resulting data values were 5, 6, 6, 8, 7, 7, 9 , 5, 4, 8, 11, 6, 7, 8, 7 Find the mean, median, mode, range, variance and standard deviation. In the answers for the mean, variance, and standard deviation leave one digit after decimal point . You are not required to show your work in finding the mean, median, mode, range, variance, and standard deviation. The data set arranged in ascending order: 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 11 a. Mean = ___6.9_____ b. Median = ____7_____ c. Mode = _____7____ d. Range = ___7_____ e. Variance = ___3.1______ f. Standard deviation = __1.8____ The data set arranged in ascending order : 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 11

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4 4. (4 marks) Fill in missing values in the following statements. The distribution of the survival times of a group of laboratory animals being infected with a particular form of bacteria is known to be skewed with the mean of 36 days and standard deviation of 3 days. a. According to Tchebysheff’s Theorem, at least 75% of the survival times will fall into the interval between ___ 30 ________ days and _____ 42 _________ days. b. At least ____ 89 ______ percent of survival times fall between 27 days and 45 days. c. At most ____ 25 ______ percent of survival times are larger than 42 days. d. At most ____ 11_ _____ percent of survival times are smaller than 27 days or larger than 45 days. You are not required to show your work to state the answers for all the questions.

5 5. (3 marks) Fill in missing values in the following statements. Breathing rates for humans can be as low as 4 breaths per minute or as high as 75 breaths per minute. The distribution of resting breathing rates for young people of 18-25 years old is mound-shaped and symmetrical with the mean of 12 and standard deviation of 2.3 breaths per minute. According Empirical rule a. __ 95 _____ percent of young people have breathing rates from 7.4 minutes to 16.6 minutes . b. _ 81.5 ____ percent of young people have breathing rates from 9.7 minutes to 16.6 minutes. c. 2.5_ ____ percent of young people have breathing rates larger than 16.6 breaths per You are not required to show your work to state the answers for all the questions.

6 6. (3 marks) Write missing values in the statements a. and b. Write proper names in the statement c. Alex and David are taking college preparatory exams. Alex takes the SAT test, and scores 1240 out of 1600, while the David takes the ACT test and scores 30 out of 36. If the means and standard deviations of scores for the SAT test are 1000 and 170 , while the means and standard deviations of scores for the ACT test are 20 and 5 , then a. Alex’s z -score is ___ 1.41 _____________; b. David’s z -score is ___ 2.00_ ____________; c. _David_ ____________ performed on exam better than ______ Alex ________ You are not required to show your work to state the answers for all the questions.

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7 7. (2 marks) Write missing values in the statements a. and b. a. If Alex has scored 505 on the MCAT, and was at the 75 th percentile, and Boris has scored 515, and was at the 90 th percentile, then approximately __ 85__ _______% of all the students who wrote MCAT had the score lower than Alex, but higher than Boris. b. If 15% of U.S. women have the height larger than 5. 7 feet, and 40% of U.S. women have the weight less than 165 pounds, then 165 is __ 40 ______ th percentile and 5.7 is ____ 85 ____ th percentile. You are not required to show your work to state the answers for all the questions.

8 8. (5 marks) The following are amounts (rounded to the nearest $) spent by a sample of 24 customers in a grocery store on a certain day in June 2020. 30 45 85 100 40, 43, 50, 52, 58, 60, 60 , 64, 64, 65 ,65, 66, 66 , 67, 68, 69 ,69, 70, 70 , 74,79, 81, 91, 102 a. Find the median, the first and the third quartiles, and the interquartile range. b. Find the left outer fence, left inner fence, right inner fence, and right outer fence. c. Find the length of left whisker and the length of right whisker in the box – whisker plot. d. List all outliers (if any) in the last row. If there are no outliers state “none”. List all extreme values (if any) in the last row. If there are no extreme values state “none”. Place calculated value in the table given below. a. Q 1 = 60 Median = 66 Q 3 = 70 IQR =10 b. LOF = 30 LIF = 45 RIF = 85 ROF = 100 c. LLW = 10 LRW = 11 d. Outliers: 40, 43, 91 Extreme values: 102 Show how LOF, LIF, RIF, and ROF are calculated. LOF = 60 - 3(10) =30 LIF = 60 – 1.5(10) =45 RIF = 70 + 1.5(10) = 85 ROF = 70 + 3(10) = 100 Show how LLW and LRW are calculated. LLW = Q1 – smallest number between inner fences = 60 -50 = 10 LRW = the largest number between inner fences – Q3 = 81 -70 = 11 1 mark will be subtracted if required work will not be shown.

9 9. (4 marks) Three children are selected at random, and their genders are recorded. Assume that boys (b) and girls (g) are equally likely. a. The sample space of random experiment is S = {bbb, bgg, gbg, ggb, gbb, bgb, bbg, ggg} List outcomes in the following events and find their probabilities. b. The event “there are two girls and one boy in the group” is B ={bgg, gbg, ggb} The probability of event B is __ 3/8 =0.3750 __________________ c. The event “ at least two boys is in the group” is C = { bbg, bgb, gbb, bbb} The probability of event C is _ 1/2 = 0.5000 ___________________ d. The event “ at most one boy is in the group” is D = {ggg, bgg, gbg, ggb} The probability of event D is _ 1/2 = 0.5000_ __________________

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10 10. (3 marks) An unbalanced six – side die is rolled once and the number on the top of the die id observed. Suppose that probability to get number 1 is twice as large to get number 2, and probability to get number 2 is twice as large as probability to get each of the numbers from 3 to 6. Let A be the event “number 3 is obtained on the top of the die” Let B be the event “number 2 is obtained on the top of the die” Let C be the event “number 1 is obtained on the top of the die” Let D be the event “even number is obtained on the top of the die”, let E be the event “the number obtained on the top of the die is smaller than 3”. a. The probability of event A is __ 0.10 ______. b. The probability of event B is __ 0.20 ____. c. The probability of event C is _ 0.40 _______. d. The probability of event D is __ 0.40 ______. e. The probability of event E is __ 0.60 ______. Show calculations for finding probability of event A and B, and state their values above horizontal line below horizontal line: S = {1,2,3,4,5,} Suppose that p 3 = p 4 = p 5 = p 6 = p. Then p 2= 2p, and p 1= 2(2p) = 4 p 4 p + 2p + p + p + p + p =10p =1, and p = 0.10 A = {3), P(A) = p 3 = p = 0.10 B = {2}, P(B) = p 2 = 2 p = 0.20 C = {1), P(B) = p 3 = 4 p = 0.40 D = {2, 4, 6}, P(D) = 2 p + p + p = 4p= 0.40 E = {1,2), P(E) = 2p + 4p = 6p = 0.60 ON THE TEST YOU WILL NOT BE REQUIRED TO FIND PROBABILITIES OF SO MANY EVENTS. This is done for training purposes.

11 11. (3 marks) A committee of 3 people is to be selected at random from 5 males and 7 females. a. Probability that there will be exactly 2 males and 1 female on the committee is __7/22 = 0.3182 b. Probability that the committee will be comprised of females only is___7/44 = 0.1591 c. Probability that no females are selected is _____1/22 = 0.0455 ON THE TEST YOU MAY BE REQUIRED TO SHOW YOUR WORK IN FINDING ONLY SOME PROBABILITIES. P(A) = 5C2x7C1/ 12C3 = 10x7/220 = 7/22 = 0.3182 P(B) = 5C0x7C3/ 12C3 = 1x35/220 = 7/44 = 0.1591 P(A) = 5C3x7C0/ 12C3 = 10x1/220 = 1/22 = 0.0455

12 12. (3 marks) Four horses # 1, # 2, # 3, and # 4 are in a 1 -kilometer race. Assume that all of the horses are equally qualified . A is the event that horse # 1 comes the first, B is the event that horse # 1 comes the first and horse # 4 comes the last, C is the event that horse # 1 comes the first, horse # 2 come the second, horse # 3 comes the third, and horse # 4 comes the last. Show calculations of P(A), P(B), and P(C), state the obtained values. a. P(A) = 1x3P3 /4P4 = 1x3!/ 4! = 1/4 b. P(B) = 1x2P2x1 /4P4 = 1x2!x1/ 4! = 2/24 = 1/12 = 0.0833 c. P(C) = 1x1x1x1 /4P4 = 1/ 4! = 1/24 = 0.0417 a. Probability that horse #1 comes the first is __1/4 = 0.2500_________. b. Probability that horse # 1 comes the first and horse # 4 comes the last is _1/12 = 0.0833. c. Probability that the horses will come in order #1, #2, #3, and #4 is_1/24 = 0.0417___

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13 13. (4 marks) A pair of balanced dice is rolled once and the numbers on the top of the dice are observed. Find the following probabilities and represent common fractions as common fractions in the lowest terms or as the decimal fractions with 4 digits after decimal point. You are not required to show your work for this problem a. Probability that the sum of numbers on the tops of the dice is 8 is__ 5/36 = 0. 1389 b. Probability that the numbers on the top of the dice are the same is _ 6/36 =1/6 = 0.1667 Probability that the numbers on the top of the dice are different is _ 30/36 =5/6 = 0.8333 c. Probability that the number on the first dice is larger than 3 and the number on the second die is larger that 4 is _ 6/36 =1/6 = 0.1667 d. Probability that the pair of numbers on the tops of the dice contains at least one digit 6 is 11/36 = 0.3056 S = { (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) ( 6,6) } .

14 14. (3 marks) a. You play card game and select 3 cards from 52 - card deck with replacement (you replace the card after each selection and shuffle the deck). Probability that 3 cards selected by you are different is _0.9430 b. You play card game and select 3 cards from 52 - card deck with replacement (you replace the card after each selection and shuffle the deck). Probability that 3 cards selected by you are the same is 0.4000 c. You play card game and select 3 cards from 52 - card deck without replacement . Probability that 3 cards selected by you are face cards is ___0. 0010 There are 3 face cards in each of 4 suits. d. You play card game and select 3 cards from 52 - card deck without replacement . Probability that 3 cards selected by you are from the same suit is _ 0.0518 There are 13 face cards in each of 4 suits e. Probability that 3 cards selected by you are hearts is _ 0. 0129 a. P(A) = 52 x 51 x 50/ 52x 52 x 52 = 51 x 50/ 52 x 52 = 0.9430 b. P(A) = 52 x 1 x 1/ 52x 52 x 52 = 1 x 1/ 52 x 52 = 0. 00037 = 0.0004 c. P(C) = 12C3 x 40C0/ 52C3 = 220 x 1/22100 = 0.00995= 0.0010 d. P(D) = 4x 13C3 / 52C3 = 4 x 286/22100 = 4 x 0.01294 = 0.0518 e. P(E) = 13C3x39C0 / 52C3 = 286/22100 = 0.0129 ON THE TEST YOU MAY BE REQUIRED TO SHOW YOUR WORK IN FINDING PROBABILITIES.

CMTH 380, SAMPLE TEST 1. The example of writing the test.

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