Please help me on my activity in my Mathematics subject about the topic "Measures of Position in Grouped Data. Please answer the last question and use the activity with my answer to fully answer the question appropriately in the third picture. Please provide all the informations and explanations needed. Please give accurate answers. Please give long answers so I that can maximize all the spaces. Thank you!

3. What concepts or properties do you need to solve the problems? B. Read the given article/ texts carefully. Then, complete the table and answer the questions that follow. Essential Text 1 Text 2 Text 3 Question Fifty people were surveyed The table shows the The computer terminals in a on the number of full-length frequency count of the school are labeled 1 to 300. movies watched in a year. employees' responses to the The technician wants to Number of Number of survey of employees know if the computers are in Movies Persons satisfaction level. A good working condition by Watched numerical rating of 5 is testing 50 of these units. He 19-21 randomly selects the 12th to 16-18 assigned to very satisfied 13-15 (VS), 4 to satisfied (S), 3 to be the first unit to be tested. 10-12 SO JUN- neutral (N), 2 to somewhat Which computer unit will be 7-9 dissatisfied (SD), and 1 to tested next? 4-6 very dissatisfied (VD) )-3 Level of Number of At least how many movies Satisfaction Employees should one watch to belong VS (5 25 to upper 5th percentile? S (4) 20 N (3) 13 SD (2) 4 VD (1) 3 How can What is the employees' Level issues and of Satisfaction with the problems on company? And at what your percent? respective Answer: Answer: Answer: community be solved? One should watch at least 16,25 or 17 movies to belong to ( The employees Level of Satisfaction with the company is The computer unit will be tested next is the 18th computer the upper 5th percentile, 3,92 or approx. (Satisfied) with 78.4% satisfaction rate. unit. Supporting Texts: Supporting Texts: Supporting Texts: A person should watch at least 17 movies to belong to the Based on the sample grouped data of 65 employees, the mean is 3.92 or approximately 4 which indicate as The computer terminals in a school are labeled 1 to 300. upper 5th percentile or the 95th percentile. This is because Satisfied (S) on the level of satis satisfaction. This is because I |The technician wants to test 50 units and randomly selects we used the formula "Pn= Ibpc + [ n*f/100 - of /fpc ] i" came up to my own computation and formula which i 12th to be the first unit to be tested. The second or the where lbpc= first class boundary in the located cumulative "weighted me an= VS(25) + S(20) + N(13) + SD(4) + next computer unit will be tested is the 18th unit. This is frequency, n= value we need to find, f= frequency of the VD(1) / total number o s" where VS= very because we use my built up 2 formula which is "numbers value (total number of persons), and of= cumulative satisfied, S= satisfied, N= cutral, SD= somewhat of planned to test computer units/ total computer terminals frequency of the class before the class with the value of dissatisfied, VD= very dissatisfied. the mean is 3.92 or approximately 4 which indicate as Satisfied (S) on the in schools" and "first unit to be tested + answer for the the given percentile. of satisfaction, and total number of employees= 65. first formula". Reason: Reason: Reason: First, we get the the cumulative frequency and class To determine the level of satisfaction, we simply get the boundaries to complete the table. now we solve for the For this problem, we use my built up 2 formulas and upper 5th percentile, Upper 5 entile is; 100-5=95th mean by getting the sum of multiplied frequency to its' substitute the values needed, There are 300 computer percentile. So we need to calculate the 95th percentile. We | numerical value and divided it by the total frequency form terminals in a school, Fifty units are needed to test if the use the formula "Pn= Ibpc + [ n*f/100 - of /fpc li". We the formula "weighted mean= VS(25) + S(20) + N(13) + first find the n*f/100= 95*50 (50 is from the total number omputers are in good working condition, Since the first unit of persons)= 4,750/100= 47.5. Then we locate the SD(4) + VD(1) / total number of employees". We to be tested is 12th, we need to divide first the number of cumulative frequency v r equal to 47.5. After substitute the given values to the formula. weighted mean= determining, we find the n=95, lbdo=15.5, of=47, fpc=2, computer terminals which is 300 to the numbers of units 5(25) + 4(20) + 3(13) + 2(4) + 1(1) / 65 then we get the and i=18.5-15.5=3. We substitute the values in the formula then interpret, P95= 15.5 + (47.5 - 47 / 2) 3, and the answer of 3.92 and if we round it off, we'll get the answer needed to test which is 50. It is equal to 6 as the interval - of 4 that shows the employees' level of satisfaction of of testing, after that we use the first unit to be tested answer is 16.25. So value is 16.25 is in the upper 5th percentile or the 95th perc und 16.25 up to the Satisfied, And to determine the level of satisfaction in which is 12 to be added to the interval of testing and the nearest whole number is 17 (This is be ause we can't have decimal values for number of movies as it is a discrete percent, we divided the mean to the highest value of level sum is 18 which will be the next computer unit to be tested, variable). So we must have at least 17 movies to watch to of satisfaction and multiply it by 100 then we get the belong to the upper 5th percentile answer of 78.4%. Common Ideas in Reasons: It is important to note that the observation are collected randomly, hence, there is no given biased/preference for one observation over another It's worth noting that the observations are gathered at random, so there's no prejudice or biased for one observation over another. Enduring Understanding/Generalization: Frequency counts, grouped data, s I starting of descriptive statistically and lysis g d/ compact analysis and it easier pared to ungrouped data. When opposed to ungro grouped data, served as a useful beginning point analysis given a data set. It provides a more structured/ condensed analysis and is easier/moreB. Read the given article/ texts carefully. Then, complete the table and answer the questions that follow. Essential Text 1 Text 2 Text 3 Question Fifty people were surveyed The table shows the The computer terminals in a on the number of full-length frequency count of the school are labeled 1 to 300. movies watched in a year. employees' responses to the The technician wants to Number of Number of survey of employees know if the computers are in Movies Persons satisfaction level. A good working condition by Watched testing 50 of these units. He 19-21 numerical rating of 5 is randomly selects the 12th to 16-18 assigned to very satisfied 13-15 (VS), 4 to satisfied (S), 3 to be the first unit to be tested. 10-12 SO JUN- neutral (N), 2 to somewhat Which computer unit will be 7-9 dissatisfied (SD), and 1 to tested next? 4-6 very dissatisfied (VD) )-3 Level of Number of At least how many movies Satisfaction Employees should one watch to belong VS (5 25 to upper 5th percentile? S (4) 20 N (3) 13 SD (2) 4 VD (1) 3 How can What is the employees' Level issues and of Satisfaction with the problems on company? And at what your percent? respective Answer: Answer: Answer: community be solved? One should watch at least 16,25 or 17 movies to belong to The employees Level of Satisfaction with the company is The computer unit will be tested next is the 18th computer the upper 5th percentile. 3,92 or approx. (Satisfied) with 78.4% satisfaction rate. unit. Supporting Texts: Supporting Texts: Supporting Texts: A person should watch at least 17 movies to belong to the Based on the sample grouped data of 65 employees, the mean is 3.92 or approximately 4 which indicate as The computer terminals in a school are labeled 1 to 300. upper 5th percentile or the 95th percentile. This is because Satisfied (S) on the level of satis satisfaction. This is because I |The technician wants to test 50 units and randomly selects we used the formula "Pn= lbpc + [ n*f/100 - of /fpc ] i" came up to my own computation and formula which i 12th to be the first unit to be tested. The second or the where lbpo= first class boundary in the located cumulative "weighted me an= VS(25) + S(20) + N(13) + SD(4) + next computer unit will be tested is the 18th unit. This is frequency, n= value we need to find, f= frequency of the VD(1) / total number o s" where VS= very because we use my built up 2 formula which is "numbers value (total number of persons), and of= cumulative satisfied, S= satisfied, N= cutral, SD= somewhat of planned to test computer units/ total computer terminals frequency of the class before the class with the value of dissatisfied, VD= very dissatisfied. the mean is 3.92 or approximately 4 which indicate as Satisfied (S) on the in schools" and "first unit to be tested + answer for the the given percentile. of satisfaction, and total number of employees= 65. first formula". Reason: Reason: Reason: First, we get the the cumulative frequency and class To determine the level of satisfaction, we simply get the boundaries to complete the table. now we solve for the For this problem, we use my built up 2 formulas and upper 5th percentile. Upper 5 entile is; 100-5=95th mean by getting the sum of multiplied frequency to its' substitute the values needed, There are 300 computer percentile. So we need to calculate the 95th percentile. We | numerical value and divided it by the total frequency form terminals in a school. Fifty units are needed to test if the use the formula "Pn= Ibpc + [ n*f/100 - of /fpc li". We the formula "weighted mean= VS(25) + S(20) + N(13) + first find the n*f/100= 95*50 (50 is from the total number omputers are in good working condition, Since the first unit of persons)= 4,750/100= 47.5. Then we locate the SD(4) + VD(1) / total number of employees". We to be tested is 12th, we need to divide first the number of cumulative frequency v r equal to 47.5. After substitute the given values to the formula. weighted mean= determining, we find the n=95, lbdo=15.5, of=47, fpc=2, computer terminals which is 300 to the numbers of units 5(25) + 4(20) + 3(13) + 2(4) + 1(1) / 65 then we get the and i=18.5-15.5=3. We substitute the values in the formula then interpret, P95= 15.5 + (47.5 - 47 / 2) 3, and the answer of 3.92 and if we round it off, we'll get the answer needed to test which is 50. It is equal to 6 as the interval answer is 16.25. So value is 16.25 is in the upper 5th - of 4 that shows the employees' level of satisfaction of of testing, after that we use the first unit to be tested percentile or the 95th perc und 16.25 up to the Satisfied. And to determine the level of satisfaction in which is 12 to be added to the interval of testing and the nearest whole number is 17 (This is be ause we can't have percent, we divided the mean to the highest value of level sum is 18 which will be the next computer unit to be tested, decimal values for number of movies as it is a discrete variable). So we must have at least 17 movies to watch to of satisfaction and multiply it by 100 then we get the belong to the upper 5th percentile answer of 78.4%. Common Ideas in Reasons: It is important to note that the observation are collected randomly, hence, there is no given biased/ preference for one observation over another It's worth noting that the observations are gathered at random, so there's no prejudice or biased for one observation over another. Enduring Understanding/Generalization: Frequency counts, grouped data, s I starting of descriptive statistically and lysis g red/ compact analysis and it e pared to ungrouped data. When opposed to ungro grouped data, served as a useful beginning point analysis given a data set. It provides a more structured/ condensed analysis and is easier/more