2. What is common, recurring, or involved in all three problems? B. Read the given article/texts carefully. Then, complete the table and answer the questions that follow,EssentialQuestionText 1Text 2Text 3The computer terminals in aschool are labeled 1 to 300,The technician wants toknow if the computers are ingood working condition bytesting 50 of these units. Herandomly selects the 12th tobe the first unit to be tested.Which computer unit will betested next?Fifty people were surveyedThe table shows theon the number of full-length frequency count of theemployees' responses to thesurvey of employees'satisfaction level. Anumerical rating of 5 isassigned to very satisfied(VS), 4 to satisfied (S), 3 toneutral (N), 2 to somewhatdissatisfied (SD), and 1 tovery dissatisfied (VD).Number ofEmployces25movies watched in a year.Number ofMoviesWatched19-2116-1813-15Number ofРегsons15710-127-984-60-3At least how many moviesshould one watch to belongto upper 5th percentile?1215Level ofSatisfactionVS (5)S (4)N (3)SD (2)VD (1)What is the employees' Levelof Satisfaction with the201343How canissues andproblems oncompany? And at whatpercent?Answer:yourrespectivecommunitybe solved?Answer:Answer:Ore should watch at legst 1625 or 17 movies to beloe to The employees Level of Satisfaction with the company is The computer unit will be tested next is the 18th comouterthe upper 5th percentile.3,92 or aprox, (Satified) with 78,4% saifaction rate. unit.Supporting Texts:Supporting Texts:Supporting Texts:A person should wateh at least 17 movies to beleng to theBased on the aample groupad data of 65 amployeea, theThe computer termnale in a achool are labeled 1 to 300.mean is 3.92 or approximately 4 which indioate asupper Sth percentile or the 95th percentile. This is beoause Sotiefied (S) on the level of sutiefaotion This les because Iwe used the formula Pr bpc +In*t/100 - of /fpe 1iThe technician wanta to teet 50 urita and randomly sokectaeome up to my awn eomputotion and formula which la12h to be the first unit to be tasted, The second or thewhere Ibpen first alass boundary in the loated cumulative weighted means VS(25) + K(20) + NC13) + SD14) +frequeney, n value we need to find, fa frequeney of thevalue (total number of persons), and ef= oumulativenext computer unit will be tested is the 18th unit This isVD1) / total number of employees where VS= verybecause we use my buit up 2 formala which is umbereatiafind, S- autitafied, N- netral, SD- aomwhatof planned to test computer unita total computer termhaledissatisfied, VD= very dissatisfled. the mean is 392 orappreximately 4 which indicate as Saliefied (S) en thelevel of sutisfuction, ond total number of employeas 65,frequency of the class before the class with the value ofn schools" and "first unit to be tested + anawer for thethe given perentie.first formula".Reason:Reason:First, we get the the cumalative frequency and classReason:To determine the level of satisfaction, we simply get theFor this problem, we use my buit up 2 formulas andbourdariea ta oomplete the table, now we solve for theupper Sth percentile, Upper Sth percentile is 100-5-95thmean by getting the sum of multiplied frequency to ita'aubstitute the values needed, There are 300 oomputerpercentile. So we reed to colesiate the 95th percentile. We numerical value and divided it by the total frequency formuse the formula "Pr Ibpo +Entf/100 - ef fpa 11", Wefirst find the n*1/100 95*5o cso is from the total numberterminale in a school Fifty units are reeded to test if thethe formula "weighted mean= VS(25) + S(20) + N(13) +computers are in good working condition. Since the first unitSD(4) + VD(1) / total number of employees". Weof persons)a 4,750/100- 47.5. Then we loeate theto be tested is 12th, we need to divide firet the number ofeumulotive frequeney value grester or equol to 47.5. After substitute the given vales to the formula, weighted meandutermining, wa find the n=95, lbde-15.5, ef-47, fpe-2eomputer terminals which is 300 to the numbers of unitsneeded to test which is 50, It is equal to 6 as the intervalof testing, ater that we use the first unit to be tested,5(25) + 4(20) + 3(13) + 2(4) + 1(1) / 65 then we get theand i=18.5-15.5-3. We subetitute the vahes in the formulaanswer of 3.92 and if we round it off, we'l get the answer- of 4 that shows the employees' level of satiefoction ofthen inturprut, P95- 15.5 (47,5- 47 / 2)3, and theanswer is 16.25. So vakue is 16.25 is in the upper 5thpercentile or the 95th parcenntile round 16.25 up te thenearest whole number is 17 Tie io because we oon't have percent, we divided the mean to the highest value of levelwhich is 12 to be added to the interval of testing and theSatisfind, And to determine the level of satiefaction insum is 18 which will be the next computer unit to be testeddecimcal valum tar number of movina axt is a dierninvartable). So we must have at least 17 movies to watch to of satisfaction and multiply it by 100 then we get thehalang tn the pypar fith peronliln,anawer 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 noprejudice or biased for one observation over another.Enduring Understanding/Generalization:Frequery couts, grouped data, serve as pood starting of desriptivestatistically analysis given a data set it gives more structured/ compactnalysis and it easinr /mure convenient to use as compared to ungroupeddata When oppused to ungroaped tlata, fruuency cuurta, BrOuperd data, servedas a uselul beginning point for deseriptive statiratically analysis given a datahet. It provideu a more structured/ condensed analysin and is uusier/more

Question

2. What is common, recurring, or involved in all three problems? B. Read the given article/texts carefully. Then, complete the table and answer the questions that follow,EssentialQuestionText 1Text 2Text 3The computer terminals in aschool are labeled 1 to 300,The technician wants toknow if the computers are ingood working condition bytesting 50 of these units. Herandomly selects the 12th tobe the first unit to be tested.Which computer unit will betested next?Fifty people were surveyedThe table shows theon the number of full-length frequency count of theemployees' responses to thesurvey of employees'satisfaction level. Anumerical rating of 5 isassigned to very satisfied(VS), 4 to satisfied (S), 3 toneutral (N), 2 to somewhatdissatisfied (SD), and 1 tovery dissatisfied (VD).Number ofEmployces25movies watched in a year.Number ofMoviesWatched19-2116-1813-15Number ofРегsons15710-127-984-60-3At least how many moviesshould one watch to belongto upper 5th percentile?1215Level ofSatisfactionVS (5)S (4)N (3)SD (2)VD (1)What is the employees' Levelof Satisfaction with the201343How canissues andproblems oncompany? And at whatpercent?Answer:yourrespectivecommunitybe solved?Answer:Answer:Ore should watch at legst 1625 or 17 movies to beloe to The employees Level of Satisfaction with the company is The computer unit will be tested next is the 18th comouterthe upper 5th percentile.3,92 or aprox, (Satified) with 78,4% saifaction rate. unit.Supporting Texts:Supporting Texts:Supporting Texts:A person should wateh at least 17 movies to beleng to theBased on the aample groupad data of 65 amployeea, theThe computer termnale in a achool are labeled 1 to 300.mean is 3.92 or approximately 4 which indioate asupper Sth percentile or the 95th percentile. This is beoause Sotiefied (S) on the level of sutiefaotion This les because Iwe used the formula Pr bpc +In*t/100 - of /fpe 1iThe technician wanta to teet 50 urita and randomly sokectaeome up to my awn eomputotion and formula which la12h to be the first unit to be tasted, The second or thewhere Ibpen first alass boundary in the loated cumulative weighted means VS(25) + K(20) + NC13) + SD14) +frequeney, n value we need to find, fa frequeney of thevalue (total number of persons), and ef= oumulativenext computer unit will be tested is the 18th unit This isVD1) / total number of employees where VS= verybecause we use my buit up 2 formala which is umbereatiafind, S- autitafied, N- netral, SD- aomwhatof planned to test computer unita total computer termhaledissatisfied, VD= very dissatisfled. the mean is 392 orappreximately 4 which indicate as Saliefied (S) en thelevel of sutisfuction, ond total number of employeas 65,frequency of the class before the class with the value ofn schools&#34; and &#34;first unit to be tested + anawer for thethe given perentie.first formula&#34;.Reason:Reason:First, we get the the cumalative frequency and classReason:To determine the level of satisfaction, we simply get theFor this problem, we use my buit up 2 formulas andbourdariea ta oomplete the table, now we solve for theupper Sth percentile, Upper Sth percentile is 100-5-95thmean by getting the sum of multiplied frequency to ita'aubstitute the values needed, There are 300 oomputerpercentile. So we reed to colesiate the 95th percentile. We numerical value and divided it by the total frequency formuse the formula &#34;Pr Ibpo +Entf/100 - ef fpa 11&#34;, Wefirst find the n*1/100 95*5o cso is from the total numberterminale in a school Fifty units are reeded to test if thethe formula &#34;weighted mean= VS(25) + S(20) + N(13) +computers are in good working condition. Since the first unitSD(4) + VD(1) / total number of employees&#34;. Weof persons)a 4,750/100- 47.5. Then we loeate theto be tested is 12th, we need to divide firet the number ofeumulotive frequeney value grester or equol to 47.5. After substitute the given vales to the formula, weighted meandutermining, wa find the n=95, lbde-15.5, ef-47, fpe-2eomputer terminals which is 300 to the numbers of unitsneeded to test which is 50, It is equal to 6 as the intervalof testing, ater that we use the first unit to be tested,5(25) + 4(20) + 3(13) + 2(4) + 1(1) / 65 then we get theand i=18.5-15.5-3. We subetitute the vahes in the formulaanswer of 3.92 and if we round it off, we'l get the answer- of 4 that shows the employees' level of satiefoction ofthen inturprut, P95- 15.5 (47,5- 47 / 2)3, and theanswer is 16.25. So vakue is 16.25 is in the upper 5thpercentile or the 95th parcenntile round 16.25 up te thenearest whole number is 17 Tie io because we oon't have percent, we divided the mean to the highest value of levelwhich is 12 to be added to the interval of testing and theSatisfind, And to determine the level of satiefaction insum is 18 which will be the next computer unit to be testeddecimcal valum tar number of movina axt is a dierninvartable). So we must have at least 17 movies to watch to of satisfaction and multiply it by 100 then we get thehalang tn the pypar fith peronliln,anawer 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 noprejudice or biased for one observation over another.Enduring Understanding/Generalization:Frequery couts, grouped data, serve as pood starting of desriptivestatistically analysis given a data set it gives more structured/ compactnalysis and it easinr /mure convenient to use as compared to ungroupeddata> When oppused to ungroaped tlata, fruuency cuurta, BrOuperd data, servedas a uselul beginning point for deseriptive statiratically analysis given a datahet. It provideu a more structured/ condensed analysin and is uusier/more

Accepted Answer

A percentile is a statistical measure used to express the percentage of observations falling below…