\n50-60<\/td>\n | 80 – 75=5<\/td>\n | 5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Modal class has maximum frequency (30) in class 30 – 40.<\/p>\n Assertion and Reason Based MCQs<\/span><\/p>\nDirections: In the following questions, A statement of Assertion (A) is followed by a statement of of A Reason (R). Mark the correct choice as. \n(A) Both A and R are true and R is the correct explanation \n(B) Both A and R are true but R is NOT the correct explanation of A \n(C) A is true but R is false \n(D) A is false and R is True<\/p>\n Question 1.<\/p>\n Assertion (A): Consider the following data:<\/h2>\n\n\n\nClass<\/td>\n | Frequency<\/td>\n<\/tr>\n | \n65-85<\/td>\n | 4<\/td>\n<\/tr>\n | \n85 -105<\/td>\n | 5<\/td>\n<\/tr>\n | \n105 -125<\/td>\n | 13<\/td>\n<\/tr>\n | \n125 -145<\/td>\n | 20<\/td>\n<\/tr>\n | \n145 -165<\/td>\n | 14<\/td>\n<\/tr>\n | \n165-185<\/td>\n | 7<\/td>\n<\/tr>\n | \n185 – 205<\/td>\n | 4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nThe difference of the upper limit of the median class and the lower limit of the modal class is 20. \nReason (R): The median class and modal class of grouped data always be different.<\/h2>\nAnswer: \n(C) A is true but R is false \nExplanation: \nIn case of assertion :<\/p>\n \n\n\nClass<\/td>\n | Frequency<\/td>\n | Cumulative frequency<\/td>\n<\/tr>\n | \n65-85<\/td>\n | 4<\/td>\n | 4<\/td>\n<\/tr>\n | \n85-105<\/td>\n | 5<\/td>\n | 9<\/td>\n<\/tr>\n | \n105-125<\/td>\n | 13<\/td>\n | 22<\/td>\n<\/tr>\n | \n125-145<\/td>\n | 20<\/td>\n | 42<\/td>\n<\/tr>\n | \n145-165<\/td>\n | 14<\/td>\n | 56<\/td>\n<\/tr>\n | \n165-185<\/td>\n | 7<\/td>\n | 63<\/td>\n<\/tr>\n | \n185-205<\/td>\n | 4<\/td>\n | 67<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Hence, n= 67 (odd) \nso, Median = \\(\\frac{67+1}{2}\\) = 34 \n34 lies in class 125 – 145 and upper limit is 145. \nNow, the maximum frequency is 20 and it lies in class 125 – 145 (modal class). \nLower limit of modal class = 125 \nHence, the reuired difference 145 – 125 = 20. \n\u2234 Assertion is correct. \nIn case of reason : \nThe median andmodal class may be same. If modal class is median class which is not always possible as the number of trequencies may be maximum in any class. \nSo, given dtatement is not true. \n\u2234 Reason is incorrect. \nHence, assertion is correct but reason is incorrect.<\/p>\n <\/p>\n Question 2.<\/p>\n Assertion (A): Consider the data:<\/h2>\n\n\n\nClass<\/td>\n | 4-7<\/td>\n | 8-11<\/td>\n | 12-15<\/td>\n | 16-19<\/td>\n<\/tr>\n | \nFrequency<\/td>\n | 5<\/td>\n | 4<\/td>\n | 9<\/td>\n | 10<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nThe mean of the above data is 12.93. \nReason (R): The following table gives the number of pages written by Sarika for completing her own book for 30 days:<\/h2>\n\n\n\nClass<\/td>\n | 16-18<\/td>\n | 19-21<\/td>\n | 22-24<\/td>\n | 25-27<\/td>\n | 28-30<\/td>\n<\/tr>\n | \nFrequency<\/td>\n | 1<\/td>\n | 3<\/td>\n | 4<\/td>\n | 9<\/td>\n | 13<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nThe mean of the above date is 26<\/h2>\nAnswer: \n(B) Both A and R are true but R is NOT the correct explanation of A \nExplanation: \nIn case of assertion : \nClass marks of these classes are same ,so no need to convert given date continuous.<\/p>\n \n\n\nClass<\/td>\n | Class marks(xi<\/sub>)<\/td>\nmarks(xi<\/sub>) di<\/sub> = (xi<\/sub> – a)<\/td>\nFrequency (fi<\/sub>)<\/td>\nfi<\/sub>di<\/sub><\/td>\n<\/tr>\n\n4-7<\/td>\n | 5.5<\/td>\n | -4<\/td>\n | 5<\/td>\n | -20<\/td>\n<\/tr>\n | \n8-11<\/td>\n | 9.5 = a<\/td>\n | 0<\/td>\n | 4<\/td>\n | 0<\/td>\n<\/tr>\n | \n12-15<\/td>\n | 13.5<\/td>\n | +4<\/td>\n | 9<\/td>\n | 36<\/td>\n<\/tr>\n | \n16-19<\/td>\n | 17.5<\/td>\n | +8<\/td>\n | 10<\/td>\n | 80<\/td>\n<\/tr>\n | \n<\/td>\n | <\/td>\n | <\/td>\n | \u03a3fi<\/sub> = 28<\/td>\n\u03a3fi<\/sub>di<\/sub> = 96<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n a = Assumed mean, di = Deviation fro mean \nMean, \\( \\bar{x}\\) = a + \\(\\frac{\\Sigma f_{i} d_{i}}{\\Sigma f_{i}}\\) = 9.5 + \\(\\frac{96}{28}\\) = 9.5 + 3.43 \n\u2234\\( \\bar{x}\\) = 12.93 \n\u2234 Hence, the mean = 12.93 \n\u2234Assertion is correct. \nIn case of reason: \nNo need to change the class intervals of continuous intervals as class marks of continuous and discontinuous classes are same. di is deviation from assumed mean.<\/p>\n \n\n\nClass interval<\/td>\n | Midvalue<\/td>\n | di<\/sub> =(xi<\/sub> – a)<\/td>\nNumber of days<\/td>\n | fi<\/sub>di<\/sub><\/td>\n<\/tr>\n\n16-18<\/td>\n | 17<\/td>\n | -6<\/td>\n | 1<\/td>\n | -6<\/td>\n<\/tr>\n | \n19-21<\/td>\n | 20<\/td>\n | \u00a0 -3<\/td>\n | 3<\/td>\n | -9<\/td>\n<\/tr>\n | \n22-24<\/td>\n | a = 23<\/td>\n | 0<\/td>\n | 4<\/td>\n | 0<\/td>\n<\/tr>\n | \n25-27<\/td>\n | 26<\/td>\n | 3<\/td>\n | 9<\/td>\n | 27<\/td>\n<\/tr>\n | \n28-30<\/td>\n | 29<\/td>\n | 6<\/td>\n | 13<\/td>\n | 78<\/td>\n<\/tr>\n | \n<\/td>\n | <\/td>\n | <\/td>\n | \u03a3fi<\/sub> = 30<\/td>\n\u03a3fi<\/sub>di<\/sub> = 90<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n a = Assumed mean = 23 \nMean, \\( \\bar{x}\\) = a + \\(\\frac{\\Sigma f_{i} d_{i}}{\\Sigma f_{i}}\\) = 23 + \\(\\frac{90}{30}\\) = 23 + 3 = 26 \n\u2234\\( \\bar{x}\\) = 26 \nHence, the mean of pages written per day is 26. \n\u2234Reason is correct. \nHence, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n Question 3.<\/p>\n Assertion (A): If the median of a series exceeds the mean by 3, then mode exceeds mean by 10. \nReason (R): If made = 12.4 and mean = 10.5, then the median is 11.13.<\/h2>\nAnswer: \n(D) A is false and R is True<\/p>\n Explanation: \nIn case of assertion: \nGiven, median = mean + 3 \nSince, Mode = 3 Median – 2 Mean \n= 3 (Mean + 3) – 2 Mean \n\u21d2 Mode = Mean + 9 \nHence, mode exceed mean by 9. \n\u2234 Assertion is correct. \nIn case of reason: \nMedian = \\(\\frac{1}{3}\\) Mode + \\(\\frac{2}{3}\\) Mean \n= \\(\\frac{1}{3}\\) (12 . 4) + \\(\\frac{2}{3}\\) (10.5) \n\\(\\frac{12.4}{3}\\) + \\(\\frac{21}{3}\\) \n= \\(\\frac{12.4+21}{3}\\) = \\(\\frac{33.4}{3}\\) \n= \\(\\frac{33.4}{3}\\) = 11.13 \n\u2234 Reason is correct. \nHence, Assertion is incorrect but reason is correct.<\/p>\n Case – Based MCQs<\/span><\/p>\nAttempt any four sub-parts from each question. Each sub-part carries 1 mark. \nI. Read the following text and answer the following question on the basis of the same: \nCOVID-19 Pandemic The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) among humans. \n \nThe following tables shows the age distribution of case admitted during a day in two different hospitals<\/p>\n \n \nQuestion 1.<\/p>\n
The averge age for which maximum cases occurred is<\/h2>\n(A) 32.24 \n(B) 34.36 \n(C) 36.82 \n(D) 42.24 \nAnswer. \n(C) 36.82 \nExplanation: \nSince, highest frequency is 23. So, modal class is 35 – 45 \nNow, Mode = l +\\(\\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}} \\) \u00d7 h \nHere, l =35, h = 10, fi<\/sub> = 23, f0 = 21, f2 = 14, \n\u21d2 Mode = 35 + \\(\\frac{23-21}{46-21-14}\\) \u00d7 10 \n= 35 + \\(\\frac{2}{11}\\) \u00d7 10 \n= 35 + \\(\\frac{20}{11}\\) \n= 35 + 1.81 \n= h 36.818 = 36.82<\/p>\nQuestion 2.<\/p>\n The upper limit of modal class is<\/h2>\n(A) 15 \n(B) 25 \n(C) 35 \n(D) 45 \nAnswer: \n(D) 45<\/p>\n Question 3.<\/p>\n The mean of the given data is<\/h2>\n(A) 26.2 \n(B) 32.4 \n(C) 33.5 \n(D) 35.4 \nAnswer: \n(D) 35.4<\/p>\n Explanation:<\/p>\n \n\n\nAge (in years)<\/td>\n | Class marks – (xi<\/sub>)<\/td>\nfrequency (fi<\/sub>)<\/td>\nDeviation di<\/sub>= (xi<\/sub> – a)<\/td>\nfi<\/sub>di<\/sub><\/td>\n<\/tr>\n\n5-15<\/td>\n | 10<\/td>\n | 6<\/td>\n | -3<\/td>\n | -15<\/td>\n<\/tr>\n | \n15-25<\/td>\n | 20<\/td>\n | 11<\/td>\n | 6<\/td>\n | \u00a066<\/td>\n<\/tr>\n | \n25-35<\/td>\n | 30<\/td>\n | 21<\/td>\n | \u00a0 16<\/td>\n | 336<\/td>\n<\/tr>\n | \n35-45<\/td>\n | 40<\/td>\n | 23<\/td>\n | 26<\/td>\n | 598<\/td>\n<\/tr>\n | \n45-55<\/td>\n | 50<\/td>\n | 14 \u2192 a<\/td>\n | 1<\/td>\n | 18<\/td>\n<\/tr>\n | \n55-65<\/td>\n | 60<\/td>\n | 5<\/td>\n | 46<\/td>\n | 46<\/td>\n<\/tr>\n | \n<\/td>\n | <\/td>\n | \u03a3fi<\/sub> = n = 80<\/td>\n<\/td>\n | \u03a3fi<\/sub>di<\/sub> = 1,716<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Now, MEan (\\( \\bar{x}\\)) a + \\( \\bar{x}\\) \n= 14 + \\( \\bar{x}\\) \n= 14 + 21.45 \n= 35. 45 \nRefer to table 2<\/p>\n <\/p>\n Question 4.<\/p>\n The mode of the given data is<\/h2>\n(A) 41.4 \n(B) 48.2 \n(C) 55.3 \n(D) 64.6 \nAnswer: \n(A) 41.4 \nQuestion 5.<\/p>\n The median of the given data is<\/h2>\n(A) 32.7 \n(B) 40.2 \n(C) 42.3 \n(D) 48.6 \nAnswer: \n(B) 40.2<\/p>\n Explanation:<\/p>\n \n\n\nAge (in years)<\/td>\n | frequency (fi<\/sub>) (No. of cases)<\/td>\nC.f.<\/td>\n<\/tr>\n | \n5-15<\/td>\n | 8<\/td>\n | 8<\/td>\n<\/tr>\n | \n15-25<\/td>\n | 16<\/td>\n | 24<\/td>\n<\/tr>\n | \n25-35<\/td>\n | 10<\/td>\n | 34<\/td>\n<\/tr>\n | \n35-45<\/td>\n | 42 (frequency)<\/td>\n | 76 (Nearest to \\(\\frac{n}{2}\\)))<\/td>\n<\/tr>\n | \n45-55<\/td>\n | 24<\/td>\n | 100<\/td>\n<\/tr>\n | \n55-65<\/td>\n | 12<\/td>\n | 112<\/td>\n<\/tr>\n | \n<\/td>\n | \u03a3fi<\/sub> = n = 12<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Now, \\(\\frac{n}{2}\\) = \\(\\frac{112}{2}\\) =56 \nl = 35 (lower mlimit of median class) \nCf = 34 (Preceding to median class) \nHere, Madian = l + \\(\\left(\\frac{\\frac{n}{2}-c f}{f}\\right)\\) \u00d7 h \n= 35 + \\(\\left(\\frac{56-34}{42}\\right)\\) \u00d7 10 \n= 35 + \\(\\left(\\frac{22}{42}\\right)\\) \u00d7 10 \n= 35 + \\(\\left(\\frac{11}{21}\\right)\\) \u00d7 10 \n= 35 + \\(\\frac{110}{21}\\) \n= 40.25<\/p>\n II. Read the following text and answer the following question on the basis of the same: \nElectricity Energy ConsumptionElectricity energy consumption is the form of energy consumption that uses electric energy. Global electricity consumption continues to increase faster than world population, leading to an increase in the average amount of electricity consumed per person (per capita electricity consumption).<\/p>\n \n\n\nTariff\u00a0\u00a0 : LT – Residential<\/td>\n | Bill Number : 384756′<\/td>\n<\/tr>\n | \nType of Supply : Single Passes<\/td>\n | Connected lead : 3 kW<\/td>\n<\/tr>\n | \nMater Reading : 31-11-13 Date<\/td>\n | Mater Reading : 65700<\/td>\n<\/tr>\n | \nPrevious Reading : 31-10-13 Date<\/td>\n | Previous Mater : 65500 Reading<\/td>\n<\/tr>\n | \n<\/td>\n | Units consumed : 289<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n A survey is conducted for 56 families of a Colony A. The following tables gives the weekly consumption of electricity of these families.<\/p>\n \n\n\nWeekly consumption (in units)<\/td>\n | 0-10<\/td>\n | 10-20<\/td>\n | 20-30<\/td>\n | 30 – 40<\/td>\n | 40-50<\/td>\n | 50-60<\/td>\n<\/tr>\n | \nNo. of families<\/td>\n | 16<\/td>\n | 12<\/td>\n | 18<\/td>\n | 6<\/td>\n | 4<\/td>\n | 0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\nWeekly consumption (in units)<\/td>\n | 0-10<\/td>\n | 10-20<\/td>\n | 20-30<\/td>\n | 30-40<\/td>\n | 40-50<\/td>\n | 50-60<\/td>\n<\/tr>\n | \nNo. of families<\/td>\n | 0<\/td>\n | 5<\/td>\n | 10<\/td>\n | 20<\/td>\n | 40<\/td>\n | 5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Refer to data received from Colony A<\/p>\n Question 1.<\/p>\n The median weekly consumption is<\/h2>\n(A) 12 units \n(B) 16 units \n(C) 20 units \n(D) None of these \nAnswer: \n(C) 20 units<\/p>\n Explanation:<\/p>\n \n\n\nWeekly consumption (in units)<\/td>\n | frequency (fi<\/sub>) (No. of families)<\/td>\nC.f.<\/td>\n<\/tr>\n | \n0 – 10<\/td>\n | 8<\/td>\n | 16<\/td>\n<\/tr>\n | \n10-20(Medianclass)<\/td>\n | 12 (frequency)<\/td>\n | 28(Nearest to \\(\\frac{n}{2}\\))<\/td>\n<\/tr>\n | \n20-30<\/td>\n | 18<\/td>\n | 46<\/td>\n<\/tr>\n | \n30-40<\/td>\n | 6<\/td>\n | 52<\/td>\n<\/tr>\n | \n40-50<\/td>\n | 4<\/td>\n | 56<\/td>\n<\/tr>\n | \n50-60<\/td>\n | 0<\/td>\n | 56<\/td>\n<\/tr>\n | \n<\/td>\n | \u03a3fi<\/sub> = n = 56<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Now, \\(\\frac{n}{2}\\) = \\(\\frac{56}{2}\\) = 28 \nl = 10, Cf = 16 ,f = 12, h = 10 \nHere, \nMedian = l + \\(\\left(\\frac{\\frac{n}{2}-c f}{f}\\right)\\) \u00d7 h \n= 10 + \\(\\left(\\frac{28-16}{12}\\right)\\) \u00d7 10 \n= 10 + \\(\\left(\\frac{12}{12}\\right)\\) \u00d7 10 \n= 10 + 10 \n= 20 \nHence, median weekiy consumption = 20 units.<\/p>\n <\/p>\n Question 2.<\/p>\n The mean weekly consumption is<\/h2>\n(A) 19.64 units \n(B) 22.5 units \n(C) 26 units \n(D) None of these \nAnswer: \n(A) 19.64 units<\/p>\n Question 3.<\/p>\n The modal class of the above data is I<\/h2>\n(A) 0 – 10 \n(B) 10 – 20 \n(C) 20 – 30 \n(D) 30-40 \nAnswer: \n(C) 20 – 30 \nRefer to data received from Colony B<\/p>\n Question 4.<\/p>\n The modal weekly consumption is<\/h2>\n(A) 38.2 units \n(B) 43.6 units \n(C) 26 units \n(D) 32 units \nAnswer: \n(B) 43.6 units<\/p>\n Question 5.<\/p>\n The mean weekly consumption is<\/h2>\n(A) 15.65 units \n(B) 32.8 units \n(C) 38.75 units \n(D) 48 units \nAnswer: \n(C) 38.75 units<\/p>\n III. Read the following text and answer the following question on the basis of the same: \nThe weights (in kg) of 50 wrestlers are recorded in the following table : \n \n<\/p>\n Question 1<\/p>\n What is the upper limit of modal class.<\/h2>\n(A) 120 \n(B) 130 \n(C) 100 \n(D) 150 \nAnswer: \n(B) 130<\/p>\n Explanation: \nModal Class = 120 – 130 \nUpper limit = 130<\/p>\n Question 2.<\/p>\n What is the mode class frequency of the given data<\/h2>\n(A) 21 \n(B) 50 \n(C) 25 \n(D) 80 \nAnswer: \n(A) 21<\/p>\n Explanation: \nMode class frequency of the given data is 21.<\/p>\n Question 3.<\/p>\n How many wrestlers weights have more than 120 kg weight?<\/h2>\n(A) 32 \n(B) 50 \n(C) 16 \n(D) 21 \nAnswer: \n(A) 32<\/p>\n Explanation: \nNo. of wrestlers with more than 120 kg weight = 21 + 8 + 3 = 32<\/p>\n <\/p>\n Question 4.<\/p>\n What is the class mark for class 130 – 140?<\/h2>\n(A) 120 \n(B) 130 \n(C) 135 \n(d) 150 \nAnswer: \n(C) 135<\/p>\n Explanation: \nFor class mark of 130 -140, \n= \\(\\frac{130+140}{2}\\) \n= \\(\\frac{270}{2}\\) = 135<\/p>\n Question 5.<\/p>\n Which method is more suitable to find the mean of the above data ?<\/h2>\n(A) Direct method \n(B) Assumed mean method \n(C) Step-Deviation method \n(D) None of these \nAnswer: \n(C) Step-Deviation method \nIV Read the following text and answer the following question on the basis of the same: \nThe maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows. \n \n<\/p>\n Question 1.<\/p>\n What is the modal class of the given data?<\/h2>\n(A) 85 – 100 \n(B) 100 – 115 \n(C) 115 – 130 \n(D) 130 – 145 \nAnswer: \n(A) 85 – 100<\/p>\n Explanation: \nModal class is the class with highest frequency i.e., 85 – 100<\/p>\n Question 2.<\/p>\n What is the value of class interval for the given data set?<\/h2>\n(A) 10 \n(B) 15 \n(C) 5 \n(D) 20 \nAnswer: \n(B) 15<\/p>\n Explanation:<\/p>\n The value of class interval = 100 – 85 = 15 \nagain =115-100 = 15 \nand = 130-115 = 145 – 130<\/p>\n Question 3.<\/p>\n What is the median class of the given data?<\/h2>\n(A) 85-100 \n(B) 100-115 \n(C) 115 – 130 \n(D) 130 – 145 \nAnswer: \n(B) 100-115<\/p>\n Explanation: \nN = Number of observations = 33 \nMedian of 33 observations = 17.5th<\/sup> observation, which in class 100 – 115<\/p>\n<\/p>\n Question 4.<\/p>\n What is the median of bowling speed?<\/h2>\n(A) 109.17 km\/hr (Approx) \n(B) 109.71 km\/hr (Approx) \n(C) 107.17 km\/hr (Approx) \n(D) 109.19 km\/hr (Approx) \nAnswer: \n(A) 109.17 km\/hr (Approx)Explanation: \nMedian = l + \\(\\frac{\\left(\\frac{n}{2}-c . f\\right)}{f}\\) \u00d7 h \nl = 100, f = 9, c.f. = , h = 100 – 85 = 15 \nMedian = l + \\(\\frac{\\left(\\frac{n}{2}-c . f\\right)}{f}\\) \u00d7 h \n= 100 + \\(\\frac{\\left(\\frac{33}{2}-11\\right)}{9}\\) \u00d7 15 \n= \\(\\frac{100+(16.5-11)}{9 \\times 15}\\) \n= 100 + \\(\\frac{5.5 \\times 15}{9}\\) \n= 100 + \\(\\frac{82.5}{9}\\) \n= 100 + 9.166 \n= 109.17 km\/h (Approx) \nHence, the median bowling sppeed is 109. 17 km\/h (Approx)<\/p>\n Question 5.<\/p>\n What is the sum of lower limit of modal class and upper limit of median class?<\/h2>\n(A) 100 \n(B) 200 \n(C) 300 \n(D) 400 \nAnswer: \n(B) 200<\/p>\n Explanation: \nLower limit of modal class = 85 and upper limit of median class =115 sum = 85 + 115 200V. Read the following text and answer the following question on the basis of the same: \n100 m RACE A stopwatch was used to find the time that it took a group of students to run 100 m. \n \n<\/p>\n Question 1.<\/p>\n Estimate the mean time taken by a student to finish the race.<\/h2>\n(A) 54 \n(B) 63 \n(C) 43 \n(D) 50 \nAnswer: \n(C) 43<\/p>\n Explanation:<\/p>\n \n\n\nTime (in sec)<\/td>\n | x<\/td>\n | f<\/td>\n | cf<\/td>\n | fx<\/td>\n<\/tr>\n | \n0 – 20<\/td>\n | 10<\/td>\n | 8<\/td>\n | 8<\/td>\n | 80<\/td>\n<\/tr>\n | \n20 \u2013 40<\/td>\n | 30<\/td>\n | 10<\/td>\n | 18<\/td>\n | 300<\/td>\n<\/tr>\n | \n40 \u2013 60<\/td>\n | 50<\/td>\n | 13<\/td>\n | 31<\/td>\n | 650<\/td>\n<\/tr>\n | \n60 \u2013 80<\/td>\n | 70<\/td>\n | 6<\/td>\n | 37<\/td>\n | 420<\/td>\n<\/tr>\n | \n80 \u2013 100<\/td>\n | 90<\/td>\n | 3<\/td>\n | 40<\/td>\n | 270<\/td>\n<\/tr>\n | \nTotal<\/td>\n | <\/td>\n | 40<\/td>\n | <\/td>\n | 1720<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Mean = \\(\\frac{1720}{40}\\) = 43<\/p>\n Question 2.<\/p>\n What will be the upper limit of the modal class?<\/h2>\n(A) 20 \n(B) 40 \n(C) 60 \n(D) 80 \nAnswer: \n(C) 60<\/p>\n Explanation: \nModal class = 40-60 \nUpper limit = 60<\/p>\n <\/p>\n Question 3.<\/p>\n The construction of cumulative frequency table is useful in determining the<\/h2>\n(A) Mean \n(B) Median \n(C) Mode \n(D) All of the above \nAnswer: \n(B) Median<\/p>\n Explanation: \nThe construction of c.f. table is useful in determining the median.<\/p>\n Question 4.<\/p>\n The sum of lower limits of median class and modal class is<\/h2>\n(A) 60 \n(B) 100 \n(C) 80 \n(D) 140 \nAnswer: \n(C) 80<\/p>\n Explanation: \nMedian class = 40-60 \nModal class = 40-60 \nTherefore, the sum of the lower limits of median and modal class = 40 + 40 = 80<\/p>\n <\/p>\n Question 5.<\/p>\n How many students finished the race within 1 minute?<\/h2>\n(A) 18 \n(B) 37 \n(C) 31 \n(D) 8 \nAnswer: \n(C) 31<\/p>\n Explanation: \nNumber of students who finished the race within 1 minute = 8 + 10 + 13 = 31<\/p>\n | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |