{"id":87835,"date":"2021-06-17T11:52:44","date_gmt":"2021-06-17T06:22:44","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=87835"},"modified":"2021-07-02T15:06:42","modified_gmt":"2021-07-02T09:36:42","slug":"1st-puc-economics-question-bank-chapter-6-part-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-economics-question-bank-chapter-6-part-1\/","title":{"rendered":"1st PUC Economics Question Bank Chapter 6 Measures of Dispersion"},"content":{"rendered":"

Karnataka 1st PUC Economics Question Bank Chapter 6 Measures of Dispersion<\/h2>\n

1st PUC Economics Measures of Dispersion TextBook Questions and Answers<\/h3>\n

I. Choose the correct answers<\/span><\/p>\n

Question 1.
\nInter quartile range is based upon middle _______of the values in a distribution.
\n(a) 30%
\n(b) 20%
\n(c) 50%
\n(d) 25%
\nAnswer:
\n(c) 50%<\/p>\n

Question 2.
\nFor QD, we need to calculate values of Q3<\/sub> and
\n(a) Q2<\/sub>
\n(b) Q4<\/sub>
\n(c) Q1<\/sub>
\n(d) Q0<\/sub>
\nAnswer:
\n(c) Q1<\/sub>.<\/p>\n

Question 3.
\nCoefficient of range = _______
\n(a) \\(\\frac{\\mathbf{L}-\\mathbf{S}}{\\mathbf{L}+\\mathbf{S}}\\)
\n(b) \\(\\frac{\\mathbf{R}-\\mathbf{L}}{\\mathbf{R}+\\mathbf{L}}\\)
\n(c) \\(\\frac{\\mathbf{L}+\\mathbf{S}}{\\mathbf{L}-\\mathbf{S}}\\)
\n(d) None of the these
\nAnswer:
\n(a) \\(\\frac{\\mathbf{L}-\\mathbf{S}}{\\mathbf{L}+\\mathbf{S}}\\)<\/p>\n

Question 4.
\nWhat is the graphical measure available to measure dispersion?
\n(a) Lorenz curve
\n(b) Marshal curve
\n(c) Production curve
\n(d) None of the these
\nAnswer:
\n(a) Lorenz curve<\/p>\n

II. Answer the following questions in a word or sentence<\/span><\/p>\n

Question 1.
\nWhat is dispersion? (S – 2018) (N – 2018)
\nAnswer:
\nDispersion refers to the extent to which values in a distribution differ from the average of the distribution<\/p>\n

Question 2.
\nHow do range and quartile deviation measure the dispersion?
\nAnswer:
\nRange and quartile deviation measure the dispersion by calculating the spread within which the values lie<\/p>\n

Question 3.
\nHow do means deviation and standard deviations measure dispersion?
\nAnswer:
\nThe mean deviation and standard deviations measure dispersion by calculating the extent to which the values differ from the average.<\/p>\n

Question 4.
\nGive the meaning of standard deviation.
\nAnswer:
\nStandard deviation is the positive square root of the mean of squared deviations from the main<\/p>\n

Question 5.
\nWrite the formula to calculate the coefficient of variation.
\nAnswer:
\nThe coefficient of variation can be calculated with the help of the following formula:
\nCoefficient of variation = \\(\\frac{\\text { Standard deviation }}{\\text { Arithmetic mean }}\\) \u00d7 100<\/p>\n

III. Answer the following questions in about four sentences<\/span><\/p>\n

Question 1.
\nWhich quartiles are used to calculate the interquartile range? Give the formula of it.
\nAnswer:
\nTire upper and lower quartiles are used to calculate interquartile range. It is calculated by using the formula.
\nInterquartile range = Q3<\/sub> – Q1<\/sub><\/p>\n

Question 2.
\nWhat is the mean deviation? Which average is used to find it? (N – 2018)
\nAnswer:
\nMean deviation is the arithmetic mean of the differences of the values from their average. The average used is either arithmetic mean or median.<\/p>\n

IV. Answer the following questions in about twelve sentences<\/span><\/p>\n

Question 1.
\nThe yield of wheat per acre for 10 districts of a state are as follows:
\n\"1st
\nCalculate quartile deviation (QD)
\nAnswer:
\nQD for Wheat
\nFirst, we need to arrange the observations in tire ascending order
\nx: 9 10 10 12 15 16 18 19 21 25
\nCalculation of lower quartile (Q1<\/sub>)
\nQ1<\/sub> = Size of \\(\\left(\\frac{N}{4}\\right)^{\\text {th }}\\) item = Size of \\(\\left(\\frac{10+1}{4}\\right)^{t h}\\) item
\n= size of \\(\\left(\\frac{11}{4}\\right)^{\\text {th }}\\) item
\ni.e. 2.75th item<\/p>\n

The size of 2.75th item is size of 2nd item + 0.75 item
\nSize of 2nd item =10
\nSize of 0.75 item = Size of 3rd item – 2nd item)
\n= 10 – 10 = 0
\n\u2234 Q1<\/sub> = size of 2nd item + 075 item is
\nQ1<\/sub> = 10 + 0.75(10 – 10)
\n= 10 + 0.75(0)= 10 + 0
\nQ1<\/sub> = 10<\/p>\n

Calculation of Upper Quartile (Q3<\/sub>)
\nQ3<\/sub> = Size of 3\\(\\left(\\frac{N+1}{4}\\right)^{\\text {th }}\\) item
\n= Size of 3\\(\\frac{3(10+1)^{\\text {th }}}{4}\\) item = Size of \\(\\left(\\frac{3(11)}{4}\\right)^{th}\\) item
\n= Size of \\(\\left(\\frac{33}{4}\\right)^{t h}\\) item = Size of 8.25th item<\/p>\n

i.e.. size of 8th item + 0.25 item (size of 9th – 8th item)
\nQ3<\/sub> = 19 + 0.25(21 – 19)
\nQ3<\/sub> = 19 + 0.25(2)
\nQ3<\/sub> = 19 + 0.5
\nQ3<\/sub> = 19.5
\nNow QD= \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{2}=\\frac{19.5-10}{2}=\\frac{9.5}{2}\\) =4.75
\nQD = 4.75 tonnes<\/p>\n

Question 2.
\nThe yield of rice per acre for 10 districts of a state is as under (Board Paper)
\n\"1st
\nCalculate quartile deviation (QD)
\nAnswer:
\nQD for rice:
\nArrange the given observations in ascending order.
\nx : 12 12 12 15 18 18 22 23 29 34
\nCalculation of lower quartile (Q1 )
\nQ1<\/sub> = Size of \\(\\left(\\frac{\\mathrm{N}+1}{4}\\right)^{\\mathrm{th}}\\) item
\n\"1st
\n= Size of \\(\\left(\\frac{10+1}{4}\\right)^{t h}\\) item = Size of \\(\\left(\\frac{11}{4}\\right)^{t h}\\) item
\n= Size of 2.75th item
\nThe size of 275th item = size of item + 0.75 item
\n(0.75 item size of 3rd item – size of 2\u2019nd item)
\n\u2234 Q1<\/sub>= 12 + 0.75(12 – 12)
\nQ1<\/sub> = 12 +0.75 (0)
\nQ1<\/sub> = 12<\/p>\n

Calculation of upper quartile (Q3<\/sub>)
\nQ3<\/sub> = Size of \\(\\left(\\frac{N+1}{4}\\right)^{t h}\\) item = Size of \\(\\frac{3(10+1)^{t h}}{4}\\) item
\n= Size of \\(\\frac{3(11)^{\\text {th }}}{4}\\) item = Size of \\(\\left(\\frac{3(11)}{4}\\right)^{\\text {th }}\\) item
\n= Size 0f 8.25th item
\nThe size of 825th item = size of 8th item + 0.25 items
\n(0.025 item = size 0f 9th item – size of 8th item)
\n\u2234 Q3<\/sub> = size of 8.25th item i.e.,
\nQ3<\/sub> = 23 + 0.25 (29 – 23) = 23 + 0.25 (6)
\n= 23 + 1.5
\nQ3<\/sub> = 24.51<\/p>\n

Quartile deviation = \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{2}=\\frac{24.5-12}{2}\\)
\n= \\(\\frac{12.5}{2}\\) = 6.25
\nQD = 6.25 tonnes<\/p>\n

Question 3.
\nThe yield of wheat per acre for 10 districts of a state is as under:
\n\"1st
\nCalculate mean Deviation from Mean value.
\nAnswer:
\n\"1st
\nMean(X\u0304) = \\(\\frac{\\Sigma x}{N}=\\frac{155}{10}\\) = 15.5 N 10
\n\u2234 X\u0304 = 15.5
\nMD from Mean = \\(\\frac{\\Sigma|\\mathrm{D}|}{\\mathrm{N}}=\\frac{43}{10}\\) = 4.3 tonnes
\n\u2234 MDX\u0304<\/sub> = 4.3 tonnes<\/p>\n

Question 4.
\nThe yield of rice per acre for 10 districts of a state is as under:
\n\"1st
\nCalculate mean deviation from the mean.
\nAnswer:
\n\"1st
\nX\u0304 = \\(\\frac{\\Sigma x}{\\mathrm{~N}}=\\frac{195}{10}\\) = 19.5
\n\u2234 X\u0304 = 19.5
\nMD from mean = \\(\\frac{\\Sigma|D|}{N}=\\frac{60}{10}\\) = 6
\n\u2234 MDX\u0304<\/sub> = 6 tonnes<\/p>\n

Question 5.
\nThe yield of wheat per acre for 10 districts of a state is as under:
\n\"1st
\nCalculate the mean deviation from the median. (S – 2018)
\nAnswer:
\nFirst, we need to write the ie1d of what in arc ascending order
\nx: 9 10 10 12 15 16 18 19 21 25
\nMedian = Size of \\(\\left(\\frac{N+1}{2}\\right)^{\\text {th }}\\) item
\n= Size of \\(\\left(\\frac{10+1}{2}\\right)^{\\text {th }}\\) item
\n= Size of \\(\\left(\\frac{11}{2}\\right)^{\\text {th }}\\) item
\n= Size of 5.5th item<\/p>\n

To get the median, we need to add the 5th and 6th items and the same divided by 2.
\nMedian = \\(\\frac{5^{\\text {th }} \\text { item }+6^{\\text {th }} \\text { item }}{2}=\\frac{15+16}{2}=\\frac{31}{2}\\) = 15.5
\nMedian = 15.5
\nCalculation of mean deviation from the median
\n\"1st
\nMean deviation from the median (MDXm<\/sub>)
\nMDM<\/sub> = \\(\\frac{\\sum|\\mathrm{D}|}{\\mathrm{N}}=\\frac{43}{10}\\) = 4.3
\n\u2234 MDM<\/sub> = 4.3 tonnes<\/p>\n

Question 6.
\nThe yield of rice per acre for 10 districts of a state is as under:
\n\"1st
\nCalculate mean deviation from the median.
\nAnswer:
\nValues in ascending order
\nx : 12 12 12 15 18 18 22 23 29 34
\nMedian = Size of \\(\\left(\\frac{\\mathrm{N}+1}{2}\\right)^{\\mathrm{th}}\\) item
\nMedian = Size of \\(\\left(\\frac{10+1}{2}\\right)^{\\text {th }}\\) item
\n= Size of \\(\\left(\\frac{11}{2}\\right)^{\\text {th }}\\) item
\n= 5.5th item<\/p>\n

To get median, we need to add 5th and 6th items and divide it by 2.
\nMedian = \\(\\frac{5^{\\text {th }} \\text { item }+6^{\\text {th }} \\text { item }}{2}=\\frac{18+18}{2}=\\frac{36}{2}\\) = 18
\n\u2234 Median = 18<\/p>\n

Calculation of mean deviation from median:
\n\"1st
\nMean deviation from median (MDM<\/sub> )
\nMDM<\/sub> = \\(\\frac{\\Sigma|\\mathrm{D}|}{\\mathrm{N}}=\\frac{57}{10}\\) = 5.7
\nMDM<\/sub> = 5.7 tonnes<\/p>\n

Question 7.
\nCalculate the Mean Deviation using the mean for the following distribution.<\/p>\n\n\n\n\n\n\n\n\n\n
Class Interval<\/strong><\/td>\nFrequency<\/strong><\/td>\n<\/tr>\n
20-40<\/td>\n3<\/td>\n<\/tr>\n
40-80<\/td>\n6<\/td>\n<\/tr>\n
80-100<\/td>\n20<\/td>\n<\/tr>\n
100-120<\/td>\n12<\/td>\n<\/tr>\n
120-140<\/td>\n9<\/td>\n<\/tr>\n
<\/td>\n50<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\nCalculation of mean deviation from mean
\n\"1st
\nX\u0304 = \\(\\frac{\\Sigma f m}{n}=\\frac{4740}{50}\\) = 94.8
\nX\u0304 = 94.8
\nMean deviation from mean (MDX\u0304)
\nMDX\u0304<\/sub> = \\(\\frac{\\Sigma \\mathrm{f}|\\mathrm{D}|}{\\mathrm{N}}=\\frac{998.4}{50}\\) = 19.97
\n\u2234 MDX\u0304<\/sub> = 19.97<\/p>\n

Question 8.
\nCalculate the QD and its coefficient for the following data.
\n78, 80, 80, 82, 82, 84, 84, 86, 86, 88, 88, 90.
\nAnswer:
\n\"1st
\nWe know that QD = \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{2}\\)
\nSo, to calculate QD we need to calculate Q3<\/sub> and Q1<\/sub>.
\nQ3<\/sub> = Size of \\(\\frac{3(\\mathrm{~N}+1)^{\\mathrm{th}}}{4}\\) item = Size of \\(\\frac{3(12+1)}{4}\\)th item
\nQ3<\/sub> = Size of \\(\\left(\\frac{3(13)}{4}\\right)^{\\text {th }}\\) item = Size of \\(\\left(\\frac{39}{4}\\right)^{\\mathrm{th}}\\) item
\nQ = Size of 9.75th item
\ni.e., size of 9th + 0.75 (10th – 9th items)
\nQ = 86 + 0.75(88 – 6) = 86 + 0.75(2) = 86 + 1.5
\nQ3<\/sub> = 87.5
\nQ1<\/sub> = Size of \\(\\left(\\frac{\\mathrm{N}+1}{4}\\right)^{\\mathrm{th}}\\) item = Size of \\(\\left(\\frac{12+1}{4}\\right)^{\\text {th }}\\) item
\n= Size of \\(\\left(\\frac{13}{4}\\right)^{\\mathrm{th}}\\) item = Size of 3.25th item
\ni.e., 3rd + 0.25 (4th – 3 item)
\n\u2234 Q1<\/sub> = 80 + 0.25 (82 – 80) = 80 + 0.25 (2)
\nQ1<\/sub> = 80 + 0.5
\nQ1<\/sub> =80.5
\nNow to get quartile deviation (QD)
\nQD = \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{2}=\\frac{87.5-80.5}{2}=\\frac{7}{2}\\)
\nQ.D = 3.5<\/p>\n

Coefficient of QD = \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{\\mathrm{Q}_{3}+\\mathrm{Q}_{\\mathrm{i}}}=\\frac{87.5-80.5}{87.5+80.5}=\\frac{7}{168}\\) = 0.0416<\/p>\n

Question 9.
\nCalculate the mean deviation from mean and its coefficient for the following data
\n1100,150, 200,360, 490, 500, 600.
\nAnswer:
\n\"1st
\nx\u0304 = \\(\\frac{2400}{7}\\)
\nx\u0304 = 342.8
\nM.Dx\u0304<\/sub> = \\(\\frac{\\Sigma|d|}{\\mathrm{N}}=\\frac{1157.2}{7}\\)
\nM.Dx\u0304<\/sub> = 165.31
\nCoefficient of MD = \\(\\frac{\\mathrm{MD}_{\\bar{x}}}{\\overline{\\mathrm{X}}}=\\frac{165.31}{342.8}\\) = 0.482<\/p>\n

Question 10.
\nCalculate MD from the median and its relative measure for the following data
\n37,45,52,46,56, 40,47,55, 43.
\nAnswer:
\n\"1st
\nMedian = Size of \\(\\left(\\frac{\\mathrm{N}+1}{2}\\right)^{\\mathrm{th}}\\) item
\n= Size of \\(\\left(\\frac{9+1}{2}\\right)^{\\text {th }}\\) item
\n= Size of \\(\\left(\\frac{10}{2}\\right)^{t h}\\) item = 5th item
\nMedian = 46
\nM.Dme<\/sub> = \\(\\frac{\\Sigma|d|}{N}=\\frac{47}{9}\\) = 5.22.
\nM.Dme<\/sub> = 5.22<\/p>\n

Relative measure i.e., co-efficient of MD from median
\nCoefficient MDme<\/sub> = \\(\\frac{\\mathrm{MD}_{\\mathrm{me}}}{\\mathrm{M}}=\\frac{5.22}{46}\\) = 0.113
\nRelative measure i.e.. co-efficient of MD from median is 0.113.<\/p>\n

Question 11.
\nFind the mean deviation from the mean for the following distribution.
\n\"1st
\nAnswer:
\n\"1st
\nx\u0304 = \\(\\frac{\\Sigma f x}{N}=\\frac{6320}{99}\\) = 63.8
\nx\u0304 = 63.8
\nM.Dx\u0304<\/sub> = \\(\\frac{\\Sigma f|d|}{\\mathrm{N}}=\\frac{121.4}{99}\\)
\nM.Dx\u0304<\/sub> = 1.22<\/p>\n

Question 12.
\nCalculate the coefficient of MD from the median for the following frequency distribution.
\n\"1st
\nAnswer:
\n\"1st
\nMedian = Size of \\(\\left(\\frac{\\mathrm{N}+1}{2}\\right)^{\\mathrm{th}}\\) item = Size of \\(\\left(\\frac{45+1}{2}\\right)^{t h}\\) item
\nMedian = Size of \\(\\left(\\frac{46}{2}\\right)^{\\text {th }}\\) item = Size of 23rd item i.e:, 7
\nMedian = 7
\nMD = \\(\\frac{\\sum f|d|}{N}=\\frac{43}{45}\\) = 0.95
\nCoefficient of MDme<\/sub> = \\(\\frac{\\mathrm{MD}_{\\mathrm{me}}}{\\mathrm{M}_{\\mathrm{me}}}=\\frac{0.95}{7}\\) = 0.13
\nMDMe<\/sub> = 0.95<\/p>\n

Question 13.
\nFind M.D. from the mean for the following distribution regarding the difference in ages (years) between husband and wife of a particular community.
\n\"1st
\nAnswer:
\n\"1st
\nx\u0304 = \\(\\frac{\\Sigma f m}{n}=\\frac{6152}{1370}\\) = 4.49 (4.5)
\nMDx\u0304<\/sub> = \\(\\frac{\\Sigma f|d|}{\\mathrm{N}}=\\frac{2562}{1370}\\) = 1.87
\nMDx\u0304<\/sub> = 1.87<\/p>\n

Question 14.
\nFind the standard deviation for the following data:
\n8, 9,15, 23, 5,11,19, 8,10, 12.
\nAnswer:
\n\"1st
\nx\u0304 = \\(\\frac{\\sum x}{N}=\\frac{120}{10}\\)
\nx\u0304 = 12
\n\u03c3 = \\(\\sqrt{\\frac{\\Sigma d^{2}}{n}}=\\sqrt{\\frac{274}{10}}=\\sqrt{27.4}\\)
\n\u03c3 = 5.23<\/p>\n

Question 15.
\nFind the standard deviation of the following data by the deviation method.
\n25, 50, 45, 30, 70,42, 36,48, 34, 60.
\nAnswer:
\n\"1st
\n\u03c3 = \\(\\sqrt{\\frac{\\sum d^{12}}{n}-\\left(\\frac{\\Sigma d^{1}}{n}\\right)^{2}}\\) \u00d7 C = \\(\\sqrt{\\frac{145.4}{10}-\\left(\\frac{27.8}{10}\\right)^{2}}\\) \u00d7 5
\n\u03c3 = \\(\\sqrt{14.54-(2.78)^{2} \\times 5}=\\sqrt{14.54-7.72 \\times 5}=\\sqrt{6.82}\\) \u00d7 5 = 2.61 \u00d7 5
\n\u03c3 = 13.05<\/p>\n

V. Answer the following in about twenty sentences (each question carries 6 marks)<\/span><\/p>\n

Question 1.
\nThe yield of wheat per acre for 10 districts of a state are as under:
\n\"1st
\nCalculate standard deviation and coefficient of variation (N – 2018)
\nAnswer:
\nStandard Deviation (actual mean method)
\n\"1st
\n\"1st<\/p>\n

Question 2.
\nThe yield of rice per acre for 10 districts of a state is as under:
\n\"1st
\nCalculate standard deviation and coefficient of variation.
\nAnswer:
\n\"1st
\n\"1st<\/p>\n

Question 3.
\nA batsman is to be selected for a cricket team. The choice is between X and Y on the basis of their scores in five previous tests which are:
\n\"1st
\nWhich batsman should be selected if we want,
\n(i) a higher run-getter, or
\n(ii) a more reliable batsman in the team?
\nAnswer:
\n(i) To determine a higher run-getter, we need to find out the average of both the batsman
\nX\u0304 = \\(\\frac{\\Sigma x}{N}\\)
\nX\u0304 = \\(\\frac{350}{5}\\) = 70 Runs
\n\"1st<\/p>\n

X\u0304 = \\(\\frac{\\Sigma X_{y}}{N}=\\frac{310}{5}\\) = 62 Runs
\nBatsman X: 70 Runs
\nBatsman Y: 62 Runs<\/p>\n

Batsman \u2018X\u2019 should be selected on the basis of runs scored as his average runs are more than that of Batsman Y.<\/p>\n

(ii) To determine a reliable batsman, we need to calculate coefficient of variation (CV)
\n\"1st<\/p>\n

Question 4.
\nTo check the quality of the two brands of light bulbs, their life in burning hours was estimated for 100 bulbs of each brand as shown below.
\n\"1st
\n(i)Which brand gives longer life?
\n(ii) Which brand is more dependable?
\nAnswer:
\nTo know which brand gives a longer life we need to calculate the average life of both the brands. To know the most dependable brand we need to find out the coefficient of variation (CV).
\nBrand \u2018A\u2019
\n\"1st
\n\"1st
\nBrand ‘B’
\n\"1st<\/p>\n

\"1st<\/p>\n

(i)Brand B has a longer life average
\n(ii) Brand B is more dependable.<\/p>\n

Question 5.
\nCalculate the standard deviation for the following distribution.<\/p>\n\n\n\n\n\n\n\n\n\n
Class Interval<\/strong><\/td>\nFrequency<\/strong><\/td>\n<\/tr>\n
20-40<\/td>\n3<\/td>\n<\/tr>\n
40-80<\/td>\n6<\/td>\n<\/tr>\n
80-100<\/td>\n20<\/td>\n<\/tr>\n
100-120<\/td>\n12<\/td>\n<\/tr>\n
120-140<\/td>\n9<\/td>\n<\/tr>\n
<\/td>\n50<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\nCalculation of SD (\u03c3)
\n\"1st
\n\"1st<\/p>\n

Question 6.
\nCalculate the coefficient of QD for the following distribution.
\n\"1st
\nAnswer:
\n\"1st
\nQD = \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{2}\\)
\nQ1<\/sub> = \\(\\left(\\frac{n}{4}\\right)^{\\text {th }}\\) item = \\(\\left(\\frac{88}{4}\\right)^{\\text {th }}\\) = 22nd item
\nQ1<\/sub> falls in 64-68
\nTo find the exact Q1<\/sub> the following formula is used:
\n\"1st
\nNow to calculate Q3<\/sub>
\nQ3<\/sub> = Size of \\(\\left(\\frac{3(n)}{4}\\right)^{t h}\\) item = Size of \\(\\frac{3(88)^{\\text {th }}}{4}\\) item
\n= Size of \\(\\left(\\frac{264}{4}\\right)^{\\text {th }}\\) item = Size of 66 item
\nSo the Q3<\/sub> falls in 76 – 80<\/p>\n

To calculate the exact Q3<\/sub> the folloing formula is used
\n\"1st
\nQD = 5.89
\nCoefficient of quartile deviation
\n= \\(\\frac{Q_{3}-Q_{1}}{Q_{3}+Q_{1}}=\\frac{78-66.22}{78+66.22}=\\frac{11.78}{144.22}\\)
\nCoefficient of QD = 0.0816<\/p>\n

Question 7.
\nCalculate QD and its relative measures for the following distribution.
\n\"1st
\nAnswer:
\n\"1st
\nQD = \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{2}\\)
\nQ1<\/sub> = Size of \\(\\left(\\frac{n}{4}\\right)^{\\text {th }}\\) item = Size of \\(\\left(\\frac{804}{4}\\right)^{th}\\) = Size of 201st item
\nSo, Q1<\/sub> lies in 30 – 39<\/p>\n

To find exact Q1<\/sub> the following formula is used.
\n\"1st
\nQ1<\/sub> = 30 + 2.81;
\nQ1<\/sub> = 32.81
\nNow to calculate Q3<\/sub>
\nQ3<\/sub> = Size of \\(\\left(\\frac{3(n)}{4}\\right)^{t h}\\)item = Size of \\(\\left(\\frac{3(804)}{4}\\right)^{t h}\\) item
\n= Size of \\(\\left(\\frac{2412}{4}\\right)^{\\mathrm{th}}\\) item = Size of 603rd item
\nSo, Q3<\/sub> lies in 40 – 49<\/p>\n

To find the exact Q3<\/sub> following formula is used.<\/p>\n

\"1st
\nA relative measure of QD
\nCoefficient of QD
\n= \\(\\frac{\\mathrm{Q}_{3}-\\mathrm{Q}_{1}}{\\mathrm{Q}_{3}+\\mathrm{Q}_{1}}=\\frac{48.14-32.81}{48.14+32.81}=\\frac{15.33}{80.95}\\)
\nCoefficient of Q.D = 0.189<\/p>\n

Question 8.
\nCalculate the standard deviation and variance for the following distribution by deviation method
\n\"1st
\nAnswer:
\n\"1st
\n\"1st<\/p>\n

Question 9.
\nCalculate the SD for the following frequency distribution of heights of 30 persons by the direct method. (S – 2018)
\n\"1st
\nAnswer:
\n\"1st
\nx\u0304 = \\(\\frac{\\Sigma f m}{n}=\\frac{5160}{30}\\) = 172
\nx\u0304 = 172<\/p>\n

Standard deviation (direct method)
\n\u03c3 = \\(\\sqrt{\\frac{\\Sigma f m^{2}}{N}-(\\bar{x})^{2}}=\\sqrt{\\frac{8,89,987.5}{30}-(172)^{2}}=\\sqrt{29,666.25-29,584}=\\sqrt{82.25}\\)
\n\u03c3 = 9.06<\/p>\n

Question 10.
\nCalculate the SD for the following frequency distribution of heights of 30 persons by the step deviation method.
\n\"1st
\nAnswer:
\n\"1st<\/p>\n

1st PUC Economics Question Bank with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

Karnataka 1st PUC Economics Question Bank Chapter 6 Measures of Dispersion 1st PUC Economics Measures of Dispersion TextBook Questions and Answers I. Choose the correct answers Question 1. Inter quartile range is based upon middle _______of the values in a distribution. (a) 30% (b) 20% (c) 50% (d) 25% Answer: (c) 50% Question 2. For …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/87835"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=87835"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/87835\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=87835"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=87835"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=87835"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}