{"id":39367,"date":"2022-12-06T12:37:14","date_gmt":"2022-12-06T07:07:14","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=39367"},"modified":"2022-12-07T12:05:54","modified_gmt":"2022-12-07T06:35:54","slug":"maths-2a-measures-of-dispersion-important-questions-long-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2a-measures-of-dispersion-important-questions-long-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2A Measures of Dispersion Important Questions Long Answer Type"},"content":{"rendered":"
Students must practice these Maths 2A Important Questions<\/a> TS Inter Second Year Maths 2A Measures of Dispersion Important Questions Long Answer Type to help strengthen their preparations for exams.<\/p>\n Question 1. <\/p>\n Solution:<\/p>\n <\/p>\n Here, \u03a3fi<\/sub> = 100 Question 2. <\/p>\n Solution: <\/p>\n Here, N = \u03a3fi<\/sub> = 100 <\/p>\n Question 3. <\/p>\n Solution: <\/p>\n Let the assumed mean be A = 35 Question 4. <\/p>\n Solution: <\/p>\n Let the assumed mean be A = 25 <\/p>\n Question 5. <\/p>\n Solution:<\/p>\n <\/p>\n Median class = Class containing \\(\\frac{N}{2}^{\\text {th }}\\) item Question 6. <\/p>\n Solution:<\/p>\n <\/p>\n Median class = Class containing \\(\\frac{\\mathrm{N}^{\\text {th }}}{2}\\) item <\/p>\n Question 7. <\/p>\n Solution: <\/p>\n \u2234 Mean, \\(\\bar{x}=\\frac{1}{N} \\sum_{i=1}^7 f_i x_i=\\frac{3100}{50}\\) = 62 Question 8. <\/p>\n Solution: <\/p>\n From the table, N = \\(\\sum_{i=1}^9\\) fi<\/sub> = 72 <\/p>\n <\/p>\n Question 9. <\/p>\n Solution: For cricketer B: The deviations of the respective observations from mean i.e., (xi<\/sub> – \\(\\overline{\\mathbf{x}}\\)) are: – 10, 32, – 7, – 38, – 24, 73, 28, – 7, – 13, – 34 Question 10. <\/p>\n Solution: Share Y: <\/p>\n Question 11. <\/p>\n Solution: <\/p>\n Let the assumed mean, A = 55 <\/p>\n Question 12. <\/p>\n Which section of students has greater variability In performance? <\/p>\n The variance of distribution of marks of section A is \u03c31<\/sub>2<\/sup> = 64 <\/p>\n Question 13. <\/p>\n Solution: <\/p>\n Mean of the given data is (\\(\\overline{\\mathrm{x}}\\)) = \\(\\frac{1}{N} \\sum_{i=1}^7 f_i x_i\\) Question 14. <\/p>\n Solution: <\/p>\n Mean of the given data is (\\(\\overline{\\mathrm{x}}\\)) = \\(\\frac{1}{N} \\sum_{i=1}^n f_i x_i\\) <\/p>\n Question 15. <\/p>\n Solution: <\/p>\n Median class = Class containing \\(\\frac{N^{t h}}{2}\\) item Question 16. <\/p>\n From the above data suggest which model to purchase. <\/p>\n From the table, N = \\(\\sum_{\\mathbf{i}=1}^5\\) fi<\/sub> = 46, \\(\\sum_{\\mathbf{i}=1}^5\\) fi<\/sub>xi<\/sub> = 212 <\/p>\n Model B: <\/p>\n From the table, N = \\(\\sum_{i=1}^5\\) fi<\/sub> = 49 Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Measures of Dispersion Important Questions Long Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2A Measures of Dispersion Important Questions Long Answer Type Question 1. Find the mean deviation about the mean for the following … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter Second Year Maths 2A Measures of Dispersion Important Questions Long Answer Type<\/h2>\n
\nFind the mean deviation about the mean for the following continuous distribution. [AP – Mar. 2015]<\/p>\n
\n\u03a3fi<\/sub>xi<\/sub> = 7100
\n\u2234 Mean \\(\\frac{\\sum_{\\mathrm{i}=1}^{\\mathrm{n}} \\mathrm{f}_{\\mathrm{i}} \\mathrm{x}_{\\mathrm{i}}}{\\mathrm{N}}=\\frac{7100}{100}\\) = 71
\n\u2234 Mean deviation about mean is M.D = \\(\\frac{\\Sigma \\mathrm{f}_{\\mathrm{i}}\\left|\\mathrm{x}_{\\mathrm{i}}-\\overline{\\mathrm{x}}\\right|}{\\mathrm{N}}=\\frac{1040}{100}\\) = 10.4.<\/p>\n
\nFind the mean deviation about the mean for the following continuous distribution.<\/p>\n
\nWe can form the following table from the given data :<\/p>\n
\n\u03a3fi<\/sub>xi<\/sub> = 12530
\n\u2234 Mean, \\(\\bar{x}=\\frac{\\sum_{i=1}^n f_i x_i}{N}=\\frac{12530}{100}\\) = 125.3
\n\u2234 Mean deviation about mean is M.D = \\(\\frac{\\Sigma \\mathrm{f}_{\\mathrm{i}}\\left|\\mathrm{x}_{\\mathrm{i}}-\\overline{\\mathrm{x}}\\right|}{\\mathrm{N}}\\)
\n= \\(\\frac{1128.8}{100}\\)
\n= 11.288 = 11.28.<\/p>\n
\nFind the mean deviation from the mean of the following data, using the step deviation method. [AP – Mar. ’18; TS – Mar. 2016]<\/p>\n
\nWe can form the following table from the given data :<\/p>\n
\nHere, C = 10
\nN = \u03a3fi<\/sub> = 50
\nMean, \\(\\bar{x}=A+\\frac{\\sum_{i=1}^n f_i d_i}{N} \\times C\\)
\n= \\(A+\\frac{\\sum_{i=1}^7 f_i d_i}{N} \\times C\\)
\n= 35 + \\(\\frac{(-8)}{50}\\) \u00d7 10
\n= 35 – 1.6 = 33.4
\n\u2234 Mean deviation about mean is M.D = \\(\\frac{\\Sigma f_{\\mathrm{i}}\\left|\\mathrm{x}_{\\mathrm{i}}-\\overline{\\mathrm{x}}\\right|}{\\mathrm{N}}=\\frac{659.2}{50}\\) = 13.185.<\/p>\n
\nFind the mean deviation about the mean for the given data using step deviation method. [AP – Mar. ’17, ’16, AP – May 2015] [TS – Mar. ’18, May ’16]<\/p>\n
\nWe can form the following table from the given data:<\/p>\n
\nHere, C = 10
\nN = \u03a3fi<\/sub> = 50
\n\\(\\bar{x}=A+\\frac{\\sum_{i=1}^n f_i d_i}{N} \\times C\\)
\n= \\(\\mathrm{A}+\\frac{\\sum_{\\mathrm{i}=1}^5 \\mathrm{f}_{\\mathrm{i}} \\mathrm{d}_{\\mathrm{i}}}{\\mathrm{N}} \\times \\mathrm{C}\\)
\n= 25 + \\(\\frac{10}{50}\\) \u00d7 10 = 27
\n\u2234 Mean deviation about mean is M.D = \\(\\frac{\\Sigma \\mathrm{f}_{\\mathrm{i}}\\left|\\mathrm{x}_{\\mathrm{i}}-\\overline{\\mathrm{x}}\\right|}{\\mathrm{N}}=\\frac{472}{50}\\) = 9.44.<\/p>\n
\nFind the mean deviation about the median for the following continuous distribution.<\/p>\n
\n= \\(\\frac{50}{2}\\) = 25 item
\n= 20 – 30 class
\nHere, l = 20, C = 10, f = 14, m = 14, N = 50
\n\u2234 Median, M = l + \\(\\frac{\\frac{N}{2}-m}{f}\\) \u00d7 C
\n= 20 + \\(\\frac{25-14}{14}\\) \u00d7 10
\n= 20 + 7.86 = 27.86
\n\u2234 Mean deviation about median is M.D = \\(\\frac{\\Sigma \\mathrm{f}_{\\mathrm{i}}\\left|\\mathrm{x}_{\\mathrm{i}}-\\mathrm{M}\\right|}{\\mathrm{N}}\\)
\n= \\(\\frac{517.16}{50}\\) = 10.34<\/p>\n
\nFind the mean deviation about median for the following data.<\/p>\n
\n= \\(\\frac{1000}{2}\\) = 500th item = 35 – 40 class
\nHere, l = 35, C = 5, f = 160, m = 420, N = 1000
\n\u2234 Median, M = l + \\(\\frac{\\frac{N}{2}-m}{f}\\) \u00d7 C
\n= 35 + \\(\\frac{500-420}{160}\\) \u00d7 5
\n= 35 + 2.5 = 37.5
\n\u2234 Mean deviation about median is M.D = \\(\\frac{\\Sigma \\mathrm{f}_{\\mathrm{i}}\\left|\\mathrm{x}_{\\mathrm{i}}-\\mathrm{M}\\right|}{\\mathrm{N}}=\\frac{8175}{1000}\\) = 8.175.<\/p>\n
\nCalculate the variance and standard deviaon,of the following continuous frequency distribution.
\n[AP – May 2016; TS – Mar. 2017; Mar. \u201814, Board Paper]<\/p>\n
\nWe can form the following table from the given data:<\/p>\n
\nVariance, \u03c32<\/sup> = \\(\\frac{1}{N} \\sum_{i=1}^7 f_i\\left(x_i-\\bar{x}\\right)^2\\)
\n= \\(\\frac{1}{50}\\) (10050) = 201
\nStandard deviation, \u03c3 = \\(\\sqrt{\\frac{1}{N} \\sum_{i=1}^7 f_i\\left(x_i-\\bar{x}\\right)^2}\\)
\n= \\(\\sqrt{\\text { variance }}=\\sqrt{201}\\)= 14.17 (approximately).<\/p>\n
\nThe following table gives the daily wages of workers in a factory. Compute the standard deviation and the coefficient of variation of the wages of the workers.<\/p>\n
\nWe shall solve this problem using the step deviation method, since the midpoints of the class intervals are numerically large.<\/p>\n
\nLet the assumed mean, A = 300
\nHere, C = 50
\nFrom the table, \\(\\sum_{i=1}^9\\) fi<\/sub>di<\/sub> = – 31
\nMean, \\(\\bar{x}=A+\\frac{\\sum_{i=1}^9 f_i d_i}{N}\\) \u00d7 C
\n= 300 + \\(\\frac{-31}{72}\\) \u00d7 50
\n= 300 – \\(\\frac{1550}{72}\\)
\n= 278.5 (approximately)<\/p>\n
\nThe scores of two cricketers A and B in 10 innings are given below. Find who is a better run getter and who is a more consistent player.<\/p>\n
\nFor cricketer A : Scores of A are 40, 25, 19. 80, 38, 8, 67, 121, 66, 76
\n\u2234 Mean, \\(\\overline{\\mathrm{x}}_{\\mathrm{A}}=\\frac{\\text { Sum of the scores }}{\\text { No. of innings }}\\)
\n= \\(\\frac{40+25+19+80+38+8+67+121+66+76}{10}\\)
\n= \\(\\frac{540}{10}\\) = 54
\nThe deviations of the respective observations from the mean i.e., xi<\/sub> – \\(\\overline{\\mathrm{x}}\\) are: – 14, – 29, – 35, 26, – 16, -46, 13, 67, 12, 22
\nVariance, \u03c32<\/sup> = \\(\\frac{1}{n} \\sum_{i=1}^{10}\\left(x_i-\\bar{x}\\right)^2\\)
\n= \\(\\frac{196+841+1225+676+256+2116+169+4489+144+484}{10}\\)
\n= \\(\\frac{10596}{10}\\) = 1059.6
\nStandard deviation, \u03c3A<\/sub> = \\(\\sqrt{1059.6}\\) = 32.55
\nC.V of A = \\(\\frac{\\sigma_{\\mathrm{A}}}{\\overline{\\mathrm{x}}_{\\mathrm{A}}}\\) \u00d7 100
\n= \\(\\frac{32.55}{54}\\) \u00d7 100 = 60.28.<\/p>\n
\nScores of B are 28, 70, 31, 0, 14, 111, 66, 31, 25, 4
\nMean \\(\\overline{\\mathrm{x}}_{\\mathrm{B}}=\\frac{\\text { Sum of the scores }}{\\text { No. of innings }}\\)
\n= \\(\\frac{28+70+31+0+14+111+66+31+25+4}{10}=\\frac{380}{10}\\)
\n= 38<\/p>\n
\nVariance, \u03c3B<\/sub>2<\/sup> = \\(\\frac{1}{n} \\sum_{i=1}^{10}\\left(x_i-\\bar{x}\\right)^2\\)
\n= \\(\\frac{100+1024+49+1444+576+5329+784+49+169+1156}{10}\\)
\n= \\(\\frac{10680}{10}\\) = 1068
\nStandard deviation. \u03c3B<\/sub> = \\(\\sqrt{1068}\\) = 32.68
\nC.V of B = \\(\\frac{\\sigma_B}{\\bar{x}_B}\\) \u00d7 100
\n= \\(\\frac{32.68}{38}\\) \u00d7 100 = 86
\nSince, \\(\\overline{\\mathbf{x}}_{\\mathrm{A}}\\) is greater than \\(\\overline{\\mathbf{x}}_{\\mathrm{B}}\\) then cricketer A is better run getter (scorer).
\nSince, C.V of A is less than C.V of B then cricketer \u00a1s also a more consistent player.<\/p>\n
\nFrom the prices of shares X and Y given below, for 10 days of trading, find out which share is more stable?<\/p>\n
\nShare X:
\nPrices of share X are 35, 54, 52, 53, 56, 58, 52, 50, 51, 49
\nMean, \\(\\overline{\\mathrm{x}}_{\\mathrm{X}}=\\frac{1}{\\mathrm{~m}} \\sum_{\\mathrm{i}=1}^{10} \\mathrm{x}_{\\mathrm{i}}\\)
\n= \\(\\frac{510}{10}\\) = 51
\nThe deviations of the respective observations from mean i.e., (xi<\/sub> – \\(\\overline{\\mathrm{x}}\\))2<\/sup> are: – 16, 3, 1, 5, 7, 1, – 10, – 2
\nVariance, \u03c3X<\/sub>2<\/sup> = \\(\\frac{1}{n} \\sum_{i=1}^{10}\\left(x_i-\\bar{x}\\right)^2\\)
\n= \\(\\frac{256+9+1+25+49+1+1+0+4}{10}\\)
\n= \\(\\frac{350}{10}\\) = 35
\nStandard deviation, \u03c3x<\/sub> = \u221a35 = 5.9 (approximately)
\nC.V of X = \\(\\frac{\\sigma_X}{\\bar{x}_X}\\) \u00d7 100
\n= \\(\\frac{5.9}{51}\\) \u00d7 100 = 11.5 (approximately)<\/p>\n
\nPrices of share Y are 108, 107, 105. 105, 106, 107, 104, 103, 104, 101
\nMean, \\(\\overline{\\mathrm{x}}_{\\mathrm{Y}}=\\frac{1}{\\mathrm{n}} \\sum_{\\mathrm{i}=1}^{10} \\mathrm{x}_{\\mathrm{i}}\\)
\n= \\(\\frac{1050}{10}\\) = 105
\nThe deviations of the respective observations from mean i.e.. (xi<\/sub> – \\(\\overline{\\mathrm{x}}\\)) are 3, 2, 0, 0, 1, 2, – 1, – 2, – 1, – 4
\nVariance, \u03c3Y<\/sub>2<\/sup> = \\(\\frac{1}{n} \\sum_{\\mathrm{i}=1}^{10}\\left(\\mathrm{x}_{\\mathrm{i}}-\\overline{\\mathrm{x}}\\right)^2\\)
\n= \\(\\frac{9+4+0+0+1+4+1+4+1+16}{10}\\)
\n= \\(\\frac{40}{10}\\) = 4
\nStandard deviation, \u03c3Y<\/sub> = \u221a4 = 2
\nC.V. of Y = \\(\\frac{\\sigma_Y}{\\bar{x}_Y}\\) x 100
\n= \\(\\frac{2 \\times 100}{105}\\) x 100 = 1.904
\nSince, C.V of X is greater than C.V of Y then the share of Y is more consistent (stable).<\/p>\n
\nFind the mean and variance using the step deviation method of the following tabular data, giving the age distribution of 542 members.<\/p>\n
\nWe form the following table From the given data:<\/p>\n
\nHere, C = 10<\/p>\nSome More Maths 2A Measures of Dispersion Important Questions<\/h3>\n
\nStudents of two sections A and B of a class show the following performance in a test (conducted 100 marks).<\/p>\n
\nSolution:
\nGiven that,<\/p>\n
\n\u2234 The standard deviation of distribution of marks of section A is \u03c31<\/sub> = 8
\nThe variance of distribution of marks of section B is \u03c32<\/sub>2<\/sup> = 81
\n\u2234 The standard deviation of distribution of marks of section B is \u03c32<\/sub> = 9
\nThe average marks in the test of section A is \\(\\bar{x}_1\\) = 45
\nThe average marks in the test of section B is \\(\\bar{x}_2\\) = 45
\n\u2235 The average marks in the test of both sections of students is the same i.e., 45.
\nThe standard deviation of the distribution of marks in section B is greater than the standard deviation of (he distribution of marks in section A.
\nHence, the section B has greater variability in the performance.<\/p>\n
\nFind the variance and standard deviation of the following frequency distribution.<\/p>\n
\nFrom the given data, we can form the following table:<\/p>\n
\n= \\(\\frac{760}{40}\\) = 19
\nVariance, \u03c32<\/sup> = \\(\\frac{1}{N} \\sum_{i=1}^7 f_i\\left(x_i-\\bar{x}\\right)^2\\)
\n= \\(\\frac{1}{40}\\) (1736) = 43.4
\nStandard deviation, \u03c3 = \\(\\sqrt{\\text { variance }}\\)
\n= \\(\\sqrt{43.4}\\) = 6.58 (approximately).<\/p>\n
\nCalculate the variance and standard deviation for a discrete frequency distribution.<\/p>\n
\nFrom the given data, we can form the following table:<\/p>\n
\n= \\(\\frac{1}{N} \\sum_{i=1}^7 f_i x_i=\\frac{420}{30}\\) = 14
\nVariance, \u03c32<\/sup> = \\(\\frac{1}{N} \\sum_{i=1}^7 f_i\\left(x_i-\\bar{x}\\right)^2\\)
\n= \\(\\frac{1}{30}\\) (1374) = 45.8
\nStandard deviation, \u03c3 =\\(\\sqrt{\\text { variance }}\\)
\n= \\(\\sqrt{45.8}\\) = 6.76 (approximately).<\/p>\n
\nFind the mean deviation about the median for the following data.<\/p>\n
\nWe can form the following table from the given data:<\/p>\n
\n= \\(\\frac{100}{2}\\) = 50th item = 40 – 50 class
\nHere, l = 40, f = 28, m = 32, C = 10, N = 100
\n\u2234 Median, M = l + \\(\\frac{\\frac{\\dot{N}}{2}-\\mathrm{m}}{\\mathrm{f}}\\) \u00d7 C
\n= 40 + \\(\\frac{50-32}{28}\\) \u00d7 10
\n= 40 + \\(\\frac{180}{28}\\)
\n= 40 + 6.42 = 46.42
\n\u2234 Mean deviation about median is M.D = \\(\\frac{\\Sigma \\mathrm{f}_{\\mathrm{i}}\\left|\\mathrm{x}_{\\mathrm{i}}-\\mathrm{M}\\right|}{\\mathrm{N}}=\\frac{1428.4}{100}\\) = 14.284<\/p>\n
\nLives of two models of refrigerators A and B obtained \u00a1n a survey are given below.<\/p>\n
\nSolution:
\nModel – A:
\nTo find the mean and variance of lives of model A refrigerators, we shall construct the following table:<\/p>\n
\nMean, \\(\\bar{x}_A=\\frac{1}{N} \\sum_{i=1}^5 f_i x_i\\)
\n= \\(\\frac{212}{46}\\) = 4.6
\nFrom the table, \\(\\sum_{i=1}^5 f_i\\left(x_i-\\bar{x}\\right)^2\\) = 244.96
\nVariance, \u03c3A<\/sub>2<\/sup> = \\(\\frac{1}{N} \\sum_{i=1}^5 f_i\\left(x_i-\\bar{x}\\right)^2\\)
\n= \\(\\frac{244.96}{46}\\) = 5.32
\nStandard deviation, \u03c3A<\/sub> = \\(\\sqrt{5.32}\\)
\n= 2.3 (approximately)
\nCoefficient of variation of model A = \\(\\frac{\\sigma_{\\mathrm{A}}}{\\mathrm{x}_{\\mathrm{A}}}\\) \u00d7 100
\n= \\(\\frac{2.3}{4.6}\\) \u00d7 100
\n= 50 (approximately).<\/p>\n
\nTo find the mean and variance of lives of model B refrigerators, we shall construct the following table:<\/p>\n
\n\\(\\sum_{i=1}^5\\) fi<\/sub>xi<\/sub> = 297
\nMean, \\(\\bar{x}_B=\\frac{1}{N} \\sum_{i=1}^5 f_i x_i\\)
\n= \\(\\frac{297}{49}\\) = 6.06
\nFrom the table, \\(\\) = 224.8164
\nVariance, \u03c3B<\/sub>2<\/sup> = \\(\\frac{1}{N} \\sum_{i=1}^5 f_i\\left(x_i-\\bar{x}\\right)^2\\)
\n= \\(\\frac{224.8164}{49}\\) = 4.58 (approximately)
\nStandard deviation, \u03c3B<\/sub> = \\(\\sqrt{4.58}\\)
\n= 2.1 (approximately)
\nCoefficient of variation of model B = \\(\\frac{\\sigma_B}{x_B}\\) \u00d7 100
\n= \\(\\frac{2.1}{6.06}\\) \u00d7 100 = 34.65 (approximately)
\nSince C.V. of model B is less than C.V. of model A then the model B is more consistent than the model A with regard to life in years.
\nHence, we suggest model B for purchase.<\/p>\n","protected":false},"excerpt":{"rendered":"