{"id":35881,"date":"2022-10-03T15:33:43","date_gmt":"2022-10-03T10:03:43","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35881"},"modified":"2022-10-03T15:33:43","modified_gmt":"2022-10-03T10:03:43","slug":"ap-inter-1st-year-economics-study-material-chapter-10","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-inter-1st-year-economics-study-material-chapter-10\/","title":{"rendered":"AP Inter 1st Year Economics Study Material Chapter 10 Economic Statistics"},"content":{"rendered":"

Andhra Pradesh BIEAP\u00a0AP Inter 1st Year Economics Study Material<\/a> 10th Lesson Economic Statistics Textbook Questions and Answers.<\/p>\n

AP Inter 1st Year Economics Study Material 10th Lesson Economic Statistics<\/h2>\n

Essay Questions<\/span><\/p>\n

Question 1.
\nWhat is the relationship between Economics and Statistics? [March 18, 16]
\nAnswer:
\nThere is a close relationship between statistics and economics. In the words of Tugwell “The science of economics is becoming statistical in its method”. All economic laws are pronounced on the basis of statistical facts and figures. The theory of the population of Malthus, the law of family expenditure of Engels, etc., were propounded after statistical tests. Statistics help economics to become an exact science.<\/p>\n

In the study of theoretical economics, the application and use of statistical methods are of great importance. Most of the doctrines of economics are based on the study of a large number of units and their analysis. This is done through statistical methods. The Law of demand was formulated because of statistical methods.<\/p>\n

The importance of statistics is felt in all branches of knowledge- in accountancy and auditing in banking, in insurance in research, and many fields. Without statistics no branch of knowledge is complete.<\/p>\n

\"AP<\/p>\n

Question 2.
\nExplain the Geometric diagrams. What are Bar diagrams and Pie diagrams ?
\nAnswer:
\nDiagrams are more suitable to illustrate the data which is discrete, while continuous data is better represented by graphs.<\/p>\n

BAR DIAGRAM and PIE DIAGRAM come in the category of geometric diagrams.
\nThe bar diagram are 3 types. Simple, Multiple and Component bar diagrams.
\n\"AP
\nSimple Bar Diagrams : Bar diagrams are very commonly used and is better for representation of quantitative data. Bars are simply
\ncorresponding numerical values.
\nEx: Maximum temperature recorded in Hyderabad in the 1st six months in a year in a bar diagram.<\/p>\n

Multiple Bar Diagram : Multiple bar diagrams are used for comparing two or more sets of data.
\nEx : The multiple bar diagram depicts the number of students in a college studying two foreign languages, French and German for the period 1960 – 2010 is given below.
\n\"AP
\nComponent Bar Diagram : Component bar diagram charts also called sub-divided bar diagram, are very useful in comparing the sizes of different component parts.
\nEx : Enrolment at primary level in a district of Bihar. Boys, girls and total children in the given age group are denoted in the component bar diagram is given below.
\n\"AP
\nPie – Diagram : This diagram enables us to show the partitioning of total into component parts. It is also called a pie chart.
\nEx : Draw the pie digram for following data.
\n\"AP
\nTotal area = 16 + 24 + 10 + 8 + 5 = 63
\nArea changed into Degrees =
\nPaddy =\\(\\frac{16 \\times 360}{63}\\) = 91\u00b0
\nWheat = \\(\\frac{24 \\times 360}{63}\\) = 137\u00b0
\nMaize = \\(\\frac{10 \\times 360}{63}\\) = 57\u00b0
\nJower = \\(\\frac{8 \\times 360}{63}\\) = 46\u00b0
\nMillets = \\(\\frac{5 \\times 360}{63}\\) = 29\u00b0
\nTotal = 360\u00b0
\n\"AP<\/p>\n

Question 3.
\nCalculate the AM in direct method of the following data.
\n\"AP
\nAnswer:
\n\"AP
\n\\(\\overline{\\mathrm{X}}=\\frac{\\Sigma \\mathrm{fx}}{\\mathrm{N}}\\)
\nWhere N is total frequency
\n\u03a3fx = 7160; N = 200
\n\\(\\overline{\\mathrm{X}}\\) = \\(\\frac{7160}{200}\\) = 35.8
\n\u2234 \\(\\overline{\\mathrm{X}}\\) = 35.8<\/p>\n

\"AP<\/p>\n

Question 4.
\nCalculate the A.M. in Direct method.
\n\"AP
\nAnswer:
\n\"AP
\n\\(\\overline{\\mathrm{X}}=\\frac{\\Sigma \\mathrm{fx}}{\\mathrm{N}}\\)
\n\\(\\overline{\\mathrm{X}}\\) = \\(\\frac{3300}{100}\\) = 33
\n\u2234 \\(\\overline{\\mathrm{X}}\\) = 33<\/p>\n

Question 5.
\nCalculate the A.M. in Deviation method or Shortcut method.
\n\"AP
\nAnswer:
\n\"AP
\nHere class interval (C) = 10
\nAssumed mean = 55
\nA.M. (or) \\(\\overline{\\mathrm{X}}\\) = A + \\(\\frac{1}{N}\\) \u03a3fi\u03bci \u00d7 C
\n= 55 + \\(\\frac{1}{90}\\) (-44) \u00d7 10
\n= 55 + (-4.8)
\n\u2234 \\(\\overline{\\mathrm{X}}\\) = 50.2<\/p>\n

\"AP<\/p>\n

Question 6.
\n\"AP
\nAnswer:
\nCalculation of Mean :
\n\"AP
\n\\(\\overline{\\mathrm{X}}\\) = \\(\\frac{\\Sigma \\text { fixi }}{\\mathrm{N}}=\\frac{4530}{100}\\) = 45.3
\n\u2234 \\(\\overline{\\mathrm{X}}\\) = 45.3
\nCalculation of Median :
\n\"AP
\nHere N = 100; \\(\\frac{\\mathrm{N}}{2}\\) = \\(\\frac{100}{2}\\) = 50
\n50th<\/sup> value included in the cumulative frequency 40-49 class intervals. But here class intervals are unequal. So we can adjust that lower limit is \\(\\frac{39+40}{2}=\\frac{79}{2}\\) = 39.5.
\nF = 31; f = 38, C = 10
\nMedian = L + \\(\\left[\\frac{\\frac{N}{2}-F}{f}\\right]\\) \u00d7 C
\nMedian = 39.5 + \\(\\left[\\frac{50-31}{38}\\right]\\) \u00d7 10
\n= 39.5 + \\(\\left[\\frac{19}{38}\\right]\\) \u00d7 10
\n= 39.5 + 0.5 \u00d7 10
\n= 39.5 + 5
\n= 44.5
\n\u2234 Median = 44.5
\nCalculation of Mode:
\n\"AP
\nHere the maximum frequency occurs in 40 – 49 class.
\nf = 38; f1<\/sub> = 16; f2<\/sub> = 15; C = 10
\nL = \\(\\frac{39+40}{2}=\\frac{79}{2}\\) = 39.5 (Lower boundary of the model class)
\nMode = L + \\(\\left[\\frac{f-f_1}{2 f-f_1+f_2}\\right]\\) \u00d7 C
\n= 39.5 + \\(\\left[\\frac{38-16}{2(38)-(16+15)}\\right]\\) \u00d7 10
\n= 39.5 + \\(\\left[\\frac{22}{76-31}\\right]\\) \u00d7 10
\n= 39.5 + 0.48 \u00d7 10
\n= 39.5 + 4.5
\nMode = 44.38<\/p>\n

Short Answer Questions<\/span><\/p>\n

Solve the problems for the following datas.<\/p>\n

Question 1.
\nCalculate the Median for the following data.
\n\"AP
\nAnswer:
\n\"AP
\nHere N = 100 ; \\(\\frac{\\mathrm{N}}{2}=\\frac{50}{2}\\) = 25
\n25th<\/sup> value included in the cumulative frequency 21-30 class intervals.
\nLower limit = \\(\\frac{20+21}{2}=\\frac{41}{2}\\) = 20.5
\nF = 15; f = 16; C = 10 N
\nMedian = L + \\(\\left[\\frac{\\frac{N}{2}-F}{f}\\right]\\) \u00d7 C
\nMedian = 20.5 + \\(\\left(\\frac{25-15}{16}\\right)\\) \u00d7 10
\n= 20.5 + \\(\\left(\\frac{10}{16}\\right)\\) \u00d7 10
\n= 20.5 + 0.62 \u00d7 10
\n= 20.5 + 6.25 = 26.75
\nMedian = 26.75<\/p>\n

\"AP<\/p>\n

Question 2.
\nCalculate the Median for the following data.
\n\"AP
\nAnswer:
\n\"AP
\nHere N = 125 ; \\(\\frac{\\mathrm{N}}{2}=\\frac{125}{2}\\) = 62.5
\n65th<\/sup> value included in the cumulative frequency 7.6
\nTherefore 30 – 40 in the median class
\n65th<\/sup> value appears in 30 – 40 class.
\nLower limit = 30; F = 40; f = 36; C = 10 N
\nMedian = L + \\(\\left[\\frac{\\frac{N}{2}-F}{f}\\right]\\) \u00d7 C
\nMedian = 30 + \\(\\left(\\frac{62.5-40}{36}\\right)\\) \u00d7 10
\n= 30 + \\(\\left(\\frac{22.5}{36}\\right)\\) \u00d7 10
\n= 30 + 0.625 \u00d7 10
\n= 30 + 6.25
\n= 36.25<\/p>\n

\"AP<\/p>\n

Question 3.
\nCalculate the Mode for the following data.
\n\"AP
\nAnswer:
\n\"AP
\nHere the maximum frequency occurs in 59 – 60 class.
\nL = \\(\\frac{59+60}{2}=\\frac{119}{2}\\) = 59.5
\nf = 31, f1<\/sub> = 20; f2<\/sub> = 17; C = 10
\nMode = L + \\(\\left[\\frac{f-f_1}{2(f)-\\left(f_1+f_2\\right)}\\right]\\) \u00d7 C
\n= 59.5 + \\(\\left[\\frac{31-20}{2(31)-20+17}\\right]\\) \u00d7 10
\n= 59.5 + \\(\\left(\\frac{11}{62-37}\\right)\\) \u00d7 10
\n= 59.5 + \\(\\left(\\frac{11}{25}\\right)\\) \u00d7 10
\n= 59.5 + 0.44 \u00d7 10
\n= 59.5 + 4.4
\n= 63.9<\/p>\n

Additional Questions<\/span><\/p>\n

Question 4.
\nMarks of 10 students 43, 45, 68, 55, 33, 57, 40, 48, 77, 60. Calculate A.M by direct method.
\nAnswer:
\n\\(\\overline{\\mathrm{X}}=\\frac{\\Sigma \\mathrm{x}}{\\mathrm{N}}\\)
\nWhere \u03a3x = 526 ; N = 10
\n\u2234 \\(\\overline{\\mathrm{X}}\\) = 52.6<\/p>\n

Question 5.
\nCalculate A.M from the following data by direct method.
\n\"AP
\nAnswer:
\n\"AP
\n\\(\\overline{\\mathrm{X}}=\\frac{\\Sigma \\text { fixi }}{\\mathrm{N}}\\) (or) \\(\\frac{\\Sigma \\mathrm{fx}}{\\mathrm{n}}\\)
\nwhere = \u03a3fixi = 1870; N = 100
\n\\(\\overline{\\mathrm{X}}=\\frac{1870}{100}\\) = 18.7<\/p>\n

\"AP<\/p>\n

Question 6.
\nFind Median from the following data.
\n10, 15, 18, 21, 25, 30, 35
\nAnswer:
\nArrange asscending order
\n10, 15, 48, 21, 25, 30, 35
\nMedian = \\(\\left(\\frac{N+1}{2}\\right)^{\\text {th }}\\) item
\n= \\(\\left(\\frac{7+1}{2}\\right)^{\\text {th }}\\) item = \\(\\left(\\frac{8}{2}\\right)^{\\text {th }}\\) item = 4th<\/sup> item
\n4th<\/sup> item = 21
\n\u2234 Median = 21<\/p>\n

Question 7.
\nFind median from the following data.
\n\"AP
\nAnswer:
\n\"AP
\nWhere \\(\\) = 32.5 it lies in the 20 – 30 class interval.
\nSo lower limit = 20; F = 15; f = 18; N = 65; C = 10
\nMedian = L + \\(\\left[\\frac{\\frac{N}{2}-F}{f}\\right]\\) \u00d7 C
\n= 20 + \\(\\left[\\frac{32.5-15}{18}\\right]\\) \u00d7 10
\n= 20 + 9.722 = 29.72
\n\u2234 Median = 29.72<\/p>\n

\"AP<\/p>\n

Question 8.
\nFind the mode when median is 125.6 and mean is 128
\nAnswer:
\nMode = 3 Median – 2 Mean
\n= (3 \u00d7 125.6) – (2 \u00d7 128)
\n= 376.8 – 256
\n= 120.8<\/p>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nWhat is meant by Arithmetic Mean ? [March 16]
\nAnswer:
\nArithmetic Mean of a statistical data is defined as the quotient of the sum of all, the items or entries divided by the number of items.<\/p>\n

Question 2.
\nFind the A.M. for the data 30, 20, 32, 16, 27. [March 18, 17, 16]
\nAnswer:
\n\\(\\overline{\\mathrm{X}}=\\frac{\\Sigma \\mathrm{x}}{\\mathrm{N}}=\\frac{30+20+32+16+27}{5}=\\frac{125}{5}\\) = 25
\n\u2234 \\(\\overline{\\mathrm{X}}\\) = 25<\/p>\n

Question 3.
\nFind the Median for the data 10, 20, 15, 29, 35, 42.
\nAnswer:
\nThe values arranged in ascending order
\n(i.e,) 10, 15, 20, 29, 35, 42
\nMedian = \\(\\left(\\frac{n+1}{2}\\right)^{\\text {th }}\\) item
\n= \\(\\left(\\frac{6+1}{2}\\right)^{\\text {th }}\\) item
\n= \\(\\left(\\frac{7}{2}\\right)^{\\text {th }}\\) item = 3.5th<\/sup> item
\n= \\(\\frac{20+29}{2}\\) = 24.5<\/p>\n

Question 4.
\nFind the value of Median for the data 19, 1, 3, 17, 6, 12, 11, 8.
\nAnswer:
\nThe values arranged in ascending order
\ni.e., 1, 3, 6, 8, 11, 12, 17, 19
\nMedian = \\(\\left(\\frac{n+1}{2}\\right)^{\\text {th }}\\) item
\n= \\(\\left(\\frac{8+1}{2}\\right)^{\\text {th }}\\) item
\n= \\(\\left(\\frac{9}{2}\\right)^{\\text {th }}\\) item = 4.5th<\/sup> value
\n= \\(\\frac{8+11}{2}=\\frac{19}{2}\\) = 9.5
\n\u2234 Median = 9.5<\/p>\n

\"AP<\/p>\n

Question 5.
\nFind the Mode for the data 17, 6, 19, 14, 8, 6, 12, 15, 6, 10.
\nAnswer:
\nMode is 6 is occuring many items i.e., 3<\/p>\n

Question 6.
\nFind the Mode for the data 12, 11, 15, 11, 12 ,12, 15, 12, 9, 12. [March 18]
\nAnswer:
\nMode is 12 is occuring many items i.e 5<\/p>\n

Question 7.
\nWhat is a Pie diagram ? [March 18, 17, 16]
\nAnswer:
\nIt is also called a pie chart. The circle is divided into many parts as there are compo-nents by drawing straight lines from the centre to the circumference.<\/p>\n

Question 8.
\nWhat is a Bar diagram ?
\nAnswer:
\nBars are simply vertical lines, where the lengths of bars are proportional to their corresponding numerical values.<\/p>\n

Question 9.
\nFind the median of the observation – 3,- 5, -8, 0, 2, -10,
\nAnswer:
\nArrange data in ascending order – 10, – 8, – 5, – 3, 0, 2
\nMedian = \\(\\left(\\frac{n+1}{2}\\right)^{\\text {th }}\\) item
\n= \\(\\left(\\frac{6+1}{2}\\right)^{\\text {th }}\\) item
\n= \\(\\left(\\frac{7}{2}\\right)^{\\text {th }}\\) item = 3.5
\n= \\(\\frac{-5+(-3)}{2}=\\frac{-8}{2}\\) = -4
\n\u2234 Median = – 4<\/p>\n

Additional Questions<\/span><\/p>\n

Question 10.
\nWhat is mode if data is Bio-modal?
\nAnswer:
\nMode = 3 Median – 2 Mean<\/p>\n

\"AP<\/p>\n

Question 11.
\nWhat is Median ?
\nAnswer:
\nThe median is a measure of central tendency, which appears in the centre of an ordered data. It is aften described as the “middle most” value. It is called a position average.<\/p>\n

Question 12.
\nWhat is Mode?
\nAnswer:
\nMode is most frequently occuring value in data.<\/p>\n

Question 13.
\nWhat is Simple Bar Diagram ?
\nAnswer:
\nBar diagram comprising a group of equi-spaced and equi-width rectangular bars for each class of data.<\/p>\n

Question 14.
\nWhat is Multiple Bar Diagrams ?
\nAnswer:
\nMultiple bar diagrams are used for comparing two or more set of data.<\/p>\n

Question 15.
\nWhat is Component Bar Diagram or Sub-Divided Bar Diagram ?
\nAnswer:
\nThese diagrams are used to represent various parts of the total.<\/p>\n","protected":false},"excerpt":{"rendered":"

Andhra Pradesh BIEAP\u00a0AP Inter 1st Year Economics Study Material 10th Lesson Economic Statistics Textbook Questions and Answers. AP Inter 1st Year Economics Study Material 10th Lesson Economic Statistics Essay Questions Question 1. What is the relationship between Economics and Statistics? [March 18, 16] Answer: There is a close relationship between statistics and economics. In the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35881"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35881"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35881\/revisions"}],"predecessor-version":[{"id":35882,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35881\/revisions\/35882"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35881"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35881"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35881"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}