{"id":43772,"date":"2022-05-29T17:00:49","date_gmt":"2022-05-29T11:30:49","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=43772"},"modified":"2023-01-25T14:18:11","modified_gmt":"2023-01-25T08:48:11","slug":"ml-aggarwal-class-6-solutions-for-icse-maths-chapter-15-ex-15-5","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-6-solutions-for-icse-maths-chapter-15-ex-15-5\/","title":{"rendered":"ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 15 Data Handling Ex 15.5"},"content":{"rendered":"
Question 1. (ii) After arranging data, we get Question 2. (b) Arranging the given data in ascending order, we get (c) In the given numbers, Question 3. (b) Arranging the given data in ascending order, we get, (c) In the given numbers 3 is repeated more than any other number ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 15 Data Handling Ex 15.5 Question 1. Find the median of the following data: (i) 3, 1, 5, 6, 3, 4, 5 (ii) 3, 1, 5, 6, 3, 4, 5, 6 Solution: (i) Arrange the data in ascending order We get, 1, 3, 3, 4, 5, … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\n
\nFind the median of the following data:
\n(i) 3, 1, 5, 6, 3, 4, 5
\n(ii) 3, 1, 5, 6, 3, 4, 5, 6
\nSolution:
\n(i) Arrange the data in ascending order
\nWe get, 1, 3, 3, 4, 5, 5, 6
\n\u2234 Median = \\(\\left(\\frac{n+1}{2}\\right)^{t h}\\) term
\n= \\(\\left(\\frac{7+1}{2}\\right)^{\\text { th }}\\) = 4th<\/sup> term<\/p>\n
\n1, 3, 3, 4, 5, 5, 6, 6
\n\u2234 Median = \\(\\frac{4+5}{2}=\\frac{9}{2}=4 \\cdot 5\\)<\/p>\n
\nCalculate the mean, the median and the mode of the numbers :
\n1, 3, 2, 6, 2, 3, 1, 3
\nSolution:
\n(a) Mean = \\(\\frac{1+3+2+6+2+3+1+3}{8}\\)
\n= \\(\\frac{21}{8}\\) = 2.625
\nHence mean = 2.625<\/p>\n
\n1, 1, 2, 2, 3, 3, 3,6
\nTotal number of observations (items) = 8 (even).
\nThere are two middle items : – 2 and 3.
\nTheir average = \\(\\frac{2+3}{2}=\\frac{5}{2}=2 \\cdot 5\\)
\nHence the median of the given numbers = 2.5.<\/p>\n
\n3 is repeated more number of time than any other number.
\n\u2234 Mode = 3.<\/p>\n
\nCalculate the mean, the median and the mode of the following numbers :
\n3, 7, 2, 5, 3, 4, 1, 5, 3, 6
\nSolution:
\n(a) Mean = \\(\\frac{3+7+2+5+3+4+1+5+3+6}{10}=\\frac{39}{10}=3 \\cdot 9\\)
\nHence the mean = 3.9<\/p>\n
\n1, 2, 3, 3, 3, 4, 5, 5, 6, 7
\nTotal numbers of obvervations (items) = 10 (even)
\nThere are two middle items – 3 and 4
\nTheir average = \\(\\frac{3+4}{2}=\\frac{7}{2}=3 \\cdot 5\\)
\nHence the median of the given numbers = 3.5<\/p>\n
\n\u2234 Mode = 3<\/p>\nML Aggarwal Class 6 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"