Example 9 - Consider frequency distribution table which - Examples

Example 9 - Chapter 15 Class 9 Probability - Part 2

Example 9 - Chapter 15 Class 9 Probability - Part 3

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Example 9 Consider the frequency distribution table which gives the weights of 38 students of a class. Find the probability that the weight of a student in the class lies in the interval 46 โ€“ 50 kg. Total number of students = 38 Number of students with weight in the interval 46 โ€“ 50 kg = 3 P(weight of a student is in the interval 46 โ€“ 50 kg) = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก๐‘  ๐‘ค๐‘–๐‘กโ„Ž ๐‘ค๐‘’๐‘–๐‘”โ„Ž๐‘ก ๐‘–๐‘› ๐‘กโ„Ž๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘Ÿ๐‘ฃ๐‘Ž๐‘™ 46 โˆ’ 50 ๐‘˜๐‘”)/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก๐‘ ) = 3/38 = 0.079 Example 9 (ii) Give two events in this context, one having probability 0 and the other having probability 1. Probability 0 There are no student weighing 30kg, Hence, we take probability of student weighing 30 kg Number of student weighing 30 kg = 0 Total number of students = 38 P(student weighing 30 kg) = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก๐‘  ๐‘ค๐‘’๐‘–๐‘”โ„Ž๐‘–๐‘›๐‘” 30 ๐‘˜๐‘”)/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก๐‘ ) = 0/38 = 0 Probability 1 All students weigh more than 30 kg, Hence, we take probability of student weighing more than 30 kg Number of student weighing more than 30 kg = 38 Total number of students = 38 P(student weighing more than 30 kg) = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก๐‘  ๐‘ค๐‘’๐‘–๐‘”โ„Ž๐‘–๐‘›๐‘” ๐‘š๐‘œ๐‘Ÿ๐‘’ ๐‘กโ„Ž๐‘Ž๐‘› 30 ๐‘˜๐‘”)/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘ ๐‘ก๐‘ข๐‘‘๐‘’๐‘›๐‘ก๐‘ ) = 38/38 = 1

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.