Example 9 - Consider frequency distribution table which - Examples

  1. Chapter 15 Class 9 Probability
  2. Serial order wise
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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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