Example 3 - Chapter 12 Class 9 Statistics

Last updated at April 16, 2024 by

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Example 3 A teacher wanted to analyze the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows: 0 - 20, 20 - 30, . . ., 60 - 70, 70 - 100. Then she formed the following table: Make histogram of the data The data has class intervals of varying width. So, we calculate adjusted frequency Adjusted frequency = (πΉπ‘Ÿπ‘’π‘žπ‘’π‘’π‘›π‘π‘¦ Γ— π‘…π‘’π‘žπ‘’π‘–π‘Ÿπ‘’π‘‘ 𝑠𝑖𝑧𝑒 π‘œπ‘“ π‘–π‘›π‘‘π‘’π‘Ÿπ‘£π‘Žπ‘™)/(πΆπ‘’π‘Ÿπ‘Ÿπ‘’π‘›π‘‘ 𝑠𝑖𝑧𝑒 π‘œπ‘“ π‘–π‘›π‘‘π‘’π‘Ÿπ‘£π‘Žπ‘™)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo