For frequency distribution of marks in statistics of 200 candidates grouped in intervals 1-5,6-10,11-15,16-20,21-25.......the mean and standard deviation werevfound to be 40 and 15 respectively .later it was discovered that score 43 was misread as 53 in obtaining the frequency distribution.find the corrected mean and standard deviation corresponding to the corrected frequency distribution
Answers
Given : frequency distribution of marks in statistics of 200 candidates grouped in intervals 1-5,6-10,11-15,16-20,21-25.......the mean and standard deviation were found to be 40 and 15 respectively . score 43 was misread as 53
To find : the corrected mean and standard deviation
Solution:
Mean = Total Marks / Total Students
=> 40 = Total Marks / 200
=> Total Marks = 8000
score 43 was misread as 53
53 Falls in 51 - 55 With Mean 53
43 Falls in 41 - 43 with mean 43
Hence total Score = 8000 - 53 + 43
= 7990
Corrected Mean = 7990/200 = 39.95
SD² = ∑x²/n - ( Mean)²
=> 15² = ∑x²/200 - 40²
=> ∑x²/200 = 1825
=> ∑x² = 200 * 1825
Corrected ∑x² = 200 * 1825 - 53² + 43²
= 200 * 1825 - 960
= 200 ( 1825 - 4.8)
= 200 ( 1820.2)
Corrected SD² = 200 ( 1820.2)/200 - (39.95)²
=> Corrected SD² = 224.1975
=> Corrected SD = 14.97
Corrected Mean = 39.95
Corrected SD = 14.97
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