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Grade: 11
        The mean and standard deviation of a group of 100 observations were found to be 20 and 3 , respectively. Later on it was found that three observations were incorrect , which were recorded as 21 , 21, and 18 . Find the mean and standard deviation if incorrect observations are omitted.
2 years ago

Answers : (1)

Arun
23036 Points
							
Dear Yashika
 
 
Hello student,
Incorrect eman = 20 and incorrect S.D. =3.
Hence, \sum_{i=1}^{n}\frac{x_{i}}{100}= 20
This gives S xi = 2000
corrected S x= 2000-21-21-18 = 1940
correct mean = 1940/97 = 20.
Incorrect S.D. = 3
Incorrect variance = 9.
\frac{1}{100}(incorrect \sum_{i=1}^{n}x_{i}^{2}) - (incorrect mean)^{2} = 9
\frac{1}{100}(incorrect \sum_{i=1}^{n}x_{i}^{2}) - 400= 9
incorrect \sum_{i=1}^{n}x_{i}^{2} = 40900
Correct Sxi2 = 40900 – 21- 21- 182
Correct Sxi2 = 39694
Hence, variance of remaining observations = 39694/ 97 – (20)2
= 409.216 – 400
= 9.216
Hence, corrected S.D. = v9.216 = 3.035
2 years ago
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