The mean and sample standard deviation of the dataset consisting of 6 observations is 19 and 9 respectively. Later it is noted that one observation 11 is wrongly noted as 7. What is the sample variance of the original dataset? (Correct to 2 decimal place accuracy)
Answers
Given : The mean and sample standard deviation of the dataset consisting of 6 observations is 19 and 9 respectively.
one observation 11 is wrongly noted as 7
To Find : sample variance of the original dataset
Solution:
Mean of 6 observation = 19
Hence sum of observations ∑X = 19 * 6 =114
Sample SD = 9
=> Sample Variance = SD² = 9² = 81
Sample Variance = [∑X² - {(∑X)²/n }]/(n - 1)
n = 6 , ∑X = 114
=> 81 = [∑X² - {(114)²/6 }]/(6 - 1)
=> 405 = ∑X² - 2166
=> ∑X² = 2571
one observation 11 is wrongly noted as 7
Sum in Original data set ∑X = 114 - 7 + 11 =118
∑X² in Original data set = 2571 - 7² + 11² = 2643
Sample Variance in Original data set = [∑X² - {(∑X)²/n }]/(n - 1)
∑X² = 2643 , ∑X = 118 , n = 6
Sample Variance in Original data set = [2643 - {(118)²/6} ]/(6-1)
= (15858 - 13924 )/30
= 1934/30
= 64.4666
sample variance of the original dataset = 64.47
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Answer:
Step-by-step explanation:
Mean of 6 observation = 19
Hence sum of observations ∑X = 19 * 6 =114
Sample SD = 9
=> Sample Variance = SD² = 9² = 81
Sample Variance = [∑X² - {(∑X)²/n }]/(n - 1)
n = 6 , ∑X = 114
=> 81 = [∑X² - {(114)²/6 }]/(6 - 1)
=> 405 = ∑X² - 2166
=> ∑X² = 2571