The mean and sample standard deviation of the dataset consisting of 10 observations is 16 and 10 respectively. Later it is noted that one observation 16 is wrongly noted as 10. What is the mean of the original dataset? (Correct to 2 decimal place accuracy)
Answers
Given : The mean and sample standard deviation of the dataset consisting of 10 observations is 16 and 10respectively.
one observation 16 is wrongly noted as 10
To Find : sample variance of the original dataset
Solution:
Mean of 10 observation = 16
Hence sum of observations ∑X = 16 * 10 =160
Sample SD = 10
=> Sample Variance = SD² = 10² = 100
Sample Variance = [∑X² - {(∑X)²/n }]/(n - 1)
n = 10 , ∑X = 160
=> 100 = [∑X² - {(160)²/10 }]/(10 - 1)
=>900 = ∑X² - 2560
=> ∑X² = 3460
one observation 16 is wrongly noted as 10
Sum in Original data set ∑X = 160 - 10 + 16 =166
∑X² in Original data set =3460 - 10² + 16² = 3616
Sample Variance in Original data set = [∑X² - {(∑X)²/n }]/(n - 1)
∑X² =3616 , ∑X = 166 , n = 10
Sample Variance in Original data set = [3616 - {( 166²/10} ]/(10-1)
= (36160 - 27556 )/90
=8,604/90
= 95.6
sample variance of the original dataset =95.6
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Answer:
Step-by-step explanation:
Given : The mean and sample standard deviation of the dataset consisting of 10 observations is 16 and 10respectively.
one observation 16 is wrongly noted as 10
To Find : sample variance of the original dataset
Solution:
Mean of 10 observation = 16
Hence sum of observations ∑X = 16 * 10 =160
Sample SD = 10
=> Sample Variance = SD² = 10² = 100
Sample Variance = [∑X² - {(∑X)²/n }]/(n - 1)
n = 10 , ∑X = 160
=> 100 = [∑X² - {(160)²/10 }]/(10 - 1)
=>900 = ∑X² - 2560
=> ∑X² = 3460
one observation 16 is wrongly noted as 10
Sum in Original data set ∑X = 160 - 10 + 16 =166
∑X² in Original data set =3460 - 10² + 16² = 3616
Sample Variance in Original data set = [∑X² - {(∑X)²/n }]/(n - 1)
∑X² =3616 , ∑X = 166 , n = 10
Sample Variance in Original data set = [3616 - {( 166²/10} ]/(10-1)
= (36160 - 27556 )/90
=8,604/90
= 95.6
sample variance of the original dataset =95.6