1) The mean and sample standard deviation of the dataset consisting of 10 observations are 15 and 11 respectively. Later it is noted that one observation 14 is wrongly noted as 11. What is the mean of the original dataset? (Correct to 2 decimal place accuracy)
2) The mean and sample standard deviation of the dataset consisting of 10 observations is 15 and 11 respectively. Later it is noted that one observation 14 is wrongly noted as 11. 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 10 observations is 15 and 11 respectively.
one observation 14 is wrongly noted as 11
To Find : sample variance of the original dataset
Solution:
Mean of 10 observation = 15
Hence sum of observations ∑X = 15 * 10 =150
Sample SD = 11
=> Sample Variance = SD² = 11² = 121
Sample Variance = [∑X² - {(∑X)²/n }]/(n - 1)
n = 10 , ∑X = 150
=> 121 = [∑X² - {(150)²/10 }]/(10 - 1)
=> 1089 = ∑X² - 2250
=> ∑X² = 3339
one observation 14 is wrongly noted as 11
Sum in Original data set ∑X = 150 - 11 + 14 =153
∑X² in Original data set =3339 - 11² + 14² = 3414
Sample Variance in Original data set = [∑X² - {(∑X)²/n }]/(n - 1)
∑X² = 3414 , ∑X = 153 , n = 10
Sample Variance in Original data set = [3414 - {( 153)²/10} ]/(10-1)
= (34140 - 23409 )/90
= 10731/90
= 119.233
sample variance of the original dataset =119.23
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Answer:
119.23 is the answer tou can check it by solving it by yourself