Math, asked by AvidLearnerHS, 1 month ago

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

Answered by amitnrw
29

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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Answered by rajsinghania4475
3

Answer:

119.23 is the answer tou can check it by solving it by yourself

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