The mean variance of a set of 10 values are know to be 17 and 33 respectively of the ten values. One value 26 was subsequently found inaccurate and was weeded out what is the resulting mean and standard deviation
Dear Student
Your question seems to be unclear. Following is the reframed question as per your query-
The mean and variance of a set of 10 values are known to be 17 and 33 respectively. Of the ten values one value 26 was subsequently found inaccurate and was weeded out. What is the resulting mean and the resulting standard deviation ?
Answer-
Mean = Sum of terms / No. of Terms
Given Mean = 17 , No. of terms = 10
Therefore,
17 = Sum of terms/ 10
Sum of Terms = 170
As, 26 is Weeded
New Sum = 170 - 26 = 144
Now Terms = 9
New Mean = 144/9 = 16
Variance = ∑x² /n - (mean)²
33 = ∑x²/10 - 17²
∑x² / 10 = 33 + 289
∑x² = 3220
New ∑x² = 3220 - 26² = 2544
New Variance = new ∑x² / 9 - (New mean)²
=> New Variance = 2544/9 - 16²
=> New Variance = 26.67
New Variance = New SD²
=> New SD = 5.16
Mean = 16
SD = 5.16
Regards
Your question seems to be unclear. Following is the reframed question as per your query-
The mean and variance of a set of 10 values are known to be 17 and 33 respectively. Of the ten values one value 26 was subsequently found inaccurate and was weeded out. What is the resulting mean and the resulting standard deviation ?
Answer-
Mean = Sum of terms / No. of Terms
Given Mean = 17 , No. of terms = 10
Therefore,
17 = Sum of terms/ 10
Sum of Terms = 170
As, 26 is Weeded
New Sum = 170 - 26 = 144
Now Terms = 9
New Mean = 144/9 = 16
Variance = ∑x² /n - (mean)²
33 = ∑x²/10 - 17²
∑x² / 10 = 33 + 289
∑x² = 3220
New ∑x² = 3220 - 26² = 2544
New Variance = new ∑x² / 9 - (New mean)²
=> New Variance = 2544/9 - 16²
=> New Variance = 26.67
New Variance = New SD²
=> New SD = 5.16
Mean = 16
SD = 5.16
Regards