The arithmetic mean and variance of a set of ten figures are known to be 17 and 33 respectively. Of the ten figures, one figure (i.e. 26) was subsequently found to be inaccurate, and was weeded out. What is the resulting (i) arithmetic mean? (ii) standard deviation
Answers
Given : The arithmetic mean and variance of a set of ten figures are known to be 17 and 33 respectively. Of the ten figures, one figure (i.e. 26) was subsequently found to be inaccurate, and was weeded out.
To find : (i) arithmetic mean? (ii) standard deviation
Solution :
Mean = Sum of Terms / Number of Terms
=> 17 = Sum of terms/ 10
=> Sum of Terms = 170
Now 26 is removed from it
Sum = 170 - 26 = 144
Terms = 9
New Mean = 144/9 = 16
Variance = ∑x² / 10 - (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 = 2544/9 - 16²
=> new Variance = 26.67
SD = √Variance
=> New SD = 5.16
New Mean = 16
New SD = 5.16
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