In an experiment with 10 observations on x the following results are available displaystyle ∑ x2=354 and displaystyle ∑ x=58 . If one observation that was 8 was found to be wrong and was replaced by the corrected value 10, then the corrected variance is

Question

In an experiment with 10 observations on x the following results are available displaystyle ∑ x2=354 and displaystyle ∑ x=58 . If one observation that was 8 was found to be wrong and was replaced by the corrected value 10, then the corrected variance is (A) 5 (B) 3 (C) 4 (D) 6

Tardigrade Admin · Accepted Answer

∑ x2=354, ∑ x=58 ∑ x prime=58-8+10=60 ∑ x prime 2=354+100-64=390 Variance =(∑(x prime)2/10)-(( Sigma(x prime)/10))2=(390/10)-((60/10))2 =39-36 =3

Q. In an experiment with $10$ observations on $x$ the following results are available $\sum x^{2} = 354$ and $\sum x = 58$ . If one observation that was $8$ was found to be wrong and was replaced by the corrected value $10,$ then the corrected variance is

$5$

$3$

$4$

$6$

$\sum x^{2} = 354, \sum x = 58$
$\sum x^{'} = 58 - 8 + 10 = 60$
$\sum x^{'2} = 354 + 100 - 64 = 390$
Variance $= \frac{\sum ( x ^{'} ) ^{2}}{10} - (\frac{Σ ( x ^{'} )}{10})^{2} = \frac{390}{10} - (\frac{60}{10})^{2}$
$= 39 - 36$
$= 3$

Q. In an experiment with 10 observations on x the following results are available ∑x2=354 and ∑x=58 . If one observation that was 8 was found to be wrong and was replaced by the corrected value 10, then the corrected variance is

5

3

4

6

∑x2=354,∑x=58 ∑x′=58−8+10=60 ∑x′2=354+100−64=390 Variance =10∑(x′)2​−(10Σ(x′)​)2=10390​−(1060​)2 =39−36 =3

Q. In an experiment with $10$ observations on $x$ the following results are available $\sum x^{2} = 354$ and $\sum x = 58$ . If one observation that was $8$ was found to be wrong and was replaced by the corrected value $10,$ then the corrected variance is

$5$

$3$

$4$

$6$

$\sum x^{2} = 354, \sum x = 58$
$\sum x^{'} = 58 - 8 + 10 = 60$
$\sum x^{'2} = 354 + 100 - 64 = 390$
Variance $= \frac{\sum ( x ^{'} ) ^{2}}{10} - (\frac{Σ ( x ^{'} )}{10})^{2} = \frac{390}{10} - (\frac{60}{10})^{2}$
$= 39 - 36$
$= 3$