Let xi(1 ≤ i ≤ 10) be ten observations of a random variable X. If displaystyle∑i=110(xi-p)=3 and displaystyle∑i=110(xi-p)2=9 where 0 ≠ p ∈ R, then the standard deviation of these observations is :

Question

Let xi(1 ≤ i ≤ 10) be ten observations of a random variable X. If displaystyle∑i=110(xi-p)=3 and displaystyle∑i=110(xi-p)2=9 where 0 ≠ p ∈ R, then the standard deviation of these observations is : (A) √(3/5) (B) (7/10) (C) (9/10) (D) (4/5)

Tardigrade Admin · Accepted Answer

Variance =( Sigma( x i - p )2/ n )-(( Sigma( x i - p )/ n ))2 =(9/10)-((3/10))2=(81/100) S.D. = (9/10)

Q. Let $x_{i} (1 \leq i \leq 10)$ be ten observations of a random variable X. If $i = 1 \sum 10 (x_{i} - p) = 3$ and $i = 1 \sum 10 (x_{i} - p)^{2} = 9$ where $0 \neq = p \in R$ , then the standard deviation of these observations is :

$\frac{3}{5}$

$\frac{7}{10}$

$\frac{9}{10}$

$\frac{4}{5}$

Variance $= \frac{Σ ( x _{i} - p ) ^{2}}{n} - (\frac{Σ ( x _{i} - p )}{n})^{2}$
$= \frac{9}{10} - (\frac{3}{10})^{2} = \frac{81}{100}$
$S . D . = \frac{9}{10}$

Q. Let xi​(1≤i≤10) be ten observations of a random variable X. If i=1∑10​(xi​−p)=3 and i=1∑10​(xi​−p)2=9 where 0=p∈R, then the standard deviation of these observations is :

53​​

107​

109​

54​