Let xi(1≤i≤10) be ten observations of a random variable X . If ∑i=110xi-p=3 and ∑i=110xi-p2=9 where 0≠p∈R , then the standard deviation of these observations is :

Given ∑i=110xi-p=3 It gives ∑xi=10P+3 Given ∑i=110xi-p2=9 It gives ∑xi2-2p∑xi+10p2=9∑xi2=2p(10p+3)-10p2+9=10p2+6p+9The standard deviation is σ2=∑xi210−∑xi102=p2+6p10+910−p2+9100+6p10=81100 Hence the standard deviation is σ=910