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 Let xi(1i10) be ten observations of a random variable X . If i=110xi-p=3 and i=110xi-p2=9 where 0pR , then the standard deviation of these observations is : 

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a
45
b
35
c
710
d
910

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detailed solution

Correct option is D

 Given i=110xi-p=3 It gives xi=10P+3 Given i=110xi-p2=9 It gives xi2-2pxi+10p2=9xi2=2p(10p+3)-10p2+9=10p2+6p+9

The standard deviation is 

σ2=xi210xi102=p2+6p10+910p2+9100+6p10=81100

 Hence the standard deviation is σ=910

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