Can you solve this for me?√(.63-4.68)^(2)+(.67-4.68)^(2)+(6.12-4.68)^(2)+(5.67-4.68)^(2)+(5.76-4.68)^(2)+(2.64-4.68)^(2)+(.99-4.68)^(2)+(7.81-4.68)^(2)+(1.31-4.68)^(2)+(8.03-4.68)^(2)+(8.26-4.68)^(2)+(8.27-4.68)^(2)/12-1
Hi, welcome to Just Answer.
It seems you are trying to compute the sample standard deviation of ( 0.63, 0.67, 6.12, 5.67, 5.76, 2.64, 0.99, 7.81, 1.31, 8.03, 8.26, 8.27).
The sample mean is 4.68.
The sum of square (the long expression in the numerator) is
(.63-4.68)^(2)+(.67-4.68)^(2)+(6.12-4.68)^(2)+(5.67-4.68)^(2)+(5.76-4.68)^(2)+(2.64-4.68)^(2)+(.99-4.68)^(2)+(7.81-4.68)^(2)+(1.31-4.68)^(2)+(8.03-4.68)^(2)+(8.26-4.68)^(2)+(8.27-4.68)^(2)
= 16.4025 + 16.0801 + 2.0736 + 0.9801 + 1.1664 + 4.1616 + 13.6161 + 9.7969 + 11.3569 + 11.2225 + 12.8164 + 12.8881
=(NNN) NNN-NNNN
The variance is(NNN) NNN-NNNN (12-1) = 10.23284.
Finally, the sample standard deviation is √10.23284 = 3.19888.
The answer is 3.19888.
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John
Experience: MS and BS in Math