cara baca uang rupiah ini Rp59.959.096

the\ standard\ deviation\ (\sigma)= \sqrt{ \frac{\big{(x_1-\overline{x})^2+(x_2-\overline{x})^2+...+(x_n-\overline{x})^2}}{\big{n}}} \\\\\\and\ \ \ \overline{x}= \frac{\big{x_1+x_2+...+x_n}}{\big{n}} \\\\---------------------\\\\x_1=0.26;\ \ \ x_2=0.23\ \ \ \Rightarrow\ \ \ \overline{x}= \frac{0.26+0.23}{2} =0.245\\\\\\\sigma=\sqrt{ \frac{\big{(0.26-0.245)^2+(0.23-0.245)^2}}{\big{2}}} =\sqrt{ \frac{\big{(0.015)^2+(-0.015)^2}}{\big{2}}} =
=\sqrt{ \frac{ \big{2\cdot (0.015)^2}}{\big{2}}} =\sqrt{ (0.015)^2} =0.015

Answered by kate200468 | 2024-06-10

To find the standard deviation of 0.26 and 0.23, we first calculated the mean, which is 0.245. We then computed the variance as 0.000225 and, finally, the standard deviation, which equals 0.015. This indicates how spread out the values are from the mean.
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