Answer: 2.76 g
Step-by-step explanation:
The formula to find the standard deviation:-
[tex]\sigma=\sqrt{\dfrac{\sum(x_i-\overline{x})^2}{n}}[/tex]
The given data values : 560 g, 562 g, 556 g, 558 g, 560 g, 556 g, 559 g, 561 g, 565 g, 563 g.
Then, [tex]\overline{x}=\dfrac{\sum_{i=1}^{10} x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{560+562+556+558+560+556+559+561+565+563}{10}\\\\\Rightarrow\ \overline{x}=\dfrac{5600}{10}=560[/tex]
Now, [tex]\sum_{i=1}^{10}(x_i-\overline{x})^2=0^2+2^2+(-4)^2+(-2)^2+0^2+(-4)^2+(-1)^2+1^2+5^2+3^2\\\\\Rightarrow\ \sum_{i=1}^{10}(x_i-\overline{x})^2=76[/tex]
Then, [tex]\sigma=\sqrt{\dfrac{76}{10}}=\sqrt{7.6}=2.76[/tex]
Hence, the standard deviation of his measurements = 2.76 g
