The sample is given
![37.3055,37.4055,37.4673,37.3520,37.5109,37.3970,37.4819,37.2700,37.3763](https://img.qammunity.org/2023/formulas/mathematics/college/r9qqmlhzy76p6nyaouv41ipy9r321t9ig9.png)
First determine the mean of the sample,
The total number of masses are 9.
Then mean is calculated as sum of all the masses divided by total number of masses.
![(37.3055+37.4055+37.4673+37.3520+37.5109+37.3970+37.4819+37.2700+37.3763)/(9)](https://img.qammunity.org/2023/formulas/mathematics/college/lsfc1tvafxabgg36vmieykr8myucf4njqe.png)
![(336.5614)/(9)=37.39572](https://img.qammunity.org/2023/formulas/mathematics/college/rz4vy607wxm18nqviuex7gi1o7g9cqwpem.png)
The mean obtained is 37.39572.
To determine the standard deviation , use the formula.
![\sigma=\sqrt[]{(1)/(N)\Sigma(x_i-\mu)^2}](https://img.qammunity.org/2023/formulas/mathematics/college/9gkn6b8xebzf4b5dujfsu5tya7x33x5u5n.png)
Now we have to evaluate the value of summation in the square root.
Now substitute the value in standard deviation formula
![\sigma=\sqrt[]{(1)/(9)(0.0520756928)}=\sqrt[]{0.0057861880888}=0.076066997](https://img.qammunity.org/2023/formulas/mathematics/college/swaqzmb3pared14jbjem17di5ee0wj49n3.png)
Hence the standard deviation of the masses given is 0.076066997.