The variation found in a set can be estimated by the standard deviation, s:

\({s_x}\, = \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\)

Calculate the standard deviation for each treatment.

Treatment	Dose (mg/kg)	Log of number of colonies	Mean (\(\overline x \,\))	\({x_i}\, - \,\overline x \)
Control	-	9.0,9.5,9.0,8.9	9.1	(-0.1), 0.4, (-0.1), (-0.2)
Vancomycin	1.0	8.5,8.4,8.2	8.36	0.14, 0.04, (-0.16)
	5.0	5.3,5.9,4.7	5.3	0, 0.6, (-0.6)
Teixobactin	1.0	8.5,6.0,8.4,6.0	7.22	1.28, (-1.22), 1.18, (-1,22)
	5.0	3.8,4.9,5.2,4.9	4.7	(-0.9), 0.2,0.5, 0.2

In order to calculate the standard deviation for control, values are substituted in the following equation:

\(\begin{aligned}{l}{s_c}\, &= \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\\\,\,\,\, &= \sqrt {\frac{1}{{4 - 1}}} \sum {{{({{( - 0.1)}^2} + {{0.4}^2} + {{( - 0.1)}^2} + {{( - 0.2)}^2})}^{}}} \\\,\,\,\, &= \sqrt {\frac{1}{3}} \times 0.22\\\,\,\,\, &= \sqrt {0.073} \,\\\,\,\,\, &= 0.270\end{aligned}\)

Thus, the standard deviation for control is 0.270.

In order to calculate the standard deviation for vancomycin at 1 mg/kg, values are substituted in the equation given:

\(\begin{aligned}{l}{s_{v1\,}} &= \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\\\,\,\,\, &= \sqrt {\frac{1}{{3 - 1}}} \sum {({{(0.14)}^2} + {{0.04}^2} + {{( - 0.16)}^2})} \\\,\,\,\, &= \sqrt {\frac{1}{2}} \times 0.0468\\\,\,\,\, &= \sqrt {0.0234} \,\\\,\,\,\, &= 0.152\end{aligned}\)

Thus, the standard deviation for vancomycin at 1mg/kg is 0.152.

In order to calculate the standard deviation for vancomycin at 5 mg/kg, values are substituted in the equation given:

\(\begin{aligned}{l}{s_{v5\,}} &= \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\\\,\,\,\, &= \sqrt {\frac{1}{{3 - 1}}} \sum {({{(0)}^2} + {{0.6}^2} + {{( - 0.6)}^2})} \\\,\,\,\, &= \sqrt {\frac{1}{2}} \times 0.72\\\,\,\,\, &= \sqrt {0.36} \,\\\,\,\,\, &= 0.6\end{aligned}\)

Thus, the standard deviation for vancomycin at 5mg/kg is 0.6.

In order to calculate the standard deviation for teixobactin at 1 mg/kg, values are substituted in the equation given:

\(\begin{aligned}{l}{s_{t1\,}} &= \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\\\,\,\,\, &= \sqrt {\frac{1}{{4 - 1}}} \sum {({{(1.28)}^2} + {{( - 1.22)}^2} + {{1.18}^2} + {{( - 1.22)}^2})} \\\,\,\,\, &= \sqrt {\frac{1}{3}} \times 6.006\\\,\,\,\, &= \sqrt {2.002} \,\\\,\,\,\, &= 1.14\end{aligned}\)

Thus, the standard deviation for teixobactin at 1 mg/kg is 1.14.

In order to calculate the standard deviation for teixobactin at 1 mg/kg, values are substituted in the equation given:

\(\begin{aligned}{l}{s_{t5\,}} &= \sqrt {\frac{1}{{n - 1}}} \sum {({x_i}} - \,\overline x {)^2}\\\,\,\,\, &= \sqrt {\frac{1}{{4 - 1}}} \sum {{{({{(0.9)}^2} + {{(0.2)}^2} + {{0.5}^2} + 0.2)}^2})} \\\,\,\,\, &= \sqrt {\frac{1}{3}} \times 1.14\\\,\,\,\, &= \sqrt {0.38} \,\\\,\,\,\, &= 0.616\end{aligned}\)

Thus, the standard deviation for teixobactin at 5 mg/kg is 0.616.