An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (1)
Calculate the standard deviation for the Control group: s ≈ 0.2708 .
Calculate the standard deviation for Vancomycin (1.0 mg/kg): s ≈ 0.1581 .
Calculate the standard deviation for Vancomycin (5.0 mg/kg): s = 0.6 .
Calculate the standard deviation for Teixobactin (1.0 mg/kg): s ≈ 1.4154 .
Calculate the standard deviation for Teixobactin (5.0 mg/kg): s ≈ 0.6164 .
The calculated standard deviations are 0.27 , 0.16 , 0.60 , 1.42 , 0.62 .
Explanation
Understanding the Problem We are given a table with the log of the number of colonies for different treatments and doses. Our goal is to calculate the standard deviation for each treatment and dose using the provided data. The formula for standard deviation is:
s = n − 1 ∑ i = 1 n ( x i − x ˉ ) 2
where x i are the individual log values, x ˉ is the mean log value, and n is the number of log values.
Calculating Control Standard Deviation First, let's calculate the standard deviation for the Control group. The log values are 9.0, 9.5, 9.0, 8.9, and the mean is 9.1. So,
s = 4 − 1 ( 9.0 − 9.1 ) 2 + ( 9.5 − 9.1 ) 2 + ( 9.0 − 9.1 ) 2 + ( 8.9 − 9.1 ) 2
s = 3 ( − 0.1 ) 2 + ( 0.4 ) 2 + ( − 0.1 ) 2 + ( − 0.2 ) 2
s = 3 0.01 + 0.16 + 0.01 + 0.04
s = 3 0.22
s ≈ 0.2708
Calculating Vancomycin (1.0 mg/kg) Standard Deviation Next, let's calculate the standard deviation for Vancomycin at 1.0 mg/kg. The log values are 8.5, 8.4, 8.2, and the mean is 8.4. So,
s = 3 − 1 ( 8.5 − 8.4 ) 2 + ( 8.4 − 8.4 ) 2 + ( 8.2 − 8.4 ) 2
s = 2 ( 0.1 ) 2 + ( 0 ) 2 + ( − 0.2 ) 2
s = 2 0.01 + 0 + 0.04
s = 2 0.05
s ≈ 0.1581
Calculating Vancomycin (5.0 mg/kg) Standard Deviation Now, let's calculate the standard deviation for Vancomycin at 5.0 mg/kg. The log values are 5.3, 5.9, 4.7, and the mean is 5.3. So,
s = 3 − 1 ( 5.3 − 5.3 ) 2 + ( 5.9 − 5.3 ) 2 + ( 4.7 − 5.3 ) 2
s = 2 ( 0 ) 2 + ( 0.6 ) 2 + ( − 0.6 ) 2
s = 2 0 + 0.36 + 0.36
s = 2 0.72
s = 0.36
s = 0.6
Calculating Teixobactin (1.0 mg/kg) Standard Deviation Let's calculate the standard deviation for Teixobactin at 1.0 mg/kg. The log values are 8.5, 6.0, 8.4, 6.0, and the mean is 7.2. So,
s = 4 − 1 ( 8.5 − 7.2 ) 2 + ( 6.0 − 7.2 ) 2 + ( 8.4 − 7.2 ) 2 + ( 6.0 − 7.2 ) 2
s = 3 ( 1.3 ) 2 + ( − 1.2 ) 2 + ( 1.2 ) 2 + ( − 1.2 ) 2
s = 3 1.69 + 1.44 + 1.44 + 1.44
s = 3 5.91
s ≈ 1.4154
Calculating Teixobactin (5.0 mg/kg) Standard Deviation Finally, let's calculate the standard deviation for Teixobactin at 5.0 mg/kg. The log values are 3.8, 4.9, 5.2, 4.9, and the mean is 4.7. So,
s = 4 − 1 ( 3.8 − 4.7 ) 2 + ( 4.9 − 4.7 ) 2 + ( 5.2 − 4.7 ) 2 + ( 4.9 − 4.7 ) 2
s = 3 ( − 0.9 ) 2 + ( 0.2 ) 2 + ( 0.5 ) 2 + ( 0.2 ) 2
s = 3 0.81 + 0.04 + 0.25 + 0.04
s = 3 1.14
s ≈ 0.6164
Summary of Results The standard deviations are:
Control: 0.2708 Vancomycin (1.0 mg/kg): 0.1581 Vancomycin (5.0 mg/kg): 0.6 Teixobactin (1.0 mg/kg): 1.4154 Teixobactin (5.0 mg/kg): 0.6164
Examples
Understanding standard deviation is crucial in many real-world scenarios, especially in fields like medicine and pharmaceuticals. For instance, when testing a new drug, researchers use standard deviation to understand the variability in patient responses. A smaller standard deviation indicates that the drug's effect is consistent across the patient population, while a larger standard deviation suggests more variability, potentially due to individual differences or other factors. This helps in making informed decisions about the drug's efficacy and safety before it's released to the public. In our case, the standard deviation helps us understand the variability in the log of the number of colonies for different treatments.
