Data for cells transformed with an enzyme-expressing plasmid a transformed cells
ID: 81110 • Letter: D
Question
Data for cells transformed with an enzyme-expressing plasmid a transformed cells: compared with non-transformed cells: 2.00 times 10^9 2.05 times 10^9 3.20 times 10^9 2.30 times 10^9 2.12 times 10^9 A. Calculate the mean, median, and standard deviation for these results. B. A second data set has a mean of 3.16 times 10^9 and a standard deviation of 8.2 times 10^8. Use this data to calculate the precision, which can be stated as the coefficient of variation (CV). Which data set (first or second) is more precise? Explain how you arrived at your answer.Explanation / Answer
Total = 11.49
Mean = 11.49/5 = 2.29 * 109
Median = It is odd - So the middle one is median
First makes in order - 2, 2.05, 2.12, 2.30, 3.20
The median is 2.12 * 109
Variance = 0.997/n-1 = 0.997/4 = 0.249
SD = 0.49 * 109 = 4.9 * 108
B. CV1 = (4.9 * 108) / ( 2.29 * 109 ) * 100 = 21 %
CV2 = (8.2 * 108) / ( 3.16 * 109 ) * 100 = 25.9 %
CV1 is more precise than CV2. The lower the % in CV we get higher precision. It tells there was no difference in the way the data sample were collected if CV value are lower. If it is high >75% there is a problem in experiments it not consisent or any error in the experiments.
If there is zero % of error if some one do's the experiment then there is a chance to get CV is 0.
X X-2.29 (X-2.29)2 2 -0.29 0.084 2.05 -0.24 0.057 2.12 -0.17 0.028 2.3 0.01 0.0001 3.2 0.91 0.828 Total 0.997Related Questions
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