Question

As part of a classic experiment on mutations, 10 aliquots of identical size were taken from the same cul-ture of the bacterium E. coli. For each aliquot, the number of bacteria resistant to a certain virus was determined. The results were as follows:

14 15 13 21 15
14 26 16 20 13
Construct a frequency distribution of these days and display it as a histogram.
Determine the mean and the median of the dad and mark their locations on the histogram.

Answer 1

a. See the attached excel file for the frequency distribution table, and the attached photo for the histogram.

b. We have:

Mean = 16.7

Median = 15

Note: See the attached photo for the locations of Mean and Median on the histogram.

Explanation:

a. Construct a frequency distribution of these days and display it as a histogram.

Note: See the attached excel file for the frequency distribution table, and the attached photo for the histogram.

b. Determine the mean and the median of the dad and mark their locations on the histogram.

From the attached excel file, we have:

Total of F = 10

Total of FX = 167

Therefore, we have:

Mean = 167 / 10 = 16.7

Median is the middle number after arranged the data in ascending or descending order. Using the ascending order, we have:

13 13 14 14 15 15 16 20 21 26

Since 15 and 15 are in the middle, their average are the median which is calculated as follows:

Median = (15 + 15) / 2 = 15

Note: See the attached photo for the locations of Mean and Median on the histogram.