## biology paper

Assignment 1.2- Looking at Data

Task 1- Describing data

A).

DECRIPTIVE STATISTICS FOR POPULATION 1

MEANMEDIAN

=24.5

MODE=24.5 STANDARD DEVIATION STANDARD DEVIATION= 0.295

DESCRIPTIVE STATISTICS FOR POPULATION 2

MEAN

MEDIAN=28.2MODE=28.3STANDARD DEVIATION STANDARD DEVIATION= 2.243

B).

In population 2, I found that the mean, median and mode values are not the same. The mean was 27.5, the median was 28.2 and the mode was 28.3. There is a greater range in the mass of birds (g) for the birds in population 2. This variety of the masses of the birds in population 2, affect the mean value making it significantly different from the system analysis and design assignment help median and mode values, because it takes into account all the values of the data set, meaning it can be affected by extreme values.

C).

In population 1, the mean mass of birds is 24.6g, the median and mode values are both 24.5g. In population 2, the mean mass of birds is 27.5g, the median is 28.2g and the mode is 28.3g. The average size of sparrows in population 2, tends to be larger than those in population 1. Evidence of this, is that the mean mass of birds in population 2 is 27.5g, whereas in population 1, it is 24.6g, therefore when all data values are taken into account, the average size of bird in population 2 is larger than the average size of birds in population 1. The median value of mass of bird in population 2 is also larger than in population 1. The median is the middle number in a list of the data values that have been written out in ascending order. This is unaffected by extreme values or outliers. This means that the middle value of all the masses of the birds in population 2 is larger than in population 1, meaning that on average the birds in population 2 are larger. The mode value, is…