# Sample Coursework Paper on Sampling

The data realized in minutes was 205, 251, 250, 249, 218, 245, 248, 245, 242, and 259.

 Days Data in Minutes 1 205 2 251 3 250 4 249 5 218 6 245 7 248 8 245 9 242 10 259 Mean 241.2 Variance 275.067 Std. Dev 16.58

I derived the mean by dividing the data for the time taken in studying by the number of observation days. This implies that the mean was:

(205 + 251 + 250 + 249 + 218 + 245 + 248 + 245 + 242 + 259) / 10 = 2412 / 10 = 241.2 ≈ 241 minutes

The variance and standard deviation were derived by applying the formula

For the variance and

for the standard deviation. The denomination in the variance reduces the number by one since this is a sample.

After calculating the mean, standard deviation, and variance, the mean remained constant. This is because of same period of time, which I took in studying. A larger sample size in connection to this data may have minimal effects. This is because this data may be taken to assume sample data in the larger sample. Hence, the results of this data may be a reflection of the larger sample. There may therefore be minimal variation when the larger data can be manipulated without the application of sample data.

Using the sample I have, it is possible to have accurate conclusion. This is because when using a large population, samples are usually practical to attain the precise results. Large samples are big and hard to manage unlike the small samples. When comparing two data sets, we derive the normality and the variation of the data. From the mean, we ascertain how accurate the data is. This is achievable by comparing the mean, mode, and median. It is also possible through analyzing the normality of the graph.