Set of Scores as the Analysis Set - Essay Example

A sample of their plots would look as shown below: The above plot has a slight positive skew. A look at the figure will show that these measures do not differ significantly in this case although they may be typically up or above one another. However, there are several cases where these measures vary significantly. Consider the case of any organization, where the salary varies depending on the designation of the employee.

Typically, as one goes up the hierarchy, the salaries increase and vary inversely to the number of employees at a particular level. In a regular case, the number of low-level employees is the most while their salaries are the least in the firm. As one moves up the hierarchy, the salary increases and reaches a maximum with the top executive management, which is usually the least in population.

When the salaries are plotted on the graph against the number of people receiving it, the resulting graph is said to be highly skewed as shown below:
The relatively large skew in the above graph results in very different values for these measures.

In this scenario, determining the mean will not present a true picture as only a fraction of the entire employees would be drawing such salaries while the maximum number of employees would be drawing much lesser salaries. However, the mode would clearly present a clear figure on the maximum number of people employed at a particular salary level. The mean would further not provide any clear picture of the highest amount being drawn by a minute section of the employees, namely the top management, leading to discrepancies in statistical estimates.

In such scenarios, the mode is the preferred choice to estimate the maximum number of employees within the population.Part 2Chapter 34. The mean, median and mode will have the same value in the case of symmetric distributions.14. mean of 7 scores= 10 Initial X=15 Final X = 1 Initial total = 7*10 = 70 Final total value = 70 – (15-1) = 70 – 14 = 56 New mean = 56 / 7 = 816. N = 10 Mean = 9 Value removed = 9 Initial total = 9*10 = 90 Final total = 90 – 9 = 81 Final N = 9 New mean = 81 / 9 = 9

Chapter 42.
mean = 100 Variance = 20 Standard deviation = sqrt. of 20 = 4.472 The standard deviation determines the average spread that measures within the population can have. In the present case, this range is 100 (+-) 4.472.4. A standard deviation of zero implies that the population consists of the same values i.e, all have the same value.6. An unbiased estimate is similar to a biased estimate except that the population size used to compute the average of squares is one less than the population sum.10. mean = 75 X = 80 When variance = 2, , 80 lies outside the distribution With variance = 10, X = 80 lies within the distribution and is therefore desired.20. c.

Population: 12, 1, 10, 6, 3, 3, 7 N = 7 Variance = 16 Standard Deviation = 426. Population = 2, 0, 0, 0, 0, 2, 0, 2, 0 N = 9 Variance = 1 Standard Deviation = 1