GEOGRAPHICAL CONCEPTS, ABILITIES AND METHODS

Geographical concepts, abilities and methods

Geographical notions, skills and methods

QUESTION 1: Calculations

Average

Infant Mortality

Life Expectancy

Mean

1.02

75.97

Median

1.20

76.17

Mode

0.41

4.09

ARITHMETIC MEAN

Infant Mortality

Arithmetic Mean = Sum of all observations

Total Observations

Arithmetic mean= 102.35/100

Arithmetic mean=1.0235 per thousand person

Life Expectancy

Arithmetic Mean = Sum of all observations

Total Observations

Arithmetic mean= 7597.55/100

Arithmetic mean=75.97 years

MEDIAN

Infant Mortality

Arranging facts and figures in ascending order

Median = (100 + 1)/ 2 = 50.5

Median = 50th value + 51th value

2

Median= 1.19 + 1.21 2

Median = 1.20

Life Expectancy

Arranging facts and figures in ascending order

Median = (100 + 1)/ 2 = 50.5

Median = 50th value + 51th value

2

Median= 76.12+76.22 2

Median = 76.17

MODE

Infant Mortality

Most Repeated Value= 1.08

Life Expectancy

Most Repeated Value= 76.6

STANDARD DEVIATION

Infant Mortality= 0.4106

Life Expectancy=4.09

Question 2

Fiqure 1: Outlier's

Outliers in this data are mainly 42nd value and 34th value. Whereas all values close to zero and above 1.2 are outliers in data. An outlier is the fact that lies outside overall pattern of the circulation (Moore and McCabe 1999, 21). Usually, occurrence of an outlier shows some sort of problem. This can be the case which does not fit model under study or the mistake in measurement. Outliers can furthermore occur when matching relationships between two sets of data. Outliers of this type can be easily recognized on the scatter plot. (Ronald, 1992, 63)

When accomplishing smallest rectangles fitting to facts and figures, it is often best to reject outliers before computing line of best fit. This is particularly true of outliers along x direction, since these points may greatly influence result. (Galore, 2007, 67)

Question 3

Table 1: NORMAL DISTRIBUTION AT 95% AND 90 % CONFIDENCE INTERVAL

(Attached in appendix 1)

When you compute the self-assurance gap, you compute mean of the experiment in order to estimate signify of population. Clearly, if you already knew community mean, there would be no need for the self-assurance interval. However, to interpret how self-assurance gaps are assembled, we are going to work rearwards and start by assuming characteristics of population. Then we will display how sample facts and figures can be used to construct the self-assurance interval.

Assume that weights of 10-year vintage children are normally distributed with the signify of 90 and the benchmark deviation of 36. What covering is trying circulation of signify for the experiment dimensions of9? Recall from section on sampling distribution of mean that mean of sampling distribution is µ and standard error of mean is

For present demonstration, trying circulation of mean has the mean of 90 and the benchmark deviation of 36/3 = 12. Note that benchmark deviation of the trying circulation is its benchmark error. Figure 1 shows this distribution. The in shade locality comprises middle 95% of distribution and stretches from 66.48 to 113.52. These limits were computed by adding and subtracting 1.96 standard deviations to/from signify of 90 as follows:

 

90 - (1.96)(12) = 66.4890 + (1.96)(12) = 113.52

The worth of 1.96 is founded on fact that 95% of locality of the usual circulation is within 1.96 benchmark deviations of signify; 12 is benchmark mistake of mean.

Figure 1 displays that 95% of means are no ...