Based off the boxplot for age we can see that it is left skewed because the mean 50.11 is less than median 55. The distribution is also greater than compared to %fat. The age data has no outlier. The %fat distribution is normal because the mean and median are nearly identical 31 and 32 respectively. We determined that %fat has two outliers. The IQR = 36.8 – 29.3 = 7.5. Outlier will be 1.5 * 7.5 = 11.25. So we can see that we have two plots in the graph that are less than the 1stQuartile(29.4) - 11.25 = 18.15. This matches up with what we have in the dataset showing two values at 10.5 and 8.8.b)
In SPSS we do Analyze => Descriptive Statistics => Descriptives. We click on save standardized values as variables. This will populate SPSS data with two new columns of z-scores associated with their respective variables.agefatPercentageZscore(age)Zscore(fatPercentage)26.0010.50-1.61828-2.1553026.0030.50-1.61828-.0588229.008.80-1.41693-2.3335129.0020.80-1.41693-1.0756140.0032.40-.67863.1403545.0026.90-.34305-.4361950.0030.40-.00746-.0693055.0030.20.32813-.0902760.0033.20.66372.2242155.0036.60.32813.5806145.0044.50-.343051.4087260.0030.80.66372-.0273755.0035.40.32813.4548261.0033.20.73084.2242162.0036.10.79795.5282063.0037.90.86507.7168875.0043.201.670481.2724566.0037.701.06642.69592c)
i) Min max normalization Equation = v’ = (v-min)/(max-min)Values v in the interval [min,max] and following the normalization method the new range will be v’ = [newmin, newmax]. This data set and in general the range is [0,1]. If you plus in the values of the dataset you will get this range. ii) Z-score normalizationEquation = z = value-mean/standard deviation.