The grid below captures relationships among seven personality dimensions: "extraversion", "true_arousal_plac", "true_arousal_caff”, "arousal_plac", "arousal_caff”, "performance_plac", and "performance caff”. The diagonal represents histograms of the seven dimensions. Left of the diagonal represents scatterplots between the dimensions while the right of the diagonal represents quantitative relationships between the dimensions. The lines in the scatterplots are closest approximation of the points. The value of the relationships to the right of the diagonal can vary from -1 to +1, with -1 being the extreme linear negative relation and +1 extreme linear positive relation. (Axes of the graph are conventionally drawn).
Histogram is a graph which depicts the frequency of occurence of values in different ranges.
For example, lets consider 50 students attended an exam. The distribution of number of students in different score ranges is depicted by the following histogram
In this question we have histograms for different personality dimensions.
However the x and y co-ordinates are not given.
Since the y co-ordinate gives us frequency/number of occurences of a certain value, we can use a histogram to calculate mean median and mode even without details of x-y co ordinate axes
Option A : "Extraversion" has 2 modes.
We can calculate mean of a histogram table by averaging all the values
We can calculate the median of a histogram table by arranging the histograms in ascending order and the bar/value which is in the middle will get the median.
Median for "arousal_plac" = -3
Avergae for "arousal_plac" = (-4-3-2+1+0-3-4)/7 = -15/7 = -2.143
Median for "arousal_caff" = 0
Average for "arousal_caff" = (-1.5-1.5+2+1.75+1.5+1.75+1+0-0.5-2-2)/11 = 0.0454
Median for "performance_plac" = 0.075
Average for "performance_plac" = (0.025+0.075+0.05+0.1+0.2)/5 = 0.09
Median for "performance_caff" = 0.075
Average for "performance_caff" = (0.05+0.25+0.09+0.075+0.08+0.075+0.085+0.1+0.1+0.15+0.18)/11 = 0.1122
Video Solution
