Mathematics 12.

In everyday life we encounter reports from public opinion surveys countless number of times, we make measurements during shopping and when our work is evaluated, or when scientific experiments are carried out. Collection, presentation and evaluation of measurement results is done following strictly determined methods, since deviation might lead to false deductions. The science of statistics provides us with these methods.

In Year \latex{ 9 } (in the Statistics chapter), we have already studied different ways to plot data, learned about the mode, median and different types of averages of a data set, as well as the data's size as a measure of the spread of the data. Now we will see some examples about classifying data while we recall the above concepts and we will also meet the concept of average deviation and standard deviation as new tools for measuring the spread of data.

Curves or line graphs are useful if we would like to present changes in data or the relation between them. Pie charts are used when we compare the proportion between the data and the whole.

Charts created from these data are presented in Figure 13.

Regarding the number of tourist arrivals, Eastern Asia and Oceania took over America, although it is not true for the income, as spending per person has not increased as much as it did in America.

Spending per person by each continent can be seen in Figure 14.

We can see that the order of spending per person differs from that of tourist arrivals and income; here instead of Europe, America takes the lead. Furthermore, in America, spending per person has increased almost twofold in the studied time period. Regarding the latter two markers, Southern Asia stood at the last place, however regarding spending per capita it stepped forward from the third to the second place.

An annual increase of \latex{4.1\%} would mean that the number of tourist arrivals would increase\latex{ 1.041 } times each year compared to the previous one, which, during \latex{ 11 } years, would mean a change of \latex{1,041^{11}=1.55} times, that is, a growth by \latex{55\%.} However the increase is \latex{11\times 4.1\%=45\%.} By computing the arithmetic mean we get the increase in tourist arrivals per year if the increase rate is constant. If we would like to know the percentage of the increase in tourist arrivals supposing this percentage is the same each year, we have to solve the equation \latex{(1+x)1^{11}=1.45,} which gives us \latex{x=0.034,} therefore the studied parameter grows by \latex{45\%} during \latex{ 11 } years if the annual increase is \latex{3.4\%.}

Since on Thursday his fever has declined from \latex{ 39.7°C } to \latex{ 38.5°C } between \latex{ 6 } pm and \latex{ 9 } pm, we could tell, based on the straight line segment connecting the points, supposing a consistent change, that his temperature was approximately \latex{ 39.3°C } at \latex{ 7 } pm. However it is also possible that he received fever pills at \latex{ 6 } pm which reduced his temperature by \latex{ 7 } pm below \latex{ 38°C } , and as the medicine's effect faded his body temperature returned to \latex{ 38.5°C } by \latex{ 9 } pm. Connecting the measured values does not mean a precise measurement of the intermediate values, more like the change of measured values.

We cannot forecast his temperature at Saturday morning; it might happen that the favourable trend, started on Friday, will keep going and in this case it might be possible that he becomes free from fever, however some kind of complication might cause his fever to recur or to stay high.

The second graph suggests a more steep decrease only because a difference of \latex{ 10\, kg } corresponds to a longer segment. If we choose the scale of the graph such that the same difference corresponds to a greater segment, then we highlight, while a smaller segment soothes the size of the change. In conclusion, we must take the scale into consideration before forming an opinion based on the trend shown by a graph.

It is a common procedure that in case of big data amounts, instead of being listed one by one, data are allocated to classes thus allowing a better handling and characterisation of it, although some details might get lost. No data shall be present in two classes, but every single data must belong to a class.

The length of a class is the difference between its upper and lower bound.

For a class, the arithmetic mean of the upper and lower bound is called the mean of the class. Every element of the class is then viewed as equal to the mean of the class.

The number of elements in each class is the cumulative frequency of the class.

It is reasonable to sort the data into classes by \latex{ 5 } centimeters. The chart created this way is presented on Figure 17.

The number of players whose height is at least \latex{ 2 } meters is \latex{5+4+2+1=12.}

Mean of the players' heights is:
 

\latex{(185+187+193+194+196+197+198+198+200+201+201+\\203+204+205+205+207+209+210+212+215)\div20=\\4,020\div 20=201 (cm.)}
 

The first class is between the lower bound of the first class and that of the second class, the mean of the class is the arithmetic mean of these values, that is, \latex{ 187.5\, cm }. Similarly the rest of the means are \latex{192.5;\, 197.5;\, 202.5;\, 207.5;\, 212.5;\, 217.5.} Using these values the mean would be

\latex{\frac{2\times 187.5+2\times 192.5+4\times 197.5+5\times 202.5+4\times 207.5+2\times 212.5+1\times 217.5}{20}=}

\latex{=\frac{4,035}{20}=201.75.}
 

There is a difference from the real mean, but it's not significant. The reason behind the difference is that every data is approximated by the mean of its class. In case of a huge amount of data only this second method is applicable as we usually do not have detailed data.

Ordering the \latex{ 20 } pieces of data in an increasing order, the two in the middle, the \latex{ 10 }th and \latex{ 11 }th, are both \latex{ 201\, cm }, thus the median is \latex{ 201\, cm }. Computing the median after sorting the data into classes, the \latex{ 10 }th and \latex{ 11 }th pieces of data are both in class \latex{200–204} which is characterized by its mean, \latex{ 202.5\, cm }. Therefore this way the median is \latex{ 202.5\, cm }.

The median characterises the height of the players quite well.

When we create a frequency chart, the bar height is representing the number of data in each class best if the length of the classes are the same. Therefore it is sensible to construct such classes (Figure 18). Since the first and the last column of the graph do not correspond to \latex{ 10 } year long periods, their height is not relevant for comparison.

Take the houses built before \latex{ 1940 }, these give \latex{9.6+5.3+6.3=21.2\%} of all of the houses. Therefore approximately \latex{20\%} of the houses had been built before World War \latex{ 2 }.

The median of the years of construction is the middle value when the years are ordered in an increasing order, in other words, the same number of houses had been built before and after the median. Therefore the mean of the class for which the the ratio of houses reach \latex{50\%} can be viewed as the median. This happens in the class of \latex{ 1960–1969 }, thus the median construction year is \latex{ 1965 }.

Central values describing a set of data:
Mode: the most frequent data in the data set (there might be more than one).

Arithmetic mean or average: it is the quotient of the sum of all data and the number of data.

Advantage:

Disadvantage:

Median: the middle element of the data after ordering all data in an increasing order. In case of odd \latex{(2n+1)} number of data points, it is the middle, the \latex{(n+1)^{th}} one, while in case of even \latex{(2n)} number of data points, it is the average of the two middle  (n and n+1) values.
Advantages:

In Year \latex{ 9 } we had seen that the average wage might be biased by the higher salaries of the top management, therefore the average wage is better characterised by the median.

The spread of data is described by the size of the data: the difference between the largest and smallest data.

The smaller the size of the data, the better the above values describe the whole set of data.

During sorting the data into classes we divided the segment representing the size of the data into smaller segments.
Advantage:

Order the sample data:

The medians are:

The means are:

Sample \latex{ B } shows that due to the outlying data, the median is better than the average in terms of describing the centre of the sample.

The size of both data is \latex{18–1= 17,} this does not differentiate between the two sets of data in terms of spread.

Average deviation from the mean is:

\latex{A:\frac{|1-9|+|4-9|+|7-9|+|9-9|+|15-9|+|18-9|}{6}=\frac{30}{6}=5;}
\latex{B:\frac{|1-7.5|+|6-7.5|+|6-7.5|+|7-7.5|+|7-7.5|+|18-7.5|}{6}=\frac{21}{6}=3.5.}

Average deviation from the mean describes the spread of the data as in the case of the second set it is smaller.

Compute the average of the deviations from the median:

\latex{A:\frac{|1-8|+|4-8|+|7-8|+|9-8|+|15-8|+|18-8|}{6}=\frac{30}{6}=5;}
\latex{B:\frac{|1-6.5|+|6-6.5|+|6-6.5|+|7-6.5|+|7-6.5|+|18-6.5|}{6}=\frac{19}{6}=3.17.}

It can be shown that if the average of the deviations from another number are used instead of the median, the former cannot be smaller than in case of using the median.

The spread of the data can be described by the average deviation from the mean or from the median.

Average of the squared deviations from the mean:

\latex{A:\frac{(1-9)^{2}+(4-9)^{2}+(7-9)^{2}+(9-9)^{2}+(15-9)^{2}+(18-9)^{2} }{6}=35;}
\latex{B:\frac{(1-7.5)^{2}+(6-7.5)^{2}+(6-7.5)^{2}+(7-7.5)^{2}+(7-7.5)^{2}+(18-7.5)^{2} }{6}=26.25.}

The standard deviation of the samples presented in the above example are:

\latex{A:5.92}  and  \latex{B:5.12.}

Outlying data increase the standard deviation more than it does the average deviation.

It can be shown that if we take the average of squared deviations from another number instead of the mean, then the result cannot be smaller than what we get using the mean.

Pete tells that he scored \latex{ 14 } points above the average, while the same is only \latex{ 11 } points for Paul.
Paul argues that while Pete scored \latex{\frac{14}{14}=1} times the standard deviation more than the average, he scored \latex{\frac{11}{10}=1.1} times the standard deviation more. Therefore, in regards of the standard deviation, his test result deviates more from the average.