Position measuresThe study carried out so far allows us to describe, in general terms, the groups of values that a variable can assume. In this way, we know where the values are concentrated and how the frequency behaves.However, in order to highlight the characteristic tendencies of each distribution, in isolation or in comparison with others, we need concepts that can be expressed through numbers, which allow us to translate this tendency.These concepts are called typical elements of the distribution:- Position measures;- Dispersion measures;- Asymmetry measures;- Kurtosis measures;Among the typical elements, this lesson will focus on position measures, indicating the position of the distribution in relation to the horizontal axis (x-axis).These measures include:- Arithmetic mean;- Median;- Mode.Other position measures may be the separatrices:- The median itself;- Quartiles;- Percentiles.Arithmetic meanMean with ungrouped dataThe simple arithmetic mean is the quotient of the sum of the variable values divided by the number of values:\[ \bar{x} = \frac{\sum{x_i}}{n} \]Sometimes the mean can be a number different from all the data in the series it represents. An example:Knowing that cow A's milk production, during one week, was 10, 12, 13, 13, 15, 16, 19. What was the average production for the week?\[ \bar{x} = \frac{10 + 13 + 13 + 15 + 16 + 19 + 12}{7} = 14 \]14 in this case is the representative number of this data series, although it is not in the original data. In this case, we usually say that the mean has no concrete existence.It should be noted that the mean gives us a position of this data set; since the mean is 14, it is imagined that the cow will not produce more than 100 liters of milk in a day or a negative number of liters of milk.Deviation from the mean> We call deviation from the mean the difference between each element of a set of values and the arithmetic mean.For each value in the data set we have:\[ d_i = x_i - \bar{x_i} \]For the previous example:\[ d_1 = x_1 - \bar{x_1} = 10 - 14 = -4 \]\[ d_2 = 12 - 14 = -2 \]\[ d_3 = 13 - 14 = -1 \]\[ d_4 = 13 - 14 = -1 \]\[ d_5 = 15 - 14 = 1 \]\[ d_6 = 16 - 14 = 2 \]\[ d_7 = 19 - 14 = 5 \]Properties of the mean1st - The algebraic sum of the deviations is zero.In the example:\[ d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7 = -4 -2-1-1+1+2+5 = 0 \]2nd - If a constant c is added to (or subtracted from) all the values of a variable, the mean of the set increases (or decreases) by that constant.For each x:\[ \bar{x} + c = \frac{\sum{x_i + c}}{n} \]In the cow example, let’s add 5 to each value, obtaining: 15, 17, 18, 18, 20, 21, 24. Therefore, the mean:\[ \bar{x} = \frac{15 + 17 + 18 + 18 +20+ 21+ 24}{7} = 19 \]3rd - If all the values of a variable are multiplied by (or divided by) a constant c, the mean of the set is multiplied by (or divided by) that constant.\[ \bar{x} · c = \frac{\sum{x_i · c}}{n} \]In the cow example, let’s multiply each value by 2, obtaining: 20, 24, 26, 26, 30, 32, 38. Therefore, the mean:\[ \bar{x} = \frac{20+ 24+ 26+ 26+ 30+ 32+ 38}{7} = 28 \]Mean with grouped data: without class intervalsThe table below represents the distribution of 34 families with 4 children, taking as variable the number of male children:Number of boysFrequency021621031244Considering the frequency distribution above, it can be seen that frequency is the number of times each data point is repeated; therefore, it works as a weighting factor, which leads us to calculate the weighted arithmetic mean. Its formula is as follows:\[ \bar{x} = \frac{\sum{x_i · f_i}}{\sum{f_i}} \]In the case of the table, the weighted mean would be:\[ \bar{x} = \frac{2 · 0 + 1 · 6 + 2 · 10 + 3 · 12 + 4 · 4}{2 + 6 + 10 + 12 + 4} = 2,3 \]The mean value 2.3 in this case suggests that the greatest number of families has 2 boys and 2 girls, while the overall tendency is a slight numerical predominance in relation to the number of boys.Mean with grouped data: with class intervalsIn this case, it is conventionally assumed that all values included in a given class interval coincide with its midpoint.Consider the distribution:Height (cm)FrequencyMidpoint151 - 1558153156 - 16011158161 - 16513163166 - 1701168From this point on, it proceeds as if there were no intervals, with the value used to calculate the mean being the midpoint:\[ \bar{x} = \frac{8 · 153 + 11 · 158 + 13 · 163 + 1 · 168}{8 + 11 + 13 + 1} = 159,06 \]Mode> We call mode the value that occurs with the highest frequency in a series of values.As an example, the modal salary of employees in an industry is the most common salary, that is, the salary received by the largest number of employees in that industry.Mode with ungrouped dataWhen dealing with ungrouped data, the mode is easily recognized: according to the definition, just look for the value that is repeated most often. Example: 7, 8, 9, 10, 10, 10, 10, 11, 12, 13 and 15.But there are cases in which there will be no modal value (amodal), that is, no value appears more times than others.In other cases there may be two or more concentration values, which is called bimodal or multimodal.Mode with grouped dataOnce the data are grouped, it is possible to determine the mode immediately: just identify the value of the variable with the highest frequency.The class with the highest frequency is called the modal class. By definition, we can state that the mode, in this case, is the dominant value that lies within the limits of the modal class.The simplest method for calculating the mode consists of taking the midpoint of the modal class, which is called the raw mode.Mode curvesWhy use mode instead of meanIt is not affected by extreme values of the distribution, as long as those values do not constitute the modal valueIt is used when we want a quick and approximate measure of positionIt is widely used in economic and industrial statisticsMedian> The median is another measure of position defined as the number found in the center of a series of numbers, arranged in order. It is such that the number of observations with values less than the median is equal to the number of observations with values greater than the median.Median with ungrouped dataGiven a series of data: 5, 13, 10, 2, 18, 15, 6, 16, 9; the first step is to arrange the values: 2, 5, 6, 9, 10, 13, 15, 16, 18.Next, take the central value that has the same number of elements to the right and left. If the series has an odd number of data, this will be possible, as can be seen in the example where the median is 10. If the series has an even number of data, the midpoint of the two central values of the series is used.Thus, if a series has the following values: 2, 6, 7, 10, 12, 13, 18, 21. Since the number of terms is even, the midpoint of the central terms is used: 10 and 12; giving a median value of 11.Median with grouped data: without class intervalFor grouped data, simply check the cumulative frequency immediately above half the sum of the frequencies. The median will be the variable value that corresponds to that cumulative frequency.Number of boysFrequencyCumulative frequency02216821018312304434\[ Md_{term} = \frac{34}{2} = 17 \]The 17th position term is in the third row because the smallest cumulative frequency that exceeds this value is 18, where the corresponding value is 2. Therefore, the median is equal to 2.It should be noted that if the term is an upper extreme of a class, in the example 18, the median will be given by the midpoint between the class of the term and the class above it.Median with grouped data: with class intervalsHeight (cm)FrequencyCumulative frequency151 - 15588156 - 1601119161 - 1651332166 - 170133Since the sum of the frequencies is 33, the median is at the value 16.5. In this case, the class that contains the 17.5th term is the 156-160 class, since it contains all terms between 8 and 19. We call this class the median class.The median is given by a rule of three from the possible values of that class. Since there are 13 elements within the class and the median will be the 9.5th term (17.5 - 8 terms from the previous class), the following formula is used:\[ Md = \text{lower extreme} + \text{class width} *\frac{\text{number of terms for the median}}{\text{number of terms in the class}} \]\[ Md = 156 + 4 · \frac{9,5}{13} = 158.92 \]Why use median and not meanWe want to obtain the point that divides the distribution into equal partsThe variable under study is salaryThere are extreme values that affect the meanHow many spiders do people eat on average per year?Other measures of positionSimilar to the median, which is based on the position of the series, there are quartiles, percentiles, and deciles; which, together with the median, are known as separators.In the case of these measures of position, the division is not by the number in the middle of the series, but by the number in the position 1/4 of the series or 1/10 of the series, etc...Quartiles are often seen in stocks, as they show how much the worst or best 25% of values influence the stock.Exercises1) Determine the deviations in relation to the mean of the following data: 6, 8, 5, 12, 11, 7,4, 15 and 7,2.What is the sum of the deviations?2) Considering the data sets:a. 3,5; 2,6 ; 5,9 ; 5,2 ; 8,6b. 20,9; 7,2; 12,7; 20; 15,7c. 51 ,6; 48,7; 50,3; 49,5; 48,9d. 15; 18; 20; 13; 10; 16 ;14Calculate:I. the mean11. the median111. the mode3) Mark the correct option. The cumulative frequency curve is used to calculate:a. the law of chanceb. the mean.c. the median.d. the mode.e. the standard deviation.4) Mark the correct option. A symmetrical curve is characterized by the following attribute:a. It is left-skewed.b. The mode is greater than the median and the mean.c. The mode, the median, and the mean are equal.d. The standard deviation is greater than the median and the mode.e. The deciles are equivalent to the mean.5) Calculate the arithmetic mean, mode, and median of the following distribution:GradesFrequency0 - 252 - 484 - 6146 - 8108 - 1076) Calculate the arithmetic mean, mode, and median of the following distribution:Heights (cm)Frequency150 - 1585158 - 16612166 - 17418174 - 18227182 - 19087) Calculate the arithmetic mean, mode, and median of the following distribution:Salaries (R$)Frequency500 - 70018700 - 90031900 - 1,100151,100 - 1,30031,300 - 1,50011,500 - 1,70011,700 - 1,90018) Calculate the arithmetic mean, mode, and median of the following distribution:Weights (kg)Frequency145 - 15110151 - 1579157 - 1638163 - 1696169 - 1753175 - 1813181 - 18719) Calculate the arithmetic mean, mode, and median of the following distribution:Customer ratingFrequency234761288104Answer key1) The sum of the deviations is always zero.2) a. x = 5.1; Md = 5.2; no modeb. x = 15.3; Md = 15.7; no modec. x = 49.8; Md = 49.5; no moded. x = 15.1; Md = 15; no mode3) c.4) c.5) x = 5.3; Md = 5.3; Mo = 56) x = 172.4; Md = 174; Mo = 1787) x = 843; Md = 810; Mo = 8008) x = 159.4; Md = 157.8; Mo = 1489) x = 6.17; Md = 6; Mo = 6
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