Statistical Concepts: Mean, Median, Mode, Variance, Standard Deviation PDF

DATA DESCRIPTION Intended Learning Outcomes: Compute the mean, median, and mode of the given data. Compute the range, variance, and standard deviation of the given data. Identify the location of the given data. In the previous chapter, we learned how to organize and presen...

DATA DESCRIPTION Intended Learning Outcomes: Compute the mean, median, and mode of the given data. Compute the range, variance, and standard deviation of the given data. Identify the location of the given data. In the previous chapter, we learned how to organize and present large data sets. However, the techniques are not helpful when we need to describe verbally the main characteristics of a data set. This chapter will show the statistical methods that can be used to summarize data. MEASURES OF CENTRAL TENDENCY OR AVERAGE Average occurs regularly in our daily life and it is an important tool in statistics. A well-chosen average consists of a single number about which a given data are centered. There can be several different types of averages or sometimes called measures of central tendency. They are the mean, the median, and the mode. Mean The mean is the most commonly used measure of central tendency. When we speak of average, we always refer to the mean. Mean It is found by adding the values of the data and dividing by the total number of values. Population Mean: Sample Mean: ∑𝑥 ∑𝑥 𝜇= 𝑥= 𝑁 𝑛 where 𝑁 – total number of observations in the population 𝑛 – total number of observations in the sample Example 1 A researcher collects data on the ages of recipients of doctoral degree in science and engineering, and his study yields the following: 37 37 24 28 43 44 36 41 33 27 Determine the average age of the recipients. Solution The mean is determined by the sum of the ages and then dividing by the total number of recipients. 37 + 37 + 24 + 28 + 43 + 44 + 36 + 41 + 33 + 27 350 𝑥= = 10 10 𝑥 = 35 Therefore, the average age is 35 years old. For grouped data, the midpoints of the classes are used for the values of 𝑥. The following are the steps in solving for the mean of grouped data.  Grouped Mean 1. Find the midpoint for each class. Place them in a column. 2. Multiply the frequency by the midpoint for each class. Place them in another column. 3. Find the sum of the resulting column in step 2. 4. Divide the sum obtained in step 3 by the total number of frequencies. That is, ∑ 𝑓 ∙ 𝑥𝑚 𝑥= 𝑛 Example 2 Eighty randomly selected lightbulbs were tested to determine their lifetimes (in hours). The following frequency distribution was obtained. Class boundaries Frequency 52.5 − 63.5 6 63.5 − 74.5 12 74.5 − 85.5 25 85.5 − 96.5 18 96.5 − 107.5 14 107.5 − 118.5 5 Determine the mean lifetimes (in hours) of these bulbs. Solution Step 1 Find the midpoints of each class, place the values in the third column. 52.5 + 63.5 𝑥𝑚 = = 58, 𝑒𝑡𝑐. 2 Step 2 Next, multiply the midpoint by the frequency for each class, and place the results in the fourth column. 58 × 6 = 348, 𝑒𝑡𝑐. Step 3 Find the sum of the fourth column. ∑ 𝑓𝑥𝑚 = 6807 Step 4 Divide the sum by 𝑛. ∑ 𝑓 ∙ 𝑥𝑚 6807 𝑥= = = 85 ℎ𝑜𝑢𝑟𝑠 𝑛 80 Hence, the mean lifetime of the bulbs is 85 ℎ𝑜𝑢𝑟𝑠. These steps are summarized in the following table. Class boundaries Frequency Midpoints 𝒇 ∙ 𝒙𝒎 52.5 − 63.5 6 58 348 63.5 − 74.5 12 69 828 74.5 − 85.5 25 80 2000 85.5 − 96.5 18 91 1638 96.5 − 107.5 14 102 1428 107.5 − 118.5 5 113 565 𝑛 = 80 ∑ 𝑓 ∙ 𝑥𝑚 = 6807 Median The median is the midpoint of the data array. Before finding this value, the data must be arranged in order, from least to greatest or vice versa. The median will either be a specific value or fall between two values. Grouped Median 1. Make a table of cumulative frequency. 2. Divide 𝑛, number of frequency by 2, to get the halfway point. 3. Locate the median class in the cumulative frequency column. 4. Substitute in the formula, 𝑛 − 𝑐𝑓 𝑚𝑒𝑑𝑖𝑎𝑛 (𝑀𝑑 ) = (2 ) (𝑤 ) + 𝐿𝑚𝑑 𝑓 where 𝑛 – sum of frequencies 𝑐𝑓 – cumulative frequency of the class preceding/before the median class 𝑓 – frequency of the median class 𝑤 – class width 𝐿𝑚𝑑 – lower boundary of the median class Example 3 Seven mothers were selected and given a blood pressure check. Their systolic pressures were recorded below. 135 121 119 116 130 121 131 Find their median. Solution Arrange the data in order. 116 119 121 121 130 131 135 Select the middle value. 116 119 121 121 130 131 135 𝑚𝑒𝑑𝑖𝑎𝑛 = 121 Example 4 Eight novels were randomly selected and the number of pages were recorded as follows: 415 398 402 400 420 415 407 425 Find the median. Solution Arrange the data in order. 398 400 402 407 415 415 420 425 Since the middle point falls halfway between 407 and 415, find the median by getting the mean of these two values. 407 + 415 𝑚𝑒𝑑𝑖𝑎𝑛 = = 411 2 Example 5 Find the median of the data in Example 2. Solution Step 1 Make a column for the cumulative frequency. Step 2 Divide 𝑛 = 80 by 2 to get the halfway point, which is 40. Step 3 Find the class that contains the 40th value by using the cumulative frequency distribution. Since 40 is less than 43, then the median class is the third class. Step 4 Using the formula, 𝑛 80 − 𝑐𝑓 − 18 𝑥𝑚𝑑 = (2 ) (𝑤) + 𝐿𝑚𝑑 = ( 2 ) (11) + 74.5 = 84.18 ℎ𝑜𝑢𝑟𝑠 𝑓 25 The median is 84.18 hours. Mode The third measure of average is the mode. It is the value that occurs most often in the data set. A data can have more than one or none at all. The mode for grouped data is the modal class. The modal class is the class with the largest frequency. Example 6 From the data in Example 2, the class with the largest frequency is the third class. Therefore, this is the modal class. The mode is the only measure of central tendency that can be used in finding the most typical case when the data are nominal or categorical.  Grouped Mode 𝑑1 𝑚𝑜𝑑𝑒 (𝑀𝑜) = 𝐿𝑀𝑜 + ( ) (𝑤 ) 𝑑1 + 𝑑2 where 𝐿𝑚𝑜 – lower boundary of the modal class 𝑤 – class width 𝑑1 – difference of the frequency of the modal class and the class preceding it 𝑑2 – difference of the frequency of the modal class and the class succeeding it Example 7 Find the mode of the data in Example 6. Solution Using the formula, 𝑑1 13 𝑀𝑜 = 𝐿𝑀𝑜 + ( ) (𝑤) = 74.5 + ( ) (11) = 81.65 ℎ𝑜𝑢𝑟𝑠 𝑑1 + 𝑑2 13 + 7 The mode is 81.65 ℎ𝑜𝑢𝑟𝑠. MEASURES OF VARIATION The previous section focused on averages or measures of central tendency. Group 1 Group 2 The averages are supposed to be central scores of a given set of data. However, 14 5 not all features of a given data set may be reflected by the averages. For example, 13 19 two different groups of 5 students are given identical quizzes in Math. The following 18 18 data at the right were the results. 14 14 11 14 The averages of each group are as follows. Group 1 Group 2 mean 14 14 median 14 14 mode 14 14 These two sets of averages have no difference. But intuitively, both groups show an obvious difference. Group 2 has a more widely scattered data than Group 1. This characteristic called variability is not reflected by averages. The three basic measures of variation are range, variance, and standard deviation. Range The range is the simplest measure of variation to calculate. It is just the difference between the largest and the smallest value in a given data set. For group 1, the range is 18 − 11 = 7. The range for group 2 is 19 − 5 = 14. A much larger range suggests greater variation or dispersion. The range has a disadvantage of being influenced by extreme values called outliers. Another is that it is based on two values only. All the other values in the set are being ignored. Standard Deviation and Variance The standard deviation is the most commonly used measure of variation. The standard deviation indicates how closely the values of a given data set are clustered around the mean. A lower value of the standard deviation means that the values of that given data set are spread over a smaller range around the mean. On the other hand, a large value of the standard deviation means that values of that data set are spread over a larger range around the mean.  The standard deviation is the positive square root of the variance. The variance calculated from a population is denoted by 𝜎 2 (sigma squared) and the standard deviation by 𝜎. The basic formulas are ∑(𝑥 − 𝜇 )2 ∑(𝑥 − 𝜇 )2 𝜎2 = 𝜎=√ 𝑁 𝑁 𝑁 – number of population 𝜇 – population mean Example 1 The final exam scores of 5 students are 80, 88, 92, 90, and 85. Determine the variance and the standard deviation. Solution Find the mean (𝜇 ). ∑ 𝑥 80 + 88 + 92 + 90 + 85 435 𝜇= = = = 87 𝑁 5 5 Subtract the mean from each individual score (𝑥 − 𝜇 ). score (𝒙 − 𝝁) 88 1 80 −7 92 5 90 3 85 −2 Square each of the differences (𝑥 − 𝜇 )2. score (𝒙 − 𝝁) (𝒙 − 𝝁)𝟐 88 1 1 80 −7 49 92 5 25 90 3 9 85 −2 4 Get the sum of (𝑥 − 𝜇 )2. ∑(𝑥 − 𝜇 )2 = 1 + 49 + 25 + 9 + 4 = 88 Divide the sum by 𝑁 = 5. 2 ∑(𝑥 − 𝜇 )2 88 𝜎 = = = 17.6 𝑁 5 ∑(𝑥 − 𝜇 )2 88 𝜎=√ = √ = √17.6 𝑁 5 𝜎 = 4.20 The variance is 17.6 and the standard deviation is 4.20. The following formulas are shortcut formulas for computing the variance and standard deviation. These are mathematically equivalent to the preceding formulas. They save time when repeated subtracting and squaring occur in the original formulas. These shortcut formulas will be used mostly.  Shortcut Formulas 𝑛 ∑ 𝑥 2 − (∑ 𝑥 )2 𝑛 ∑ 𝑥 2 − (∑ 𝑥 )2 𝑠2 = 𝑠=√ 𝑛(𝑛 − 1) 𝑛(𝑛 − 1) Example 2 The weights of nine basketball players are recorded as follows (in pounds). 206 215 305 297 265 282 301 255 261 Find the variance and the standard deviation. Solution Step 1 Find the sum of the values. ∑ 𝑥 = 206 + 215 + 305 + 297 + 265 + 282 + 301 + 255 + 261 = 2387 Step 2 Square each value and find the sum. ∑ 𝑥 2 = 2062 + 2152 + 3052 + 2972 + 2652 + 2822 + 3012 + 2552 + 2612 = 643391 Step 3 Substitute in the formula. 2 𝑛 ∑ 𝑥 2 − (∑ 𝑥 )2 9(643391) − (2387)2 𝑠 = = 𝑛(𝑛 − 1) 9(9 − 1) 5 790 519 − 5 697 769 92750 𝑠2 = = = 1288.19 72 72 𝑛 ∑ 𝑥 2 − (∑ 𝑥 )2 92750 𝑠=√ =√ = 35.89 𝑛(𝑛 − 1) 72 Variance and Standard Deviation for Grouped Data The procedure is similar to that of finding the mean for grouped data, and it uses the midpoints of each class. 𝑛 ∑ 𝑓 ∙ 𝑥𝑚 2 − (∑ 𝑓 ∙ 𝑥𝑚 )2 𝑛 ∑ 𝑓 ∙ 𝑥𝑚 2 − (∑ 𝑓 ∙ 𝑥𝑚 )2 𝑠2 = 𝑠=√ 𝑛(𝑛 − 1) 𝑛(𝑛 − 1) Example 3 For 108 randomly selected high school students, the following IQ frequency distribution was made. Class Limits Frequency 90 − 98 6 99 − 107 22 108 − 116 43 117 − 125 28 126 − 134 9 Find the variance and the standard deviation. Solution Step 1 Make a table. Find the midpoints of each class. Multiply the midpoints by the frequency for each class. Class Limits Frequency 𝒙𝒎 𝒇 ∙ 𝒙𝒎 90 − 98 6 94 564 99 − 107 22 103 2266 108 − 116 43 112 4816 117 − 125 28 121 3388 126 − 134 9 130 1170 Step 2 Multiply the frequency by the square of the midpoint for each class. 1 2 3 4 5 6 Class Limits Frequency 𝒙𝒎 𝒙𝒎 𝟐 𝒇 ∙ 𝒙𝒎 𝒇 ∙ 𝒙𝒎 𝟐 90 − 98 6 94 8836 564 53016 99 − 107 22 103 10609 2266 233398 108 − 116 43 112 12544 4816 539392 117 − 125 28 121 14641 3388 409948 126 − 134 9 130 16900 1170 152100 Step 3 Find the sum of columns 2, 5, and 6. Substitute in the formula for 𝑠 2. 1 2 3 4 5 6 Class Limits Frequency 𝒙𝒎 𝒙𝒎 𝟐 𝒇 ∙ 𝒙𝒎 𝒇 ∙ 𝒙𝒎 𝟐 90 − 98 6 94 8836 564 53016 99 − 107 22 103 10609 2266 233398 108 − 116 43 112 12544 4816 539392 117 − 125 28 121 14641 3388 409948 126 − 134 9 130 16900 1170 152100 sum 𝑛 = 108 ∑ 𝒇 ∙ 𝒙𝒎 = 12204 ∑ 𝒇 ∙ 𝒙𝒎 𝟐 = 1387854 2 𝑛 ∑ 𝑓 ∙ 𝑥𝑚 2 − (∑ 𝑓 ∙ 𝑥𝑚 )2 108(1387854) − (12204)2 𝑠 = = = 82.26 𝑛(𝑛 − 1) 108(108 − 1) 𝑛 ∑ 𝑓 ∙ 𝑥𝑚 2 − (∑ 𝑓 ∙ 𝑥𝑚 )2 108(1387854) − (12204)2 𝑠=√ =√ = 9.07 𝑛(𝑛 − 1) 108(108 − 1) MEASURES OF POSITION There are times when we want to know the position of a value relative to the other observations in a data set. For instance, you took a 100-item test. You might want to know how your score of 88 compares to the scores of others. Standard Scores or Z Scores A z score measures the distance between an observation and the mean, measured in units of standard deviation. Suppose that a student got a grade of 78 in her Math test and 55 in her Science test. The scores cannot be compared directly since the exams may not be equivalent in terms of number of questions, values of each question, and so on. But the relative position of the scores can be made using z scores.  The standard score is obtained by subtracting the mean from the value/observation and dividing the result by the standard deviation. The formula is 𝑣𝑎𝑙𝑢𝑒 − 𝑚𝑒𝑎𝑛 𝑥−𝑥 𝑧= = 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑠 If the z score is positive, the score is above the mean. If the z score is 0, the score is the same as the mean. If the z score is negative, the score is below the mean. Example 1 An IQ test has a mean of 105 and a standard deviation of 20. Find the corresponding z score for each IQ. a) 88 b) 122 c) 110 Solution 𝑥−𝑥 88−105 𝑥−𝑥 110−105 a) 𝑧 = 𝑠 = 20 = −0.85 c) 𝑧 = 𝑠 = 20 = 0.25 𝑥−𝑥 122−105 b) 𝑧 = = = 0.85 𝑠 20 Example 2 Which of the following exam grades has a better relative position? A grade of 43 on an Algebra test with 𝑥 = 40 and 𝑠 = 3. or A grade of 75 on a Geometry test with 𝑥 = 72 and 𝑠 = 5. 𝑥−𝑥 43−40 Solution For a grade of 43: 𝑧= = =1 𝑠 3 𝑥−𝑥 75−72 For a grade of 75: 𝑧= = = 0.6 𝑠 5 Since the z score for the Algebra test is larger, the position in the Algebra test is higher than the position in the Geometry test. Quartiles, Percentiles, and Deciles A quartile is a measure of relative standing. Let 𝑥1 , 𝑥2 , … , 𝑥𝑛 be a set of 𝑛 measurements arranged in order of magnitude. The first quartile, 𝑄1, is the value of 𝑥 that exceeds one-fourth of the measurements and is less than the remaining three-fourths. The second quartile, 𝑄2 , is the median. The third quartile, 𝑄3 , is the value of 𝑥 that exceeds three-fourths of the measurement and is less than one-fourth. Rules When the measurements are arranged in order of magnitude, that is increasing or decreasing; 𝑛+1 𝑄𝑘 = 𝑘 ( ) 4 where 𝑘 – the position of the quartile 𝑛 – number of observations When 𝑄1, 𝑄2 , and 𝑄3 are not integers, the quartile can be found by interpolation. Example 3 Find 𝑄1, 𝑄2 , and 𝑄3 of the following set of data. 19 12 16 0 14 9 6 1 12 13 10 19 7 5 8 Solution Arrange the data from lowest to highest.  0 1 5 6 7 8 9 10 12 12 13 14 16 19 19 Using the formula, where 𝑛 = 15. 𝑛+1 15+1 𝑄1 = 1 ( ) = 1( )=4 4𝑡ℎ 𝑑𝑎𝑡𝑎: 6 4 4 𝑛+1 15+1 𝑄2 = 2 ( ) = 2( )=8 8𝑡ℎ 𝑑𝑎𝑡𝑎: 10 4 4 𝑛+1 15+1 𝑄3 = 3 ( ) = 3( ) = 12 12𝑡ℎ 𝑑𝑎𝑡𝑎: 14 4 4 Percentiles are position measures used in educational and health-related fields to indicate the position of an individual in a group. It is symbolized by 𝑃1, 𝑃2 , 𝑃3 , …, 𝑃99 and divide the distribution into 100 groups. The percentile corresponding to a given value 𝑥 is computed by using the formula: 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠 𝑏𝑒𝑙𝑜𝑤 𝑥 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒 = ∙ 100% 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠 Example 4 Find the percentile rank of a test score of 49 in the data set. 12 28 35 42 47 49 50 Solution Arrange the data in order from lowest to highest. Then substitute in the formula. 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠 𝑏𝑒𝑙𝑜𝑤 𝑥 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒 = ∙ 100% 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠 5 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑖𝑙𝑒 = ∙ 100% = 71.43% 7 The next example illustrates how to find a value corresponding to a given percentile. 𝑛+1 𝑃𝑘 = 𝑘 ( ) 100 where 𝑘 – the position of the percentile 𝑛 – number of observations When 𝑘 th-Percentile is not an integer, it can be found by interpolation. Example 5 The following are scores in a Statistics test: 2 3 5 6 8 10 12 15 18 20 Find the value corresponding to the 25th Percentile. Solution Arrange the data set from lowest to highest. Using the formula, where 𝑛 = 10. 10 + 1 𝑃25 = 25 ( ) = 2.75 → 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 100 Since 𝑃25 is not a whole number, we will proceed to the interpolation. 𝑃25 = 2𝑛𝑑 𝑣𝑎𝑙𝑢𝑒 + 0.75(3𝑟𝑑 𝑣𝑎𝑙𝑢𝑒 − 2𝑛𝑑 𝑣𝑎𝑙𝑢𝑒) 𝑃25 = 3 + 0.75(5 − 3) 𝑃25 = 4.5 → 𝑣𝑎𝑙𝑢𝑒 Therefore, the value of 𝑃25 is 4.5 which can be found in the 2.75th position (between 2nd and 3rd positions) of the data.  Deciles divide the distribution into tenths or 10 equal parts. A data set has nine deciles which is denoted by 𝐷1 , 𝐷2 , 𝐷3 , …, 𝐷9. Basically, the first decile, 𝐷1 , is the number that divides the bottom 10% of the data from the top 90%. To obtain the deciles, use the formula below. 𝑛+1 𝐷𝑘 = 𝑘 ( ) 10 where 𝑘 – the position of the decile 𝑛 – number of observations When 𝑘 th-Decile is not an integer, it can be found by interpolation. Note that the second quartile, fifth decile, and fiftieth percentile of a data set are all the same and all equal to the median. 𝑀𝑒𝑑𝑖𝑎𝑛 = 𝑄2 = 𝐷5 = 𝑃50 Similarly, 𝑄1 = 𝑃25, 𝐷1 = 𝑃10, and 𝑄3 = 𝑃75. Example 6 Find the value corresponding to the sixth decile for the given set. 80 68 53 58 76 73 85 88 91 79 Solution Arrange the data from lowest to highest. 53 58 68 73 76 79 80 85 88 91 Using the formula, where 𝑛 = 10. 10 + 1 𝐷6 = 6 ( ) = 6.6 → 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 10 Since 𝐷6 is not a whole number, we will proceed to the interpolation. 𝐷6 = 6𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 + 0.6(7𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 − 6𝑡ℎ 𝑣𝑎𝑙𝑢𝑒) 𝐷6 = 79 + 0.6(80 − 79) 𝐷6 = 79.6 → 𝑣𝑎𝑙𝑢𝑒 Therefore, the value of 𝐷6 is 79.6 which can be found in the 6.6th position (between 6th and 7th positions) of the data. Grouped Data For grouped data, the quartiles, deciles, or percentiles can be determined using the following formulas. 𝑘𝑛 𝑘𝑛 𝑘𝑛 − 𝑐𝑓𝑝 − 𝑐𝑓𝑝 − 𝑐𝑓𝑝 𝑄𝑘 = 𝐿𝑄𝑘 + ( 4 ) (𝑤 ) 𝐷𝑘 = 𝐿𝐷𝑘 + ( 10 ) (𝑤 ) 𝑃𝑘 = 𝐿𝑃𝑘 + ( 100 ) (𝑤 ) 𝑓 𝑓 𝑓 where 𝑘 – the position of the quartile, decile, or percentile class 𝐿 – lower boundary of the quartile, decile, or percentile class 𝑛 – total number of observations 𝑤 – class width 𝑐𝑓𝑝 – cumulative frequency of the preceding class 𝑓 – frequency of the quartile, decile, or percentile class 𝑘𝑛 Note: Determine first the observation/value where the quartile, decile, or percentile is located by using 4 𝑘𝑛 𝑘𝑛 for the quartile, for the decile, and for the percentile. After that, look at the cumulative frequency 10 100 column to find which class contains the computed observation/value. Example 7 Find the third quartile, 4th decile, and 70th percentile for the given frequency distribution below.  Class Boundaries Frequency 𝒄𝒇 52.5 − 63.5 6 6 63.5 − 74.5 12 18 74.5 − 85.5 25 43 85.5 − 96.5 28 71 96.5 − 107.5 14 85 107.5 − 118.5 5 90 3(90) Solution For 𝑄3 , the class that contains = 67.5th observation is the class with interval 85.5 − 96.5. 4 (The class 85.5 − 96.5 contains the 44th to 71st observations in which the 67.5th observation falls.) 𝑘𝑛 − 𝑐𝑓𝑝 67.5 − 43 𝑄3 = 𝐿𝑄3 + ( 4 ) (𝑤) = 85.5 + ( ) (11) = 95.125 𝑓 28 4(90) For 𝐷4 , the class that contains 10 = 36th observation is the class with the interval 74.5 − 85.5. (The class 74.5 − 85.5 contains the 19th to 43rd observations in which the 36th observation falls.) 𝑘𝑛 − 𝑐𝑓𝑝 36 − 18 𝐷4 = 𝐿𝐷4 + (10 ) (𝑤) = 74.5 + ( ) (11) = 82.42 𝑓 25 70(90) For 𝑃70, the class that contains 100 = 63rd observation is the class with the interval 85.5 − 96.5. (The class 85.5 − 96.5 contains the 44th to 71st observations in which the 63rd observation falls.) 𝑘𝑛 − 𝑐𝑓𝑝 63 − 43 𝑃70 = 𝐿𝑃70 + ( 100 ) (𝑤) = 85.5 + ( ) (11) = 93.36 𝑓 28 TECHNOLOGY APPLICATION https://www.edupristine.com/blog/statistical-functions-in-excel https://www.youtube.com/watch?v=KYALCR6Bdxk People usually have love-hate relationship with statistics. When you get your formulas right you are in love with it and when your answers go wrong, your feelings take the opposite route. But when you are working on statistics in Excel, things are simpler and less complicated. So today, we bring to you the most used statistical functions of Excel. Average Function in Excel (Mean) One of the most used statistical functions in Excel is Average. Calculating the Average in Excel is much simpler than it was in the school. Simply use the Average function and select the range which needs to be averaged. In the example below we want to know the average of the marks obtained by the students so we use = 𝐴𝑉𝐸𝑅𝐴𝐺𝐸(𝐵2: 𝐵12). Median in Excel Median is a function which is used to find the middle number in a given range of numbers. When you are finding median manually, you need to sort the data in an ascending order but in Excel, you can simply use the Median function and select the range and you will find your median. We take the same example as above to find the median of marks obtained by students. So we use = 𝑀𝐸𝐷𝐼𝐴𝑁(𝐵2: 𝐵12). Mode in Excel Mode helps you to find out the value that occurs most number of times. When you are working on a large amount of data, this function can be a lot of help. To find the most occurring value in Excel, use the MODE function and select the range you want to find the mode of. In our example below, we use = 𝑀𝑂𝐷𝐸(𝐵2: 𝐵12) and since 2 students have scored 55 we get the answer as 55.