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ARITHMETIC MEAN, MEASURES OF CENTRAL TENDENCY

The document discusses measures of central tendency and arithmetic mean. It provides formulas to calculate the arithmetic mean for raw data, frequency data, and grouped frequency data. It includes examples such as finding the average marks of students, average weight of children, and average salary of employees. The arithmetic mean is used to solve problems involving changes to the original data or addition of new data points.

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MEASURES OF CENTRAL TENDENCY ARITHMETIC MEAN
MEASURES OF CENTRAL TENDENCY Suppose that we have frequency distribution of marks of two class divisions as follows DIVISION A MARKS NO. OF STUDENTS 0-20 5 20-40 10 40-60 28 60-80 12 80-100 5 DIVISION B MARKS NO. OF STUDENTS 0-20 2 20-40 15 40-60 22 60-80 18 80-100 3
MEASURES OF CENTRAL TENDENCY ï‚Ą AVERAGES: They are typical values with in the range of data. These values generally lie in the center part of the distribution, individual values of the variable usually cluster around it. So averages are called as measures of central tendency. ï‚Ą TYPES OF AVERAGES: Averages are classified as two (i) MATHEMATICAL AVERAGES: They are computed averages. They are based on all the observations. Eg: Arithmetic mean, Geometric mean, Harmonic mean (ii) POSITIONAL AVERAGES: These are positional values. These averages are not based on all observations Eg: Median, Mode
ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą ARITHMETIC MEAN FOR RAW DATA: If đ‘„ is a variable that takes 𝑛 values say đ‘„1, đ‘„2, 
 , đ‘„đ‘› then arithmetic mean is given by đ‘„ = đ‘„1 + đ‘„2 + ⋯ + đ‘„đ‘› 𝑛 = đ‘„ 𝑛 Eg: Find arithmetic mean for the following data 10, 12, 14, 16, 18, 20 đ‘„ = đ‘„ 𝑛 = 10+12+14+16+18+20 6 = 90 6 đ‘„ = 15
ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Find arithmetic mean for the following data 80, 85, 90, 75, 70, 88, 94, 98 đ‘„ = đ‘„ 𝑛 = 80+85+90+75+70+88+94+98 8 = 680 8 đ‘„ = 85
ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą ARITHMETIC MEAN FOR FREQUENCY DATA: If đ‘„ is a variable that takes values say đ‘„1, đ‘„2, 
 , đ‘„đ‘› with corresponding frequencies 𝑓1, 𝑓2, 
 , 𝑓𝑛then arithmetic mean is given by đ‘„ = 𝑓1đ‘„1 + 𝑓2đ‘„2 + ⋯ + đ‘“đ‘›đ‘„đ‘› 𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛 = đ‘“đ‘„ 𝑓
ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Find the average weight of 20 children from the following table. SOLUTION: đ‘„ = đ‘“đ‘„ 𝑓 = 256 20 = 12.8𝑘𝑔 WEIGHT (IN Kg) 11 12 13 14 15 NO. OF CHILDREN 3 5 7 3 2 WEIGHT (IN Kg)(𝒙) NO. OF CHILDREN(𝒇) 𝒇𝒙 11 3 33 12 5 60 13 7 91 14 3 42 15 2 30 𝑓=20 đ‘“đ‘„=256
ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Find the average height of students from the following table. SOLUTION: đ‘„ = đ‘“đ‘„ 𝑓 = 3771 23 = 163.956𝑐𝑚 HEIGHT (IN cm) 159 162 164 166 170 NO. OF STUDENTS 3 5 7 6 2 HEIGHT (IN cm)(𝒙) NO. OF CHILDREN(𝒇) 𝒇𝒙 159 3 477 162 5 810 164 7 1148 166 6 996 170 2 340 𝑓=23 đ‘“đ‘„=3771
ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą ARITHMETIC MEAN FOR FREQUENCY DATA: If class intervals and corresponding frequencies are given, find class mark for each interval. Let it be đ‘„1, đ‘„2, 
 , đ‘„đ‘› with corresponding frequencies 𝑓1, 𝑓2, 
 , 𝑓𝑛then arithmetic mean is given by đ‘„ = 𝑓1đ‘„1 + 𝑓2đ‘„2 + ⋯ + đ‘“đ‘›đ‘„đ‘› 𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛 = đ‘“đ‘„ 𝑓 NOTE: Convert inclusive intervals to exclusive intervals.
ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Find the A.M. for the following data. SOLUTION: đ‘„ = đ‘“đ‘„ 𝑓 = 1240 50 = 24.8 Class intervals 0-10 10-20 20-30 30-40 40-50 Frequency 7 11 15 10 7 Class intervals Class mark(x) Frequency(f) 𝒇𝒙 0-10 5 7 35 10-20 15 11 165 20-30 25 15 375 30-40 35 10 350 40-50 45 7 315 𝑓=50 đ‘“đ‘„=1240
ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Find the A.M. for the following data. SOLUTION: Lower limit of successive class – Upper limit of previous class Correction factor(C.F.)= 2 Here CF= 20−19 2 = 0.5 New class intervals will be (l.l –C.F)-(u.l+ C.F) Class intervals 10-19 20- 30-39 40-49 50-59 60-69 70-79 Frequency 6 5 8 15 7 6 3
ARITHMETIC MEAN(A.M.)-đ‘„ . đ‘„ = đ‘“đ‘„ 𝑓 = 2145 50 = 42.9 Class intervals Class intervals(exclusive) Class mark(x) Frequency(f) 𝒇𝒙 10-19 9.5-19.5 14.5 6 87 20-29 19.5-29.5 24.5 5 122.5 30-39 29.5-39.5 34.5 8 276 40-49 39.5-49.5 44.5 15 667.5 50-59 49.5-59.5 54.5 7 381.5 60-69 59.5-69.5 64.5 6 387 70-79 69.5-79.5 74.5 3 223.5 𝑓=50 đ‘“đ‘„=2145
ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Find the A.M. for the following data representing daily wages of a group of employees. Given minimum salary is 700. SOLUTION: đ‘„ = đ‘“đ‘„ 𝑓 = 136500 150 = 910 Salary <800 <900 <1000 <1100 <1200 No. of people 30 70 120 140 150 Class intervals Class mark(x) Frequency(f) 𝒇𝒙 700-800 750 30 22500 800-900 850 40 34000 900-1000 950 50 47500 1000-1100 1050 20 21000 1100-1200 1150 10 11500 𝑓=150 đ‘“đ‘„=136500
ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Following data gives the consumption of electricity. Find average consumption. SOLUTION: 𝑓 = 170 đ‘“đ‘„ = 136408 đ‘„ = đ‘“đ‘„ 𝑓 = 136408 170 = 802.4 Units <200 <400 <600 <80 <100 <1200 <140 <160 No. of consumers 7 25 53 88 118 142 160 170
ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Find the missing frequency in the following data given that A.M is 31. Solution: Let the missing frequency be k. Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 5 10 - 35 15 5 Marks No. of Class mark(x) 𝒇𝒙 0-10 5 5 25 10-20 10 15 150 20-30 k 25 25k 30-40 35 35 1225 40-50 15 45 675 50-60 5 55 275 Σ𝑓 = 70 + 𝑘 ÎŁđ‘“đ‘„ = 2350 + 25𝑘
ARITHMETIC MEAN(A.M.)-đ‘„ We know đ‘„ = đ‘“đ‘„ 𝑓 And it is given that A.M for the data is 31. Substituting all values, we get, 31 = 2350+25𝑘 70+𝑘 31 70 + 𝑘 = 2350 + 25𝑘 2170 + 31𝑘 = 2350 + 25𝑘 31𝑘 − 25𝑘 = 2350 − 2170 6𝑘 = 180 𝑘 = 180 6 𝑘 = 30
ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Calculate the missing frequency in the following data given that A.M is 28. Solution: Let the missing frequency be k. Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 12 18 27 - 17 6 Marks No. of Class mark(x) 𝒇𝒙 0-10 12 5 60 10-20 18 15 270 20-30 27 25 675 30-40 k 35 35k 40-50 17 45 765 50-60 6 55 330 Σ𝑓 = 80 + 𝑘 ÎŁđ‘“đ‘„ = 2100 + 35𝑘
ARITHMETIC MEAN(A.M.)-đ‘„ We know đ‘„ = đ‘“đ‘„ 𝑓 And it is given that A.M for the data is 28. Substituting all values, we get, 28 = 2100+35𝑘 80+𝑘 28 80 + 𝑘 = 2100 + 35𝑘 2240 + 28𝑘 = 2100 + 35𝑘 35𝑘 − 28𝑘 = 2240 − 2100 7𝑘 = 140 𝑘 = 140 7 𝑘 = 20
ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Mean of 200 observations was 50. Later on it was discovered that one observation was wrongly noted as 92 instead of 192. Find the correct mean. Solution: Given 𝑛 = 200, đ‘„ = 50 đ‘„ = ÎŁđ‘„ 𝑛 50 = ÎŁđ‘„ 200 ÎŁđ‘„ = 50 × 200 = 10000 Σ𝑋 = 10000 − 92 + 192 = 10100 𝑋 = Σ𝑋 𝑁 𝑋 = 10100 200 ∎ 𝑋 = 50.5
ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Mean of 20 observations was 44. Later on it was discovered that two observations was wrongly noted as 17, 40 instead of 70, 14. Find the correct mean. Solution: Given 𝑛 = 20, đ‘„ = 44 đ‘„ = ÎŁđ‘„ 𝑛 44 = ÎŁđ‘„ 20 ÎŁđ‘„ = 44 × 20 = 880 Σ𝑋 = 880 − 17 − 40 + 70 + 14 = 907 𝑋 = Σ𝑋 𝑁 𝑋 = 907 20 ∎ 𝑋 = 45.35
ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą The mean of a certain number of observations is 40. if two more items with values 50 & 64 are added to the data the mean rises to 42. Find the number of observations in the original data. Solution: Given đ‘„ = 40 đ‘„ = ÎŁđ‘„ 𝑛 40 = ÎŁđ‘„ 𝑛 ÎŁđ‘„ = 40𝑛 Σ𝑋 = ÎŁđ‘„ + 50 + 64 = 40n + 114 𝑁 = 𝑛 + 2 𝑋 = Σ𝑋 𝑁 42 = 40𝑛 + 114 𝑛 + 2
ARITHMETIC MEAN(A.M.)-đ‘„ 42 = 40𝑛 + 114 𝑛 + 2 42 𝑛 + 2 = 40𝑛 + 114 42𝑛 + 84 = 40𝑛 + 114 42𝑛 − 40𝑛 = 114 − 84 2𝑛 = 30 𝑛 = 30 2 𝑛 = 15
ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą The mean height of a group of 49 students is 63 inch. To this 6 students are added. The heights of 5 students are known to be 58, 59, 59, 69, 69. Find the height of the 6th student if mean height of all 55 students is 63.1 inch. Solution: Given 𝑛 = 49, đ‘„ = 63 đ‘„ = ÎŁđ‘„ 𝑛 63 = ÎŁđ‘„ 49 ÎŁđ‘„ = 63 × 49 = 3087 Σ𝑋 = ÎŁđ‘„ + 58 + 59 + 59 + 69 + 69 + h Σ𝑋 = 3087 + 58 + 59 + 59 + 69 + 69 + h = 3401 + h 𝑁 = 𝑛 + 6 = 55 𝑋 = Σ𝑋 𝑁
ARITHMETIC MEAN(A.M.)-đ‘„ Also its given 𝑋 = 63.1 ∎ 𝑋 = Σ𝑋 𝑁 âŸč 63.1 = 3401 + ℎ 55 63.1 × 55 = 3401 + ℎ 3470.5 = 3401 + h ℎ = 3470.5 − 3401 ℎ = 69.5 inch
COMBINED ARITHMETIC MEAN If there are two groups- Group I and Group II containing 𝑛1 and 𝑛2 number of observations respectively and the mean of these group are đ‘„1 and đ‘„2 then the combined mean for these two groups with 𝑛1 + 𝑛2 number of observation is given by đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 Note: If there are three groups then the formula will be đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 + 𝑛3đ‘„3 𝑛1 + 𝑛2 + 𝑛3
COMBINED ARITHMETIC MEAN ï‚Ą Mean salary paid to 30 male employees is 750 and mean salary paid to 20 female employees is 650. Find the mean of all 50 employees taken together. Solution: Let Group I-Male employees and Group II- Female employees 𝑛1 = 30, đ‘„1 = 750 𝑛2 = 20, đ‘„2 = 650 đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 đ‘„ = 30 × 750 + 20 × 650 30 + 20 = 22500 + 13000 50 = 35500 50 đ‘„ = 710
COMBINED ARITHMETIC MEAN ï‚Ą The average marks in Accountancy of 45 boys and 70 girls from a class are 55 and 63 respectively. Find the average marks of entire class. Solution: Let Group I- Boys and Group II- Girls 𝑛1 = 45, đ‘„1 = 55 𝑛2 = 70, đ‘„2 = 63 đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 đ‘„ = 45 × 55 + 70 × 63 45 + 70 = 2475 + 4410 115 = 6885 115 đ‘„ = 59.87
COMBINED ARITHMETIC MEAN ï‚Ą The average marks of a class of students are 76. The average marks of boys and girls are 69 and 83 respectively. If there are 75 boys, find the number of girls. Solution: Let Group I- Boys and Group II- Girls Given đ‘„ = 76 𝑛1 = 75, đ‘„1 = 69 𝑛2 = 𝑘, đ‘„2 = 83 đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 76 = 75 × 69 + 𝑘 × 83 75 + 𝑘 76 = 5175 + 83𝑘 75 + 𝑘
COMBINED ARITHMETIC MEAN 76 75 + 𝑘 = 5175 + 83𝑘 5700 + 76𝑘 = 5175 + 83𝑘 83𝑘 − 76𝑘 = 5700 − 5175 7𝑘 = 525 𝑘 = 525 7 𝑘 = 75 ∮Number of girls in the class=75
COMBINED ARITHMETIC MEAN ï‚Ą The average weight of 500 articles manufactured in a factory is 34gm. Out of these, average weight of 30 defective articles is 25gm. Find the average weight of non defective articles. Solution: Let Group I- Defective articles and Group II- Non-defective articles Given đ‘„ = 34, 𝑛1 + 𝑛2 = 500 𝑛1 = 30, đ‘„1 = 25 𝑛2 = 500 − 30 = 470, đ‘„2 = 𝑘 đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 34 = 30 × 25 + 470 × 𝑘 500 34 = 750 + 470𝑘 500
COMBINED ARITHMETIC MEAN 34 × 500 = 750 + 470𝑘 470𝑘 = 17000 − 750 470𝑘 = 16250 𝑘 = 16250 470 𝑘 = 34.57 ∮ Average weight of non defective articles =34.57gm
COMBINED ARITHMETIC MEAN ï‚Ą The average weight of 500 students of FYBCom is 57kg. The average weight of 475 SYBCom students is 61kg. and 525 TYBCom students are 59kg. Find the average weight of all students together. Solution: 𝑛1 = 500, đ‘„1 = 57 𝑛2 = 475, đ‘„2 = 61 𝑛3 = 525, đ‘„3 = 59 đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 + 𝑛3đ‘„3 𝑛1 + 𝑛2 + 𝑛3 đ‘„ = 500 × 57 + 475 × 61 + 525 × 59 500 + 475 + 525 đ‘„ = 88450 1500 đ‘„ = 58.97kg
WEIGHTED ARITHMETIC MEAN If 𝑛 values đ‘„1, đ‘„2, 
 , đ‘„đ‘› of the variable đ‘„ are assigned the weights đ‘€1, đ‘€2, 
 , đ‘€đ‘› respectively then weighted A.M is defined as đ‘„ = đ‘€1đ‘„1 + đ‘€2đ‘„2 + ⋯ + đ‘€đ‘›đ‘„đ‘› đ‘€1 + đ‘€2 + ⋯ + đ‘€đ‘› = đ‘€đ‘„ đ‘€
WEIGHTED ARITHMETIC MEAN ï‚Ą A candidate appeared for an examination and scored following marks Physics-45, Chemistry-52, Maths-40, English-55. It was considered that weight should be given as 3,2,3,2 respectively. Calculate weighted arithmetic mean. Solution: đ‘„ = đ‘€đ‘„ đ‘€ = 469 10 = 46.9 SUBJECT MARKS(x) WEIGHT(w) 𝒘𝒙 Physics 45 3 135 Chemistry 52 2 104 Maths 40 3 120 English 55 2 110 ÎŁđ‘€ = 10 ÎŁđ‘€đ‘„ = 469
WEIGHTED ARITHMETIC MEAN ï‚Ą A student X scores 50,55,60 and 45 in four subjects. Another student Y scores 53, 50, 47 and 59 in the same subjects. If the weights are 2, 4, 3 and 1 respectively for these subjects, decide which student performed better? Solution: đ‘„ = đ‘€đ‘„ đ‘€ = 545 10 = 54.5 𝑩 = đ‘€đ‘Š đ‘€ = 506 10 = 50.6 ∎X performed better SUBJECT MARKS (x) MARKS (y) WEIGHTS (w) 𝒘𝒙 𝒘𝒚 1 50 53 2 100 106 2 55 50 4 220 200 3 60 47 3 180 141 4 45 59 1 45 59 ÎŁđ‘€ = 10 ÎŁđ‘€đ‘„ = 545 ÎŁđ‘€đ‘Š = 506
PROPERTIES OF ARITHMETIC MEAN ï‚Ą The sum of deviation of individual values of variable đ‘„ from arithmetic mean đ‘„ is zero. ÎŁ đ‘„đ‘– − đ‘„ = 0 ï‚Ą If arithmetic mean đ‘„ ,n are known, the sum of observations can be calculated. ÎŁđ‘„ = đ‘›đ‘„ ï‚Ą The sum of squares of deviation from arithmetic mean is less than that from any other average A. ÎŁ(đ‘„đ‘– − đ‘„)2< ÎŁ(đ‘„đ‘– − 𝐮)2 ï‚Ą If there are two groups- Group I and Group II containing 𝑛1 and 𝑛2 number of observations respectively and the mean of these group are đ‘„1 and đ‘„2 then the combined mean for these two groups with 𝑛1 + 𝑛2 number of observation is given by đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 ï‚Ą If each of the values of x is increased by any constant, then arithmetic mean will also increase by the same constant
MERITS OF ARITHMETIC MEAN ï‚Ą It is easy to understand and easy to calculate. ï‚Ą It is rigidly defined. ï‚Ą It is based on all observation ï‚Ą Further mathematical treatment is possible ï‚Ą It is a better representative than any other average ï‚Ą It is less affected by sampling fluctuations.
DEMERITS OF ARITHMETIC MEAN ï‚Ą If some values are not known, AM cannot be calculated ï‚Ą It may be a value which may not be present in the data ï‚Ą Sometimes, it may give absurd results ï‚Ą It is affected by extreme values. ï‚Ą In case of open end class intervals, AM can not be calculated.

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ARITHMETIC MEAN, MEASURES OF CENTRAL TENDENCY

  • 1.
    MEASURES OF CENTRALTENDENCY ARITHMETIC MEAN
  • 2.
    MEASURES OF CENTRALTENDENCY Suppose that we have frequency distribution of marks of two class divisions as follows DIVISION A MARKS NO. OF STUDENTS 0-20 5 20-40 10 40-60 28 60-80 12 80-100 5 DIVISION B MARKS NO. OF STUDENTS 0-20 2 20-40 15 40-60 22 60-80 18 80-100 3
  • 3.
    MEASURES OF CENTRALTENDENCY ï‚Ą AVERAGES: They are typical values with in the range of data. These values generally lie in the center part of the distribution, individual values of the variable usually cluster around it. So averages are called as measures of central tendency. ï‚Ą TYPES OF AVERAGES: Averages are classified as two (i) MATHEMATICAL AVERAGES: They are computed averages. They are based on all the observations. Eg: Arithmetic mean, Geometric mean, Harmonic mean (ii) POSITIONAL AVERAGES: These are positional values. These averages are not based on all observations Eg: Median, Mode
  • 4.
    ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą ARITHMETICMEAN FOR RAW DATA: If đ‘„ is a variable that takes 𝑛 values say đ‘„1, đ‘„2, 
 , đ‘„đ‘› then arithmetic mean is given by đ‘„ = đ‘„1 + đ‘„2 + ⋯ + đ‘„đ‘› 𝑛 = đ‘„ 𝑛 Eg: Find arithmetic mean for the following data 10, 12, 14, 16, 18, 20 đ‘„ = đ‘„ 𝑛 = 10+12+14+16+18+20 6 = 90 6 đ‘„ = 15
  • 5.
    ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Findarithmetic mean for the following data 80, 85, 90, 75, 70, 88, 94, 98 đ‘„ = đ‘„ 𝑛 = 80+85+90+75+70+88+94+98 8 = 680 8 đ‘„ = 85
  • 6.
    ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą ARITHMETICMEAN FOR FREQUENCY DATA: If đ‘„ is a variable that takes values say đ‘„1, đ‘„2, 
 , đ‘„đ‘› with corresponding frequencies 𝑓1, 𝑓2, 
 , 𝑓𝑛then arithmetic mean is given by đ‘„ = 𝑓1đ‘„1 + 𝑓2đ‘„2 + ⋯ + đ‘“đ‘›đ‘„đ‘› 𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛 = đ‘“đ‘„ 𝑓
  • 7.
    ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Findthe average weight of 20 children from the following table. SOLUTION: đ‘„ = đ‘“đ‘„ 𝑓 = 256 20 = 12.8𝑘𝑔 WEIGHT (IN Kg) 11 12 13 14 15 NO. OF CHILDREN 3 5 7 3 2 WEIGHT (IN Kg)(𝒙) NO. OF CHILDREN(𝒇) 𝒇𝒙 11 3 33 12 5 60 13 7 91 14 3 42 15 2 30 𝑓=20 đ‘“đ‘„=256
  • 8.
    ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Findthe average height of students from the following table. SOLUTION: đ‘„ = đ‘“đ‘„ 𝑓 = 3771 23 = 163.956𝑐𝑚 HEIGHT (IN cm) 159 162 164 166 170 NO. OF STUDENTS 3 5 7 6 2 HEIGHT (IN cm)(𝒙) NO. OF CHILDREN(𝒇) 𝒇𝒙 159 3 477 162 5 810 164 7 1148 166 6 996 170 2 340 𝑓=23 đ‘“đ‘„=3771
  • 9.
    ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą ARITHMETICMEAN FOR FREQUENCY DATA: If class intervals and corresponding frequencies are given, find class mark for each interval. Let it be đ‘„1, đ‘„2, 
 , đ‘„đ‘› with corresponding frequencies 𝑓1, 𝑓2, 
 , 𝑓𝑛then arithmetic mean is given by đ‘„ = 𝑓1đ‘„1 + 𝑓2đ‘„2 + ⋯ + đ‘“đ‘›đ‘„đ‘› 𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛 = đ‘“đ‘„ 𝑓 NOTE: Convert inclusive intervals to exclusive intervals.
  • 10.
    ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Findthe A.M. for the following data. SOLUTION: đ‘„ = đ‘“đ‘„ 𝑓 = 1240 50 = 24.8 Class intervals 0-10 10-20 20-30 30-40 40-50 Frequency 7 11 15 10 7 Class intervals Class mark(x) Frequency(f) 𝒇𝒙 0-10 5 7 35 10-20 15 11 165 20-30 25 15 375 30-40 35 10 350 40-50 45 7 315 𝑓=50 đ‘“đ‘„=1240
  • 11.
    ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Findthe A.M. for the following data. SOLUTION: Lower limit of successive class – Upper limit of previous class Correction factor(C.F.)= 2 Here CF= 20−19 2 = 0.5 New class intervals will be (l.l –C.F)-(u.l+ C.F) Class intervals 10-19 20- 30-39 40-49 50-59 60-69 70-79 Frequency 6 5 8 15 7 6 3
  • 12.
    ARITHMETIC MEAN(A.M.)-đ‘„ . đ‘„ = đ‘“đ‘„ 𝑓 = 2145 50 =42.9 Class intervals Class intervals(exclusive) Class mark(x) Frequency(f) 𝒇𝒙 10-19 9.5-19.5 14.5 6 87 20-29 19.5-29.5 24.5 5 122.5 30-39 29.5-39.5 34.5 8 276 40-49 39.5-49.5 44.5 15 667.5 50-59 49.5-59.5 54.5 7 381.5 60-69 59.5-69.5 64.5 6 387 70-79 69.5-79.5 74.5 3 223.5 𝑓=50 đ‘“đ‘„=2145
  • 13.
    ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Findthe A.M. for the following data representing daily wages of a group of employees. Given minimum salary is 700. SOLUTION: đ‘„ = đ‘“đ‘„ 𝑓 = 136500 150 = 910 Salary <800 <900 <1000 <1100 <1200 No. of people 30 70 120 140 150 Class intervals Class mark(x) Frequency(f) 𝒇𝒙 700-800 750 30 22500 800-900 850 40 34000 900-1000 950 50 47500 1000-1100 1050 20 21000 1100-1200 1150 10 11500 𝑓=150 đ‘“đ‘„=136500
  • 14.
    ARITHMETIC MEAN(A.M.)-đ‘„ Eg: Followingdata gives the consumption of electricity. Find average consumption. SOLUTION: 𝑓 = 170 đ‘“đ‘„ = 136408 đ‘„ = đ‘“đ‘„ 𝑓 = 136408 170 = 802.4 Units <200 <400 <600 <80 <100 <1200 <140 <160 No. of consumers 7 25 53 88 118 142 160 170
  • 15.
    ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Findthe missing frequency in the following data given that A.M is 31. Solution: Let the missing frequency be k. Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 5 10 - 35 15 5 Marks No. of Class mark(x) 𝒇𝒙 0-10 5 5 25 10-20 10 15 150 20-30 k 25 25k 30-40 35 35 1225 40-50 15 45 675 50-60 5 55 275 Σ𝑓 = 70 + 𝑘 ÎŁđ‘“đ‘„ = 2350 + 25𝑘
  • 16.
    ARITHMETIC MEAN(A.M.)-đ‘„ We knowđ‘„ = đ‘“đ‘„ 𝑓 And it is given that A.M for the data is 31. Substituting all values, we get, 31 = 2350+25𝑘 70+𝑘 31 70 + 𝑘 = 2350 + 25𝑘 2170 + 31𝑘 = 2350 + 25𝑘 31𝑘 − 25𝑘 = 2350 − 2170 6𝑘 = 180 𝑘 = 180 6 𝑘 = 30
  • 17.
    ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Calculatethe missing frequency in the following data given that A.M is 28. Solution: Let the missing frequency be k. Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 12 18 27 - 17 6 Marks No. of Class mark(x) 𝒇𝒙 0-10 12 5 60 10-20 18 15 270 20-30 27 25 675 30-40 k 35 35k 40-50 17 45 765 50-60 6 55 330 Σ𝑓 = 80 + 𝑘 ÎŁđ‘“đ‘„ = 2100 + 35𝑘
  • 18.
    ARITHMETIC MEAN(A.M.)-đ‘„ We knowđ‘„ = đ‘“đ‘„ 𝑓 And it is given that A.M for the data is 28. Substituting all values, we get, 28 = 2100+35𝑘 80+𝑘 28 80 + 𝑘 = 2100 + 35𝑘 2240 + 28𝑘 = 2100 + 35𝑘 35𝑘 − 28𝑘 = 2240 − 2100 7𝑘 = 140 𝑘 = 140 7 𝑘 = 20
  • 19.
    ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Meanof 200 observations was 50. Later on it was discovered that one observation was wrongly noted as 92 instead of 192. Find the correct mean. Solution: Given 𝑛 = 200, đ‘„ = 50 đ‘„ = ÎŁđ‘„ 𝑛 50 = ÎŁđ‘„ 200 ÎŁđ‘„ = 50 × 200 = 10000 Σ𝑋 = 10000 − 92 + 192 = 10100 𝑋 = Σ𝑋 𝑁 𝑋 = 10100 200 ∎ 𝑋 = 50.5
  • 20.
    ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Meanof 20 observations was 44. Later on it was discovered that two observations was wrongly noted as 17, 40 instead of 70, 14. Find the correct mean. Solution: Given 𝑛 = 20, đ‘„ = 44 đ‘„ = ÎŁđ‘„ 𝑛 44 = ÎŁđ‘„ 20 ÎŁđ‘„ = 44 × 20 = 880 Σ𝑋 = 880 − 17 − 40 + 70 + 14 = 907 𝑋 = Σ𝑋 𝑁 𝑋 = 907 20 ∎ 𝑋 = 45.35
  • 21.
    ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Themean of a certain number of observations is 40. if two more items with values 50 & 64 are added to the data the mean rises to 42. Find the number of observations in the original data. Solution: Given đ‘„ = 40 đ‘„ = ÎŁđ‘„ 𝑛 40 = ÎŁđ‘„ 𝑛 ÎŁđ‘„ = 40𝑛 Σ𝑋 = ÎŁđ‘„ + 50 + 64 = 40n + 114 𝑁 = 𝑛 + 2 𝑋 = Σ𝑋 𝑁 42 = 40𝑛 + 114 𝑛 + 2
  • 22.
    ARITHMETIC MEAN(A.M.)-đ‘„ 42 = 40𝑛+ 114 𝑛 + 2 42 𝑛 + 2 = 40𝑛 + 114 42𝑛 + 84 = 40𝑛 + 114 42𝑛 − 40𝑛 = 114 − 84 2𝑛 = 30 𝑛 = 30 2 𝑛 = 15
  • 23.
    ARITHMETIC MEAN(A.M.)-đ‘„ ï‚Ą Themean height of a group of 49 students is 63 inch. To this 6 students are added. The heights of 5 students are known to be 58, 59, 59, 69, 69. Find the height of the 6th student if mean height of all 55 students is 63.1 inch. Solution: Given 𝑛 = 49, đ‘„ = 63 đ‘„ = ÎŁđ‘„ 𝑛 63 = ÎŁđ‘„ 49 ÎŁđ‘„ = 63 × 49 = 3087 Σ𝑋 = ÎŁđ‘„ + 58 + 59 + 59 + 69 + 69 + h Σ𝑋 = 3087 + 58 + 59 + 59 + 69 + 69 + h = 3401 + h 𝑁 = 𝑛 + 6 = 55 𝑋 = Σ𝑋 𝑁
  • 24.
    ARITHMETIC MEAN(A.M.)-đ‘„ Also itsgiven 𝑋 = 63.1 ∎ 𝑋 = Σ𝑋 𝑁 âŸč 63.1 = 3401 + ℎ 55 63.1 × 55 = 3401 + ℎ 3470.5 = 3401 + h ℎ = 3470.5 − 3401 ℎ = 69.5 inch
  • 25.
    COMBINED ARITHMETIC MEAN Ifthere are two groups- Group I and Group II containing 𝑛1 and 𝑛2 number of observations respectively and the mean of these group are đ‘„1 and đ‘„2 then the combined mean for these two groups with 𝑛1 + 𝑛2 number of observation is given by đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 Note: If there are three groups then the formula will be đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 + 𝑛3đ‘„3 𝑛1 + 𝑛2 + 𝑛3
  • 26.
    COMBINED ARITHMETIC MEAN ï‚ĄMean salary paid to 30 male employees is 750 and mean salary paid to 20 female employees is 650. Find the mean of all 50 employees taken together. Solution: Let Group I-Male employees and Group II- Female employees 𝑛1 = 30, đ‘„1 = 750 𝑛2 = 20, đ‘„2 = 650 đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 đ‘„ = 30 × 750 + 20 × 650 30 + 20 = 22500 + 13000 50 = 35500 50 đ‘„ = 710
  • 27.
    COMBINED ARITHMETIC MEAN ï‚ĄThe average marks in Accountancy of 45 boys and 70 girls from a class are 55 and 63 respectively. Find the average marks of entire class. Solution: Let Group I- Boys and Group II- Girls 𝑛1 = 45, đ‘„1 = 55 𝑛2 = 70, đ‘„2 = 63 đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 đ‘„ = 45 × 55 + 70 × 63 45 + 70 = 2475 + 4410 115 = 6885 115 đ‘„ = 59.87
  • 28.
    COMBINED ARITHMETIC MEAN ï‚ĄThe average marks of a class of students are 76. The average marks of boys and girls are 69 and 83 respectively. If there are 75 boys, find the number of girls. Solution: Let Group I- Boys and Group II- Girls Given đ‘„ = 76 𝑛1 = 75, đ‘„1 = 69 𝑛2 = 𝑘, đ‘„2 = 83 đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 76 = 75 × 69 + 𝑘 × 83 75 + 𝑘 76 = 5175 + 83𝑘 75 + 𝑘
  • 29.
    COMBINED ARITHMETIC MEAN 7675 + 𝑘 = 5175 + 83𝑘 5700 + 76𝑘 = 5175 + 83𝑘 83𝑘 − 76𝑘 = 5700 − 5175 7𝑘 = 525 𝑘 = 525 7 𝑘 = 75 ∮Number of girls in the class=75
  • 30.
    COMBINED ARITHMETIC MEAN ï‚ĄThe average weight of 500 articles manufactured in a factory is 34gm. Out of these, average weight of 30 defective articles is 25gm. Find the average weight of non defective articles. Solution: Let Group I- Defective articles and Group II- Non-defective articles Given đ‘„ = 34, 𝑛1 + 𝑛2 = 500 𝑛1 = 30, đ‘„1 = 25 𝑛2 = 500 − 30 = 470, đ‘„2 = 𝑘 đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 34 = 30 × 25 + 470 × 𝑘 500 34 = 750 + 470𝑘 500
  • 31.
    COMBINED ARITHMETIC MEAN 34× 500 = 750 + 470𝑘 470𝑘 = 17000 − 750 470𝑘 = 16250 𝑘 = 16250 470 𝑘 = 34.57 ∮ Average weight of non defective articles =34.57gm
  • 32.
    COMBINED ARITHMETIC MEAN ï‚ĄThe average weight of 500 students of FYBCom is 57kg. The average weight of 475 SYBCom students is 61kg. and 525 TYBCom students are 59kg. Find the average weight of all students together. Solution: 𝑛1 = 500, đ‘„1 = 57 𝑛2 = 475, đ‘„2 = 61 𝑛3 = 525, đ‘„3 = 59 đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 + 𝑛3đ‘„3 𝑛1 + 𝑛2 + 𝑛3 đ‘„ = 500 × 57 + 475 × 61 + 525 × 59 500 + 475 + 525 đ‘„ = 88450 1500 đ‘„ = 58.97kg
  • 33.
    WEIGHTED ARITHMETIC MEAN If𝑛 values đ‘„1, đ‘„2, 
 , đ‘„đ‘› of the variable đ‘„ are assigned the weights đ‘€1, đ‘€2, 
 , đ‘€đ‘› respectively then weighted A.M is defined as đ‘„ = đ‘€1đ‘„1 + đ‘€2đ‘„2 + ⋯ + đ‘€đ‘›đ‘„đ‘› đ‘€1 + đ‘€2 + ⋯ + đ‘€đ‘› = đ‘€đ‘„ đ‘€
  • 34.
    WEIGHTED ARITHMETIC MEAN ï‚ĄA candidate appeared for an examination and scored following marks Physics-45, Chemistry-52, Maths-40, English-55. It was considered that weight should be given as 3,2,3,2 respectively. Calculate weighted arithmetic mean. Solution: đ‘„ = đ‘€đ‘„ đ‘€ = 469 10 = 46.9 SUBJECT MARKS(x) WEIGHT(w) 𝒘𝒙 Physics 45 3 135 Chemistry 52 2 104 Maths 40 3 120 English 55 2 110 ÎŁđ‘€ = 10 ÎŁđ‘€đ‘„ = 469
  • 35.
    WEIGHTED ARITHMETIC MEAN ï‚ĄA student X scores 50,55,60 and 45 in four subjects. Another student Y scores 53, 50, 47 and 59 in the same subjects. If the weights are 2, 4, 3 and 1 respectively for these subjects, decide which student performed better? Solution: đ‘„ = đ‘€đ‘„ đ‘€ = 545 10 = 54.5 𝑩 = đ‘€đ‘Š đ‘€ = 506 10 = 50.6 ∎X performed better SUBJECT MARKS (x) MARKS (y) WEIGHTS (w) 𝒘𝒙 𝒘𝒚 1 50 53 2 100 106 2 55 50 4 220 200 3 60 47 3 180 141 4 45 59 1 45 59 ÎŁđ‘€ = 10 ÎŁđ‘€đ‘„ = 545 ÎŁđ‘€đ‘Š = 506
  • 36.
    PROPERTIES OF ARITHMETICMEAN ï‚Ą The sum of deviation of individual values of variable đ‘„ from arithmetic mean đ‘„ is zero. ÎŁ đ‘„đ‘– − đ‘„ = 0 ï‚Ą If arithmetic mean đ‘„ ,n are known, the sum of observations can be calculated. ÎŁđ‘„ = đ‘›đ‘„ ï‚Ą The sum of squares of deviation from arithmetic mean is less than that from any other average A. ÎŁ(đ‘„đ‘– − đ‘„)2< ÎŁ(đ‘„đ‘– − 𝐮)2 ï‚Ą If there are two groups- Group I and Group II containing 𝑛1 and 𝑛2 number of observations respectively and the mean of these group are đ‘„1 and đ‘„2 then the combined mean for these two groups with 𝑛1 + 𝑛2 number of observation is given by đ‘„ = 𝑛1đ‘„1 + 𝑛2đ‘„2 𝑛1 + 𝑛2 ï‚Ą If each of the values of x is increased by any constant, then arithmetic mean will also increase by the same constant
  • 37.
    MERITS OF ARITHMETICMEAN ï‚Ą It is easy to understand and easy to calculate. ï‚Ą It is rigidly defined. ï‚Ą It is based on all observation ï‚Ą Further mathematical treatment is possible ï‚Ą It is a better representative than any other average ï‚Ą It is less affected by sampling fluctuations.
  • 38.
    DEMERITS OF ARITHMETICMEAN ï‚Ą If some values are not known, AM cannot be calculated ï‚Ą It may be a value which may not be present in the data ï‚Ą Sometimes, it may give absurd results ï‚Ą It is affected by extreme values. ï‚Ą In case of open end class intervals, AM can not be calculated.