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Short-cut Method— Change of Origin and Scale, , When the data given involve very big values, the calculation of, arithmetic mean becomes tedious and for simplification of calculation,, Ge Xo, , , , short-cut method is used. We define a new variable u = where xo, , and c are constants. xo, generally is a value taken by x and lies nearly at, the centre of the range of values of x. c is usually the common width the, class interval. The values of u will be small compared to the values of the, , original variable x. Then the arithmetic mean of u, i.e., i is calculated., , We can get the value of ¥ asx =cil + Xo., This relation can be proved as follows:, X— Xo, , c, , , , =e SX = CUD

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44 rw Vipul’s™ Quantitative Methods - |,, , 0: os = =, x = =c. ee WX SC + Xo., 7 F 7, EX.6: ‘, We consider the same example as in Ex. 4. Calculate the ari!, 3 mean using short cut method., , , , , , , , , , , , , , , , , , , , , , , , No. of defective articles | 20-30 | 30-40 } 40-50 | 50-60 | 60-79, No. of workers is 8 10 12 5, Solution:, No. of defective | No.of workers | Mid value Bs x-45 h, articles f x 10 M, 20-30 5 25 -2 ~10, 30-40 8 35 = -8, 40-50 10 45 0 0, 50-60 12 55 1 12, 60-70 5 65 2 10, Total 40 _t, , , , , , , , , , , , , , x takes 5 values, 45 is the middle one., , -. We take xo = 45. The intervals are equal and of magnitude 200,, , , , , , c =100, . x-45, | ES u= 9, Using this relation we get 5 values of u corresponding to 5 values, x, 25-4!, eg.for x =25,u= 70 : =-2,, , We get wu taking 5 values from — 2 to 2, with frequencies 5, 8, «, respectively. We calculate i first., , 4, , 7 Se £, u= Lf =7 =0.1, , x is obtained using the following relation., , X =c+x9=10(0.1)+45=14 45 =46 |, | Average number of defective articles is 46. 4, «Ex.7: :, Calculate arithmetic mean of monthly incomes of 400 employees i, company.

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= eee i, , , , , , , , , , , , , , , , , , , , , , , , , , , , Averages rr 45, [Income in Rs. { 3000-4000 | 4000-5000 | 5000-7000 | 7000-10000 | 10000-15000, No. of 75 125 80 70 50, employees ., Solution:, Income v No. of Mid value x - 8500, in Rs. employees (f) oe u=""500 in., 3000-4000 75 3500 -10 - 750, 4000-5000 125 4500 -8 - 1000, 5000-7000 80 6000 -5 - 400, 7000-10000 70 “8500 0 0, 10000-15000 50 12500 8 400, Total 400 - 1750, , , , , , , , , , , , , , , , Here the class intervals are unequal. So the mid values of the class, intervals x, will not differ by a constant. We take 8500 as Xo., , We get the deviations of x from 8500. e.g. 3500 - 8500 = — 5000, 4500 —, 8500 = — 4000 and so on. The common factor for these deviations is 500, and so we take c = 500. Divide the deviations by 500 and we get u as, 3500 — 8500 _ 10 4500 - 8500, , 500 == 10; 500 =-8 and so on., - Ddfu -1750, ua Shs 0S -4.375, , X =C.U +X0,c¢ = 500, xo = 8500 |, x = 500 (- 4.375) + 8500 = - 2187.5 + 8500 = Rs. 6312.5, , Properties of Arithmetic Mean, (1) The sum of the deviations of all the values of x from their arithmetic, , mean is zero., n rt, ie. > fi(ai- x) = 0., 1, , (2) The product of the arithmetic mean and the number of items gives, the total of the items., , n n, ie. Nx = > fai where N = > fi, 1 i, , (3) If the arithmetic means %1, X2 of two distributions containing N; and |, , N» items respectively are known, the arithmetic mean x of the, distribution combining the two can be calculated as

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raw Vipul’s™ Quantitative Methods - \ 3, , Nixi + Nox2, Ni + No, If the arithmetic means of marks of two tutorial bate |, containing 25 and 30 students respectively are given to be 30 an,, respectively, the arithmetic mean of marks of 55 students Combin,, the two tutorial batches is, , 25 (30) + 30 (33) 750+990 1740, , 25430 °= 55 ="55 = 31.64 marks., , This formula can be extended to any finite number of Series,, (4) The sum of the squares of the deviations from the arithmetic Mea, , is less than that from any number A other than x., N N i, te. DY (a-x? <Y (xi- Az, Azz, i=l 1, (5) The arithmetic means of sums (differences) of the corresponding, , observations in two distributions containing equal number, observations is equal to the sum (difference) of their means. i.e. if, , z=x+ythenz =x +7., Merits of Arithmetic Mean, (1) It is rigidly defined. Its value is always definite., (2) It is easy to calculate and easy to understand. It is very popular., , (3) It is based on all the observations. So that it becomes a good, representative. |, , (4) It is capable of further algebraic treatment. It is possible to find the, , sum of observations, if the arithmetic mean and the number o, observations is known. It is also possible to combine arithmeti¢, means of two or more groups and get a combined arithmetic mean, for all the groups taken together., , (5) It is not affected much by sampling fluctuations., , Demerits, , (1) It is affected by extreme values., e.g. consider the A.M. of 10, 15, 25, 500., , 2s B50,, x = =1375, , Here x goes up because of a single value 500. In such a cast, arithmetic mean is not a good representative. :

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(2) Itis a value which may not be present in the data. There may not be, , even a single item which takes the value x., (3) Sometimes it gives absurd result like 4.4 children per family., , (4) We ae calculate it when open-end class intervals are present in, the data., , Combined Mean, , If two groups containing m and 1 items are considered, and their, means are x; and x2 then the mean of the combined group of m + m2, items taken together can be calculated. It is given by, , en mxX1+mX2, v= at Ma, Ex. 8: :, Arithmetic mean of heights of 100 boys is 150 cm. and that of 50 girls, is 144 cm. Find combined arithmetic mean of 150 items., Solution:, With standard notations given information can be written as, , , , , , , , , , , , , , Group of Boys | Group of Girls, Number of observations m1 = 100 n2= 50, Arithmetic mean of Height _| ¥: = 150 cms X2 = 144 cms, , , , , , Combined A.M. of heights is given by:, , — mx tnx., Le et, , 1+ 12, : _ 100 x 150 + 50 x 144, , a 100 + 50, , _ 15,000 + 7,200, , Ls 150, , a 22,200, , ee ey), , x =148 cms., , Thus arithmetic mean of 150 observations is 148 cms., , Ex, 9:, Arithmetic mean of daily wages of men is Rs. 120 and that of women, , s. 100. If arithmetic mean of daily wages of men and women taken, s. 112, find the ratio ot ni!, fe a