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Q.Find the mean deviation from the mean of the frequency distribution given below: Class: 0-100, 100-200, 200-300, 300-400, 400-500, 500-600, 600-700, 700-800; Frequency: 4, 8, 9, 10, 7, 5, 4, 3. OR The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, then find the other two observations.

Assam AhsecAHSEC Higher Secondary (HS) 1st Year Examination 2019Subjective· 6mImportance★★★★★
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Compute the mean of the distribution, then the mean of the absolute deviations from it. [OR alternative: use the mean and variance formulas to set up and solve a pair of equations for the two unknown observations.]

Main question. Classes (mid-points xix_i) and frequencies fif_i:

Classxix_ifif_ifixif_ix_i∣xi−xˉ∣\lvert x_i-\bar x\rvertfi∣xi−xˉ∣f_i\lvert x_i-\bar x\rvert
0–1005042003081232
100–200150812002081664
200–30025092250108972
300–400350103500880
400–5004507315092644
500–60055052750192960
600–700650426002921168
700–800750322503921176

N=∑fi=50N=\sum f_i = 50, ∑fixi=17900\sum f_ix_i = 17900.

Mean: xˉ=1790050=358\bar x = \dfrac{17900}{50}=358.

∑fi∣xi−xˉ∣=1232+1664+972+80+644+960+1168+1176=7896\sum f_i\lvert x_i-\bar x\rvert = 1232+1664+972+80+644+960+1168+1176 = 7896.

Mean deviation about the mean =∑fi∣xi−xˉ∣N=789650=157.92= \dfrac{\sum f_i\lvert x_i-\bar x\rvert}{N} = \dfrac{7896}{50}=157.92.


OR (alternative question): Mean of 5 observations is 4.44.4, variance is 8.248.24; three of them are 1,2,61,2,6. Find the other two.

Let the other two be a,ba,b. Sum of all 5 =5×4.4=22= 5\times4.4=22. Sum of known three =1+2+6=9=1+2+6=9, so: …

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