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Q.Find mean and standard deviation (S.D.) of the following frequency distribution: Class-Interval 70-75, 75-80, 80-85, 85-90, 90-95, 95-100, 100-105 with Frequency 3, 4, 7, 6, 5, 3, 2 respectively.

Haryana BsehBoard of School Education Haryana (Senior Secondary Part-I / Class 11) 2019Subjective· 6mImportance★★★★★
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Use the step-deviation method with assumed mean A=87.5A=87.5 and class width h=5h=5: find ∑fidi\sum f_id_i and ∑fidi2\sum f_id_i^2, then apply the mean and variance formulas.

Class midpoints and deviations di=xi−Ahd_i=\dfrac{x_i-A}{h} with A=87.5A=87.5, h=5h=5:

ClassMidpt xix_ifif_idid_ifidif_id_ifidi2f_id_i^2
70-7572.53-3-927
75-8077.54-2-816
80-8582.57-1-77
85-9087.56000
90-9592.55155
95-10097.532612
100-105102.523618

n=∑fi=30n=\sum f_i = 30, ∑fidi=−7\sum f_id_i = -7, ∑fidi2=85\sum f_id_i^2 = 85

Mean:

xˉ=A+h⋅∑fidin=87.5+5×−730=87.5−1.16‾≈86.33\bar x = A+h\cdot\dfrac{\sum f_id_i}{n} = 87.5+5\times\dfrac{-7}{30} = 87.5-1.1\overline{6} \approx 86.33

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