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Question

# Calculate arithemtic eman from the foolowing data: Marks No. Students More than 0 150 More than 10 140 More than 20 100 More than 30 80 More than 40 80 More than 50 70 More than 60 30 More than 70 14

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Solution

## Converting more than cumulative frequency distribution into a simple frequency distribution Marks No. of students (c.f) More than 0 More than 10 More than 20 More than 30 More than 40 More than 50 More than 60 More than 70 150 140 100 80 80 70 30 14 Class Interval Mid-Values (m) No. of students (f) fm 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 5 15 25 35 45 55 65 75 150-140=10 140-100=40 100-80=20 80-80=0 80-70=10 70-30=40 30-14=16 14-0=14 50 600 500 0 450 2200 1040 1050 $\mathbit{\Sigma }\mathbit{f}$=150 $\mathbit{\Sigma }\mathbit{f}\mathbit{m}$=5890 Calculating the mean marks by using direct method: $\overline{)X}=\frac{\Sigma fm}{\Sigma f}\phantom{\rule{0ex}{0ex}}or,\overline{)X}=\frac{5890}{150}\phantom{\rule{0ex}{0ex}}⇒\overline{)X}=39.27marks$ Hence, the mean marks of the above series is 39.27

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