1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Calculate mean and standard deviation of the following data by short-cut method: Class Interval 10−20 20−30 30−40 40−50 50−60 60−70 70−80 Frequency 5 10 15 20 25 18 7

Open in App
Solution

## C.I. Mid Value (m) Frequency (f) Deviation from Assumed mean dx = m − A (A = 45) fdx $f×{{d}_{x}}^{2}=f{{d}_{x}}^{2}$ 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 15 25 35 $\overline{)45=\mathrm{A}}$ 55 65 75 5 10 15 20 25 18 7 −30 −20 −10 0 10 20 30 −150 −200 −150 0 250 360 210 4500 4000 1500 0 2500 7200 6300 Σf = N = 100 Σfdx = 320 Σfdx2 = 26000 $\mathrm{Calculating}\mathrm{Mean}\mathrm{and}\mathrm{standard}\mathrm{deviation}\mathrm{through}\mathrm{Short}-\mathrm{cut}\mathrm{method}:\phantom{\rule{0ex}{0ex}}\mathrm{Mean}\left(\overline{)\mathrm{X}}\right)=\mathrm{A}+\frac{\mathrm{\Sigma }f{d}_{x}}{\mathrm{\Sigma }f}\phantom{\rule{0ex}{0ex}}\mathrm{or},\overline{)\mathrm{X}}=45+\frac{320}{100}\phantom{\rule{0ex}{0ex}}\mathrm{or},\overline{)\mathrm{X}}=45+3.2\phantom{\rule{0ex}{0ex}}⇒\overline{)\mathrm{X}}=48.2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}S\mathrm{tan}darddeviation\left(\mathrm{\sigma }\right)=\sqrt{\frac{\mathrm{\Sigma }f{{d}_{x}}^{2}}{N}-{\left(\frac{\mathrm{\Sigma }f{d}_{x}}{N}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{\sigma }=\sqrt{\frac{26000}{100}-{\left(\frac{320}{100}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{\sigma }=\sqrt{260-{\left(3.2\right)}^{2}}\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{\sigma }=\sqrt{260-10.24}\phantom{\rule{0ex}{0ex}}\mathrm{or},\mathrm{\sigma }=\sqrt{249.76}\phantom{\rule{0ex}{0ex}}⇒\mathrm{\sigma }=15.80$

Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
Mean of Ungrouped Data
STATISTICS
Watch in App
Explore more
Join BYJU'S Learning Program