# The mean and S.D. of a group of 48 observations are 40 and 8 respectively. If two more observations 60 and 65 are added to the set, find the mean and S.D. of 50 items. - Mathematics and Statistics

Sum

The mean and S.D. of a group of 48 observations are 40 and 8 respectively. If two more observations 60 and 65 are added to the set, find the mean and S.D. of 50 items.

#### Solution

n = 48, bar"x" = 40,  sigma_"x" = 8 ........[Given]

bar"x" = 1/"n" sum"x"_"i"

∴ sum"x"_"i" = "n"bar"x" = 48 × 40 = 1920

New  sum"x"_"i" = sum"x"_"i" + 60 + 65
= 1920 + 60 + 65
= 2045
∴ New mean = 2045/50 = 40.9
Now, sigma_"x" = 8
∴ sigma_"x"^2 = 64
Since, sigma_"x"^2 = 1/"n"(sum"x"_"i"^2) - (bar"x")^2

∴ 64 = 1/48(sum"x"_"i"^2) - (40)^2

∴ 64 = 1/48(sum"x"_"i"^2) - 1600

∴ (sum"x"_"i"^2)/48 = 64 + 1600 = 1664

∴ sum"x"_"i"^2 = 48 × 1664 = 79872

New sum"x"_"i"^2 = sum"x"_"i"^2 + (60)^2  (65)^2

= 79872 + 3600 + 4225
= 87697
∴ New S.D. = sqrt(("New" sum"x"_"i"^2)/"n" - ("New mean")^2

= sqrt(87697/50 - (40.9)^2

= sqrt(1753.94 - 1672.81)

= sqrt(81.13)

Concept: Standard Deviation for Combined Data
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board
Chapter 2 Measures of Dispersion
Miscellaneous Exercise 2 | Q 12 | Page 35