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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.

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#### 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)`