Adding One Observation: Sum of Squares

Given:

Number of observations,

$$n=10$$

Mean,

$$\bar x=15$$

Variance,

$$\sigma^2=16$$

Using the formula

$$\sigma^2=\frac{\sum x_i^2}{n}-\bar x^2,$$

we get

$$16=\frac{\sum x_i^2}{10}-15^2$$

$$16=\frac{\sum x_i^2}{10}-225$$

$$\frac{\sum x_i^2}{10}=241$$

$$\sum x_i^2=2410$$

Also,

$$\sum x_i=10\times15=150$$

Now after adding the new observation $$k,$$ the mean becomes $$16.$$

Hence,

$$\frac{150+k}{11}=16$$

$$150+k=176$$

$$k=26$$

Now the new sum of squares becomes

$$2410+26^2$$

$$=2410+676$$

$$=3086$$

Variance of the $$11$$ observations:

$$\sigma^2=\frac{3086}{11}-16^2$$

$$=\frac{3086}{11}-256$$

$$=\frac{3086-2816}{11}$$

$$=\frac{270}{11}$$

$$\approx24.54$$

Closest integeris

$$\boxed{25}$$

Given mean $$=4$$ for $$10$$ observations,

$$\sum x_i=40$$

Given variance

$$=9,$$

$$9=\frac{\sum x_i^2}{10}-16$$

$$\sum x_i^2=250$$

After adding $$4,$$

new sum becomes $$44,$$

hence new mean remains $$4$$

Now,

$$x_1^2+x_2^2=50$$ is replaced by $$3^2+5^2=34$$

Also include

$$4^2=16$$

Hence new sum of squares:

$$250-50+34+16=250$$

Therefore,

$$\sigma^2=\frac{250}{11}-16=\frac{74}{11}$$

Thus,

$$11\sigma^2=74$$