Question

Consider the following data and corresponding weights.

Answer 1

Answer:

a) tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]

And replacing we got:

$\bar X= \frac{3.2+2.0+2.5+5.0}{4}=3.175$

b) $\bar X = \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i}$

And replacing we got:

$\bar X= \frac{3.2*6+2.0*3+2.5*2+5.0*8}{6+3+2+8}=\frac{70.2}{19}=3.695$

Step-by-step explanation:

Part a

For this case we can calculate the the unweigthed with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X= \frac{3.2+2.0+2.5+5.0}{4}=3.175$

Part b

For this case the weigthed mena is given :

$\bar X = \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i}$

And replacing we got:

$\bar X= \frac{3.2*6+2.0*3+2.5*2+5.0*8}{6+3+2+8}=\frac{70.2}{19}=3.695$