Solved – Why dose the assumed mean and the step deviation methods work while calculating the mean of grouped data

In the assumed mean method of calculating the mean of grouped data, you can first assume a mean, $ a$ and then calculate the deviation, $ d_i $ of the class marks $ x_i $ and use the formula $$ bar x = a + frac {sum f_i d_i}{sum f_i} $$

to get the mean. In the step-deviation method, you do something similar, but you also scale down the values by a factor h, so for example, you get a variable $$ u_i = frac {x_i – a} {h} $$

and then use the formula, $$ bar x = a + h bar u $$

My question is, why does this method work? I'm guessing that we're manipulating the numbers by using some property of the mean, but what are they and why exactly do these methods make sense?

According to the definition, sample mean $bar x$ should be $$ bar x = frac {sum f_i x_i}{sum f_i} $$ where $f_i$ is the frequency of $x_i$. Let $x_i=d_i + a$ (equivalent to $d_i = x_i – a$), we have

$ bar x = frac {sum f_i x_i}{sum f_i} = frac {sum f_i (d_i+a)}{sum f_i} = frac {sum f_i d_i}{sum f_i} +frac {sum f_i a}{sum f_i} = frac {sum f_i d_i}{sum f_i} +a = a+bar d$

Similarly

$ u_i = frac {x_i – a} {h} $ is equivalent to $x_i = hu_i + a$.

So

$ bar x = frac {sum f_i x_i}{sum f_i} = frac {sum f_i (hu_i + a)}{sum f_i} = frac {sum f_i hu_i}{sum f_i} +frac {sum f_i a}{sum f_i} = hfrac {sum f_i u_i}{sum f_i} +a = a + h bar u $

Similar Posts:

Rate this post

Leave a Comment