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I don't know how to resolve this (easy) exercise. I've calculated the first output. But I don't know if it's correct.

Calculate arithmetic mean, variance (standard deviation^2), concentration and asymmetry

INPUT:

Class           Relative Frequency
10-25           0.48
25-40           0.25
40-60           0.15
60-100          0.1
100-200         0.02

I started finding the middle value of the classes:

Class           Middle value
10-25           17.5
25-40           32.5
40-60           50
60-100          80
100-200         150

The arithmetic mean should be Σ(middle values * relative frequency)

Middle value    Relative Frequency      Weighted value
17.5            0.48                    8.4
32.5            0.25                    8.125
50              0.15                    7.5
80              0.1                     8
150             0.02                    3
                                        SUM = 35.025

Arithmetic mean should be 35.025. Or 35.025 / n (which is 7.005)? I don't know if / n is necessary since the values are already weighted. And the variance? I found a formula which is Σ(middle value^2*relative frequency)-mean^2. It outputs 649.311. Is that correct? Same for asymmetry.. formula is Σ(middle value^3*relative frequency)-mean^3. Is that correct?

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    $\begingroup$ Beware calling that last thing asymmetry -- moment skewness doesn't correspond to necessarily measure asymmetry; e.g. it can be zero while being asymmetric. $\endgroup$ – Glen_b Sep 22 '14 at 14:38
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    $\begingroup$ I read that to be more accurate I should divide "my last thing" (:D) by standard deviation^3. Should output the beta of fisher $\endgroup$ – MultiformeIngegno Sep 22 '14 at 14:40
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    $\begingroup$ "More accurate" doesn't make sense to me in this context; they're not comparable. "More interpretable" perhaps. Yes, if you standardize it you get the actual moment-skewness coefficient - which you attribute to Fisher, and wikipedia attributes to Pearson. Consistent with Stigler's law, credit doesn't belong to either, of course. $\endgroup$ – Glen_b Sep 22 '14 at 14:50
  • $\begingroup$ Thanks for the explanation. And I appreciate your rigour! $\endgroup$ – MultiformeIngegno Sep 22 '14 at 14:53
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Your calculations will provide only estimates, unless observations within classes are distributed uniformly.

You're right in arithmetic mean, as relative frequencies are used, you shouldn't divide by n. It is done when absolute frequencies instead of relative ones are used.

Others are also correct. Variance can also be calculated in the following way, which is more intuitive: $$ \Sigma_{i=1}^k ((middle value_i-\bar x)^2*relativefrequency_i) $$ where $k$ is number of classes.

Showing $\Sigma_{i=1}^k ((mv_i-\bar x)^2*rf_i) = (\Sigma_{i=1}^k mv_i^2*rf_i)-\bar x^2$, $$ \Sigma_{i=1}^k ((mv_i-\bar x)^2*rf_i)=\Sigma_{i=1}^k ((mv_i^2-2*mv_i*\bar x+\bar x^2)*rf_i) $$ $$ =(\Sigma_{i=1}^k mv_i^2*rf_i)-(\Sigma_{i=1}^k 2*mv_i*\bar x *rf_i)+(\Sigma_{i=1}^k \bar x^2*rf_i )$$ now inspecting 2nd term, since you can take constant multipliers out of summation: $$ \Sigma_{i=1}^k 2*mv_i*\bar x *rf_i =2*\bar x *\Sigma_{i=1}^k mv_i*rf_i$$ and recalling $\Sigma_{i=1}^k mv_i*rf_i$ is the formula for mean ($\bar x$): $$\Sigma_{i=1}^k 2*mv_i*\bar x *rf_i = 2*\bar x^2$$ inspecting 3rd term, similarly: $$\Sigma_{i=1}^k \bar x^2*rf_i =\bar x^2* \Sigma_{i=1}^k rf_i$$ and recalling $\Sigma_{i=1}^k rf_i=1$ since adding up relative frequencies of all classes should equal 1 by definition of relative frequency: $$\Sigma_{i=1}^k \bar x^2*rf_i =\bar x^2$$ plugging in relations for 2nd and 3rd term in the original expression: $$(\Sigma_{i=1}^k mv_i^2*rf_i)-2*\bar x^2+\bar x^2=(\Sigma_{i=1}^k mv_i^2*rf_i)-\bar x^2$$

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  • $\begingroup$ Thanks. :) How are the 2 formulas for the variance equivalent? What's the relation between them? $\endgroup$ – MultiformeIngegno Sep 22 '14 at 14:23
  • $\begingroup$ There is a relatively simple proof, I'm adding it to my answer. $\endgroup$ – Macond Sep 22 '14 at 14:31
  • $\begingroup$ Last question: Would it change something if the text says: This is the frequency distribution of 100 families"? (same distribution) $\endgroup$ – MultiformeIngegno Sep 22 '14 at 15:00
  • $\begingroup$ I mean, still providing relative frequencies but specifying the total n of people. $\endgroup$ – MultiformeIngegno Sep 22 '14 at 15:11
  • $\begingroup$ No, it won't change anything. You could multiply each rf with 100 and use absolute frequencies, then divide by n when necessary, it would lead to same results. You could also easily transform classical formula for mean and variance which use abs. frequencies to those you used in this case, just by simple algebra. $\endgroup$ – Macond Sep 22 '14 at 15:14

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