I have an assignment where I'm asked to find the the average shoe number for men and women from given data:

Sum of men and women: 312

Amount of men: 239

Standard deviation of men: 1.63

Mean of shoe number for men: 43.5

Amount of women: 239

Standard deviation of women: 1.4

Mean of shoe number for women: 38.2

Variance squared: 2321

Sum of all shoe numbers: 13183

But I have no idea how to find that using this data. Please help. Thanks in advance!

  • $\begingroup$ ?! Mean of men: 43.5, Mean of women: 38.2 ?! $\endgroup$ Feb 6, 2019 at 13:55
  • $\begingroup$ It is the mean of the shoe number for men and women $\endgroup$ Feb 6, 2019 at 13:56
  • $\begingroup$ Isn't that what you want, the average shoe number? $\endgroup$ Feb 6, 2019 at 13:56
  • $\begingroup$ Yes, I want the average shoe number specifically for men and specifically for women. $\endgroup$ Feb 6, 2019 at 13:58
  • 1
    $\begingroup$ @AlanRostem average and mean are the same thing. $\endgroup$ Feb 6, 2019 at 14:08

1 Answer 1


When you want to know the mean and variance of a population given you have these values for different groups of this population, you need to compute what is called pooled mean and variance.

Given that you already have mean and variance for both groups, I have to assume your question is really for these pooled values, for which the formulas are:

  • pooled mean: $\frac{\sum_i^n N_i \mu_i}{\sum_i^n N_i}$
  • pooled variance: $\frac{\sum_i^n (N_i-1) \sigma_i^2}{\sum_i^n (N_i-1)}$

$N_i$, $\sigma^2_i$ and $\mu_i$ stand for the number of observations, variance and mean of group $i$, respectively.

  • $\begingroup$ Sorry for the inconvenience. Due to a language barrier I had misunderstood the question which made it much harder for myself than it should have. Thanks for trying to help though I learned a lot today! $\endgroup$ Feb 6, 2019 at 14:23
  • $\begingroup$ @AlanRostem it is OK, did my answer help you? $\endgroup$ Feb 6, 2019 at 14:30
  • $\begingroup$ Yes. I used that formula to find the variance then the standard deviation. Thanks! $\endgroup$ Feb 6, 2019 at 14:40

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