title: Mean of data with different propabilities
body: <p>I'm currently facing the following problem:
  During an experiment I gathered the counts of events per channel (detector). Plotting the counts against the channels gives almost a normal distribution. I'd now like to calculate the mean of this data set. The problem is that not all of the events that generated the data occur with the same probability, but the probability for each channel is known. </p>
  
  <p>To me this situation seems very similar to wanting to calculate the mean of a histogram, therefore I would take the middle value of the channels, multiply it by the corresponding value of the channel, sum all the values up and then divide by the total number of channels. </p>
  
  <p>My implementation for this is:</p>
  
  <pre><code>import numpy as np
  import matplotlib.pyplot as plt
  counts = ... # see at the end of the post for the data set in question
  channels = np.arange(1,len(counts)+1)
  channel_probability = .... # probability for different parts of channels
  
  mean = sum((channels+1)/2 * counts)/len(counts)
  
  plt.figure()
  plt.plot(counts, channels)
  plt.stem([mean], [100])
  plt.xlabel("channels")
  plt.ylabel("counts")
  plt.show()
  </code></pre>
  
  <p>The problem is that this assumes the same probability for all the events... Therefore I tried the naive approach of just multiplying the probability as well</p>
  
  <pre><code>mean = sum((channels+1)/2 * counts * channels_probability)/len(counts)
  </code></pre>
  
  <p>But this of course only led to completely unreasonable results... So, can someone maybe explain how I would find the mean of such a distribution and how to calculate it?</p>
  
  <hr>
  
  <p>As mentioned above, here is the something similar to the data set I'm using:</p>
  
  <pre><code>counts = np.array([2.05209753  2.07860064  2.06269877  2.0706497   2.07595033  2.03619567
    2.03619567  2.06269877  2.02029381  2.00439194  2.01499318  1.9937907
    1.98583977  1.99909132  1.99909132  2.00439194  1.98583977  1.98849008
    1.99644101  2.01499318  2.00439194  2.0176435   2.02824474  1.99909132
    2.00174163  2.03354536  2.05474784  2.05474784  2.04944722  2.11305467
    2.07330002  2.13955778  2.18461305  2.19256399  2.21906709  2.25617144
    2.23496895  2.25617144  2.31182796  2.32772982  2.36483417  2.3992882
    2.42844162  2.49734969  2.56890807  2.56095714  2.59541118  2.59541118
    2.63516583  2.68817204  2.6272149   2.66961987  2.6272149   2.66961987
    2.60336211  2.62191428  2.56890807  2.5503559   2.53975466  2.52385279
    2.45229441  2.42844162  2.39133727  2.29592609  2.27737392  2.26147206
    2.21906709  2.14220809  2.17666212  2.09185219  2.03619567  2.02824474
    2.05209753  2.00439194  1.97788884  1.97788884  1.9672876   1.96463729
    1.96993791  1.95403604  1.94608511  1.9434348   1.9434348   1.93548387
    1.93813418  1.9434348   1.94078449  1.93813418  1.94078449  1.9434348])
  </code></pre>