0
$\begingroup$

Based on my understandings, I would say for the below histogram that the mode is zero, the mean is between 0 & 1 and the median is 1. I'm I right?

Histogram

$\endgroup$
1
  • 1
    $\begingroup$ It's not a histogram, it's a bar plot. Anyway, I would say all except mean are 0. $\endgroup$ Nov 4, 2021 at 7:10

1 Answer 1

1
$\begingroup$
  • As noticed in the comment, this is a bar plot. Histogram would pack the values into bins. For discrete data we don't use histograms in general, maybe unless there is a huge number of categories what would make the bar plot less unreadable.
  • Mode is the most frequent value, so the highest bar.
  • Mean would be the sum of $x$-axis values multiplied by their frequencies, i.e. $y$-axis, i.e. ${\sum_i x_i y_i} \Big/ {\sum_i y_i}$. You could calculate it from the plot with some degree of precision after reading the heights from $y$-axis.
  • Median is the value "in the middle" if you sorted the values. It would be 1 if after stacking the bars things on right would be slightly higher than things on left. The plots have $y$-axis on the logarithmic scale, so again, you would need to read the height of the bars to figure out if $y_0 \approx y_1 + y_2 + y_3$, but you can visually verify that this is not the case, so the median is also zero.
$\endgroup$
2
  • $\begingroup$ It would be helpful, I suspect, to indicate that the mean will be approximately $0.001.$ Your remarks about "measure the height of the bars" are puzzling, considering their values are clearly marked on the vertical axis (on a logarithmic scale). $\endgroup$
    – whuber
    Nov 4, 2021 at 11:55
  • 1
    $\begingroup$ @whuber I meant that you'd need to use a ruler to make more precise reading of the values. Nonetheless, I improved the wording. $\endgroup$
    – Tim
    Nov 4, 2021 at 12:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.