0
$\begingroup$

Out of curiosity... there was a study where 93% of a sample group (of 161) believed they were above average drivers. What would be the chance that they actually WERE above average drivers?

More: http://en.wikipedia.org/wiki/Illusory_superiority#Driving_ability

$\endgroup$
  • $\begingroup$ Define average. $\endgroup$ – Emily Apr 6 '15 at 15:54
  • $\begingroup$ How do you quantify driving ability to average it? $\endgroup$ – Mark Bennet Apr 6 '15 at 15:54
  • 3
    $\begingroup$ It is possible for $90\%$ of a sample to be above average, if the distribution is highly skewed to the left and the average is taken to be the mean. For example, if nine people scored $100$ on a test and one scored $0$, the mean is $90$ and nine people are above the mean. If the average is taken to be the median, however, it is not possible for more than $50\%$ of a sample to be above the median. $\endgroup$ – Rahul Apr 6 '15 at 16:04
  • 1
    $\begingroup$ If you are using the mean as your average, over 93% of all people have more than the average number of arms. $\endgroup$ – Ross Millikan Apr 6 '15 at 16:04
  • 1
    $\begingroup$ @Rahul While the precise probability will depend on which average is being used, it's possible to obtain a sample with 100% being above average (regardless of which average is used) because the average being used is from the population, not the sample. As long as the sample size is no more than 50% of the population size, this will hold even if we're using the median. $\endgroup$ – rnrstopstraffic Apr 6 '15 at 16:52
0
$\begingroup$

I think this is a questions to get you to think, not calculate. Believing you are an above-average driver and actually being above average are very different things. (Some of the scariest drivers I know would probably claim to be above average in such a survey.)

If you have a scale for judging actual driving ability or performance, and if that scale is normal (or otherwise symmetrical with a mean that exists), then the mean equals the median, and exactly half of the drivers are above the mean on that scale.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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