The high temperature population variance is less than the sample variance. Calculating the high temperature population variance would take quite a bit of time – so look for a shortcut! For the population of high temperatures, we are given the mode and the CV. Here, there are two tricks: remember that the CV = standard deviation / mean and that for a normal distribution, the mean = median = mode. We can manipulate the CV equation as standard deviation = mean × CV, or 13 × 0.165 = 2.145. Squaring this result gives a variance of 2.145²  = 4.60. Thus, the high temperature population variance is less than the sample variance.

Both remaining statements are true. The strategy here is to work from the “easiest” to the most difficult calculations. (Note: all units are ^oC unless stated otherwise.)

• The question tells us that the low temperatures are positively skewed with a mean of 6.9. For a positively skewed distribution, we know that the mean > median > mode. Thus, the mode is less than the mean, or 6.9.
• The population with the lowest CV has the least dispersion. Thus, the population of high temperatures (CV of 0.165) is less dispersed than in the population of low temperatures (CV of 0.328).
• Again, calculating the low temperature population standard deviation would take quite a bit of time – so look for a shortcut! For the population of high temperatures, we are given the mean and the CV. Here, there is only one trick (we are given the mean): manipulate the CV equation as standard deviation = mean × CV, or 6.9 × 0.328 = 2.26. Unfortunately, there is no real shortcut for the low temperature sample standard deviation, which is calculated as follows:
  • Sample mean is given at 7.2.
  • Variance = [(3 – 7.2)² + (6 – 7.2)² + (10 – 7.2)² + (9 – 7.2)² + (8 – 7.2)²] / (5 – 1) = 7.7
  • Standard Deviation = 7.7^1/2 = 2.78.

Thus, the population standard deviation of 2.26 is less than the sample standard deviation of 2.78.