A group of 894 women aged 70-79 had their height and weight measured. The mean height was 159 cm with a standard deviation of 5 cm and the mean weight was 65.9kg with a standard deviation of 12.7kg. Both sets of data are fairly normal?

The key information is the mean, the standard deviation (variation about the mean) and that the data is "normal".
A. You can pick any range about the mean that you like. The number of standard deviations from the mean reflects the percentage of the population expected to be in that range. #3sigma# includes 99.5% of the population, so a range of 145 - 174 cm for height and 27.8 - 104kg for weight can be used.

B. "Variability" of the data is directly measure by the standard deviation. The higher standard deviation value indicates more variability in the weight measurements.

C. Go back to the "Standard Normal Distribution Curve" (or use values from the Z-tables) to find the population value corresponding to the #sigma# value. 166cm is#(166 - 159)/5 = 1.4sigma#. The area to the right on the curve (one-tailed) is the percentage of the population that would be expected to exceed that value.

D. Similar to C. - Find the "area" corresponding to the #sigma# range. In this case it is two-tailed (both sides of the mean). It is almost even about the mean (65 vs 65.9) so you cane either calculate a more precise percentage on each side, or just that the approximation of the two-tailed percentage at the corresponding #sigma# value. #sigma = (75 - 55)/12.7 = 1.57#

E. This is a "left-tailed" test that could also be stated as "at what weight is 15% of the population below the level?" Again, use the Z-value tables to find 0.85 and the corresponding #sigma# value. Multiply that value by the known #sigma# for this sample and subtract it from the mean.