Stat 3011 Midterm 1 (Computer Part)
Rweb:> pnorm(0, mean=25, sd=10) [1] 0.006209665
Rweb:> pnorm(30, mean=25, sd=10) - pnorm(25, mean=25, sd=10) [1] 0.1914625
Rweb:> qnorm(0.90, mean=25, sd=10) [1] 37.81552
Rweb:> 1 - pnorm(6.5, mean = 5.31, sd = 0.58) [1] 0.02009824
Rweb:> pnorm(6.5, mean = 5.31, sd = 0.58) [1] 0.9799018
Rweb:> 1 - pnorm(6.5, mean = 11.74, sd = 3.50) [1] 0.932822
Rweb:> pnorm(6.5, mean = 11.74, sd = 3.50) [1] 0.06717805
You want either a histogram or a stem and leaf plot. They show pretty much the same thing. The histogram is just prettier. The R default stem and leaf plot is
Rweb:> "your name here" [1] "your name here" Rweb:> stem(sally) The decimal point is at the | -0 | 5 0 | 3 1 | 16789 2 | 111224444566789999 3 | 000223333344555555566777789999 4 | 0000111122244445555666667778888999 5 | 000001111112222333333333444555667777788888899 6 | 000111112222333445566778999 7 | 0001333333444444555667789 8 | 000234579 9 | 059 10 | 11237 11 | 057799 12 | 01456679999 13 | 13444566679999 14 | 00111112334444556666889999 15 | 003466889 16 | 1223344678899 17 | 01245677789 18 | 00269 19 | 20 | 3 21 | 22 | 6
Skewed. Long right tail (positive skewness). Bimodal. Maybe the 20.3 and 22.6 are outliers, but this is better described as just the long right tail.
Rweb:> mean(sally) [1] 8.396367
Rweb:> median(sally) [1] 6.39
Rweb:> sd(sally) [1] 5.001011
Rweb:> IQR(sally) [1] 8.8875
The mean goes with the standard deviation like ham and eggs. Similarly the median goes with the IQR.
Thus one sensible answer is mean (estimate of center) and standard deviation (estimate of spread). And another sensible answer is median (estimate of center) and IQR (estimate of spread). An argument can be made for either of these. No argument can be made for any other answer.
For this particular problem, neither is really satisfactory. A bimodal distribution really has two ``centers'' so no single measure of ``center'' can be satisfactory.