---------- ---------- ---------- ---------- ---------- ---------- ---------- ---------- ## The heights of 19 sixteen year-old female students:data=c(150,155,156,157,157,157,159,160,162,162,163,163,164,165,165,166,167,168,170) stem(data)# The decimal point is 1 digit(s) to the right of the | # 15 | 0 # 15 | 567779 # 16 | 022334 # 16 | 55678 # 17 | 0length(data); Median(data); median(data) # 19 162 162## The quantity is odd so the two ways of calculating the median coincide ## We suppose the values are rounded to the nearest integer. So:interv = seq(150,171,1)-1/2 # 149.5, 150.5, ... ,170.5 freq = c(1,0,0,0,0,1,1,3,0,1,1,0,2,2,1,2,1,1,1,0,1) noClass=1; histoclas(interv,freq)# The mean (brown dot) is about 161.3681 # For percentages use PERC For other statistics use the morestat()morestat() # Min. 1st Qu. Median Mean 3rd Qu. Max. # 149.5 157.1 162.2 161.4 165.1 170.5 # The brown dots are 5^ and 95^ percentiles # The red dot is the mean ## To have a larger amount of digits (of median, that is the 50^ percentile): percentile(50) # 162.2498 ## I can take162.25as an estimate of the median ## In this case (where the data represents lengths) it makes sense to take more digits.## The approximations by default and those for excess partially compensate. That is why ## I can take the median with one or two more digits.Back