This document provides information to summarize a sample of ages of Canadian tourists flying from Toronto to Hong Kong. It includes: 1) Computing the mean age as 43.2 years by taking the average of the 10 ages. 2) Computing the range as the difference between the highest (72 years) and lowest (17 years) ages, which is 55 years. 3) Computing the standard deviation as 17.62 years by calculating the variance from the mean and taking the square root. 4) Computing the interquartile range as 39.5 years by finding the middle 50% of the data. 5) Stating there are no outliers in the data set.

1. The ages of a sample of Canadian tourists flying from Toronto to Hong Kong were: 32, 21, 60,
47, 54, 17, 72, 55, 33 and 41. a)Compute the mean. b)Compute the range. c)Compute the
standard deviation. d)Compute the interquartile range. e)Are they any outliers?
Solution
a. The range is simply the largest value minus the smallest value of a
sample. In this case:
Range = 72-17 = 55
b. In order to calculate the mean deviation and the standard
deviation, we must find the mean of this sample. This is:
32+21+60+47+54+17+72+55+33+41
----------------------------- = 43.2
Therefore, using the formula in the link, the mean deviation is:
|32-43.2| + |21-43.2| + |60-43.2| + ... + |41-43.2|
---------------------------------------------------- =
11.2 + 22.2 + 16.8 + ... + 2.2
---------------------------------------------------- = 14.4
So the mean deviation is 14.4
c. Finally, the sample variance is:
(32.2-43.2)^2 + (21-43.2)^2 + ... + (41-43.2)^2
----------------------------------------------- = 310.62
So the sample standard deviation is the square root of this number, that is, 17.62.
d.39.5
e. No outliers