MATH1725 Introduction to Statistics: Worked examples

¯x = −4
100 = −0.04′ .
s 2 = 1 99 (292 − 100 × (−0.04)) = 2.9479, so s = √ (s 2 ) = √ 2.9479 = 1.717 ′ .
Worked Example: Lectures 1–2
The time between arrival of 60 patients at an intensive care unit were recorded to the nearest hour.
The data are shown below.
Time (hours) 0–19 20–39 40–59 60–79 80–99 100–119 120–139 140–159 160–179
Frequency 16 13 17 4 4 3 1 1 1
Determine the median and semi-interquartile range. Explain why this pair of statistics might be
preferred to the mean and standard deviation for these data.
Answer:
Time (hours) 0.0 19.5 39.5 59.5 79.5 99.5 119.5 139.5 159.5 179.5
Cumulative frequency 0 16 29 46 50 54 57 58 59 60
Median lies in “40–59” class, corresponding to cumulative frequency 30.
Lower quartile is in “0–19” class, corresponding to cumulative frequency 15. Notice that this
class has width 19.5 hours, not 20 hours.
Upper quartile is in “40–59” class, corresponding to cumulative frequency 45.
Median = 39.5 +
30 − 29
× 20 = 40.7 hours.
46 − 29
Lower quartile = 0.0 + 15 − 0 × 19.5 = 18.3 hours.
16 − 0
45 − 29
Upper quartile = 39.5 + × 20 = 58.3 hours.
46 − 29
Semi-interquartile range = 1 (58.3 − 18.3) = 20.0 hours.
2
The histogram for these data is positively skew, so the median and semi-interquartile range might
be preferred to the mean and standard deviation as measures of location and dispersion respectively.
Freq. per 20 hour class
0 5 10 15 20
0 50 100 150 200
Inter−arrival time (hours)
3