Question

Table #3.2.17 contains data collected on the time it takes in seconds of each passage of play in a game of rugby. ("Time of passages," 2013) Table #3.2.17: Times (in seconds) of rugby plays 39.2 2.7 9.2 14.6 1.9 17.8 15.5 53.8 17.5 27.5 4.8 8.6 22.1 29.8 10.4 9.8 27.7 32.7 32 34.3 29.1 6.5 2.8 10.8 9.2 12.9 7.1 23.8 7.6 36.4 35.6 28.4 37.2 16.8 21.2 14.7 44.5 24.7 36.2 20.9 19.9 24.4 7.9 2.8 2.7 3.9 14.1 28.4 45.5 38 18.5 8.3 56.2 10.2 5.5 2.5 46.8 23.1 9.2 10.3 10.2 22 28.5 24 17.3 12.7 15.5 4 5.6 3.8 21.6 49.3 52.4 50.1 30.5 37.2 15 38.7 3.1 11 10 5 48.8 3.6 12.6 9.9 58.6 37.9 19.4 29.2 12.3 39.2 22.2 39.7 6.4 2.5 34 a.) Using technology, find the mean and standard deviation. b.) Use Chebyshev’s theorem to find an interval centered about the mean times of each passage of play in the game of rugby in which you would expect at least 75% of the times to fall. c.) Use Chebyshev’s theorem to find an interval centered about the mean times of each passage of play in the game of rugby in which you would expect at least 88.9% of the times to fall.

Answer 1

from above:

a) mean=21.240

std deviation =14.946 ~14.95

b)

for 75% of data falls within 2 std deviation from Chebyshev’s theorem

an interval centered about the mean times of each passage of play in the game of rugby in which you would expect at least 75% of the times to fall =21.24-/+2*14.95 =-8.66 to 51.14

c)

for 88.9% of data falls within 3 std deviation from Chebyshev’s theorem

an interval centered about the mean times of each passage of play in the game of rugby in which you would expect at least 88.9% of the times to fall =21.24-/+3*14.95 =-23.61 to 66.09