Internet speed (made-up data)Speed is how much data arrives per second. The unit Mbps (megabits per second) = how many Mb (megabits) arrive in 1 second.
time (s) = file size (Mb) ÷ speed (Mbps)
Five measurements in the same place gave 82.4 / 91.1 / 68.7 / 88.2 / 79.6 Mbps. It differs each time with how busy the network is, so we "level it out" with an average.
Today file sizes are in Mb too. MB (megabyte) is 1 MB = 8 Mb and appears only in question 7.
The commonest mistake"Average = halfway between the fastest and the slowest."
(91.1 + 68.7) ÷ 2 = 79.9, but the average of the five is 82.0. The two ends say nothing about where the three in between are. An average adds all of them and divides by how many.
1 · The average of five
1Find the average of the five measurements (Mbps).
2Find each of the five measurements' difference from the average (+ if faster, − if slower). What do the five differences add up to, and why?
3Throw away the slowest run (68.7) and find the average of the other four, rounded to one decimal place. Did the average go up or down, and why?
4A sixth run is made, and the six-run average should be exactly 83 Mbps. What speed must the sixth run be?
average × count = total.
2 · From speed to time taken
5A file is 205 Mb.
(1) At the average of 82.0 Mbps, how many seconds does it take?
(2) At the slowest run, 68.7 Mbps, how many seconds? Round to one decimal place.
What do you multiply by to make the divisor whole?
6A game update of 369 Mb arrived in exactly 4.5 seconds.
(1) How many Mb per second (how many Mbps) is that?
(2) At the same speed, how many Mb arrive in 1.8 seconds?
7A video file is 45 MB. Take 1 MB = 8 Mb.
(1) How many Mb is that?
(2) At 90 Mbps, how many seconds?
(3) A friend answered "45 ÷ 90 = 0.5 seconds". What is wrong?
Small b is a bit, capital B is a byte. A byte is 8 times bigger.
3 · Does the average match what it feels like?
8An advert says "up to 100 Mbps". The five-run average of 82.0 Mbps is how many times the advertised speed? What percentage is that?
how many times = compared amount ÷ base amount.
9One day, measuring once an hour, six runs gave 94.0, 92.5, 90.5, 4.5, 93.0 and 96.5 Mbps.
(1) What is the average of the six, in Mbps?
(2) What is the average of the five without the 4.5?
(3) Someone says "it always feels like 90 Mbps or more". Which of (1) and (2) matches that feeling? And what should be done about the 4.5 run?
10Home A's five-run average is 82.0 Mbps; home B's four-run average is 84.5 Mbps.
(1) Someone found the "average of the two averages", (82.0 + 84.5) ÷ 2 = 83.25. Why is that not right?
(2) Find the average of all nine runs, rounded to one decimal place.
Go back from each average to its total.