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The average time spent on Facebook per day of 40 students in a Statistics class is 1.4 hours, while the average of the 32 students in the Calculus class across the hall is 1.9 hours. Two students from the Calculus class, who both do not use Facebook, decide to drop the course and enter into the Statistics class. What are the new daily average time spent on Facebook of the Statistics and Calculus classes, respectively?

The new averages were 1.3 and 2.0 I just can't figure out how they got those numbers

S.Elle
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  • Please show what attempts you have made on the problem and what specific issues you are having trouble with. Stack Overflow does not exist to do your homework for you. – Paul Childs Oct 02 '18 at 22:31
  • Hint: The statistics class spends $40\cdot 1.4$ hours per day in total whereas the calculus class spends $32\cdot 1.9$ hours per day. Those two figures do not change, but the number of students to average over change. – String Oct 02 '18 at 22:33
  • I'm trying to figure out if I need to standardise but there is no standard deviation. not my homework just a question I couldnt answer and was curious how to solve it – S.Elle Oct 02 '18 at 22:34
  • Also note that the word average is a bit confusing here, since it can be assumed that we are dealing with two types of averages, namely averaging over the number of students, but on the other hand averaging over the days in period of collecting the data. – String Oct 02 '18 at 22:35
  • @String the answers that came with this question showed that the new averages were 1.3 and 1.9 I just can't figure out how to get those numbers... – S.Elle Oct 02 '18 at 22:39
  • I can confirm the first figure, whereas I have the last as 2.0 hours. Did you try my hint yet? – String Oct 02 '18 at 22:43
  • @String oh sorry yes the second one is 2.0. how did you get that figure? just 40x1.4? – S.Elle Oct 02 '18 at 22:47
  • Not quite. You have 40 students spending 1.4 hours each (on average). How many hours do those 40 students spend collectively? Now average this total over $40+2$ students, since the two newcomers add no hours to the pool, only students. – String Oct 02 '18 at 22:51
  • @string would this way be correct? 40x1.4=56, 56/42=1.33 – S.Elle Oct 02 '18 at 22:53
  • Also note that both 1.3 and 2.0 are rounded off. – String Oct 02 '18 at 22:53
  • Yes, that is correct! – String Oct 02 '18 at 22:53
  • @string thank you!! – S.Elle Oct 02 '18 at 22:54
  • No problem. Note that Paul Child's comment is correct in principle, namely you must show as much of an effort as you can explaining exactly at what step you are stuck. People in here respond negatively otherwise. Best of luck! – String Oct 02 '18 at 23:01

1 Answers1

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$40$ Stats students total time $40\cdot 1.4 = 56$ hrs

New average $= \frac{56}{42} = 1.3333$ hrs

$32$ Calc students total time $32\cdot 1.9 = 60.8$ hrs

New average $= \frac{60.8}{30} = 2.0267$ hrs

Phil H
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