anonymous
  • anonymous
Score Number of Students 0 2 1 1 2 3 3 6 4 2 5 3 6 2 Based on the table, what is the mean hockey score?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[\frac{ total number }{number of scores ? }\]
anonymous
  • anonymous
Oops, does 2,1,3,6,2,3,2 in the number of student category?
anonymous
  • anonymous
belong in the #* ??

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anonymous
  • anonymous
anonymous
  • anonymous
total score=(number # of students with score 0)(0)+(number # of students with score 1)(1)+(# of students with score 2)(2)+(# of students with score 3)(3)+.....
anonymous
  • anonymous
total score/number of students for mean
anonymous
  • anonymous
mean=total score/number of students.
anonymous
  • anonymous
anonymous
  • anonymous
i got 2.7
anonymous
  • anonymous
let me see
anonymous
  • anonymous
Nope
anonymous
  • anonymous
What was the total score for yours?
anonymous
  • anonymous
um 19
anonymous
  • anonymous
No. I mentioned earlier you were supposed to multiply the amount of student by the scores. Mean means average. Total of all scores by including every student and scores/total number of student This is how you find the average score.
anonymous
  • anonymous
114
anonymous
  • anonymous
anonymous
  • anonymous
No. Just pay attention. 1st part: Total score Two students with score of 0 2x0=0 One student with score of 1 1x1=1 Three students with score of 2 3x2=6 Six students with score of 3 6x3=18 Two students with a score of 4 2x4=8 Three students with a score of 5 3x5=15 Two students with a score of 6 2x6=12 Add all the scores up 0+1+6+18+8+15+12=60 Total score is 60 Number of students 2+1+3+6++2+3+2=17 Mean=total scores/number of results or number of students Mean=60/17 =3.52 Understand now?
anonymous
  • anonymous
these are the choices though 2.7 2.9 3.2 5.2
anonymous
  • anonymous
anonymous
  • anonymous
oops 60/19
anonymous
  • anonymous
number of student is 19 lol. my mistake
anonymous
  • anonymous
then you round it.
anonymous
  • anonymous
so 3.2 ? @Shalante
anonymous
  • anonymous
anonymous
  • anonymous
yes.

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