anonymous
  • anonymous
Stath Question..
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
@ⒶArchie☁✪
campbell_st
  • campbell_st
(a) is asking for the most frequent.... or the mode (b) is looking at the day between the quartiles.... if quartiles divide a data set into 4 equal parts what percentage of the data lies between Q1 and Q3? Once you know that you can use that information to find the number .... %*150 = (c) requires you so show the values 3 standard deviations above the mean and 3 below. You need to know the percentage for a normal distribution that these limits represent.

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anonymous
  • anonymous
a) that's what I thought b)what percentage of the data lies between Q1 and Q3= 50% right
campbell_st
  • campbell_st
thats correct... so find 50% of the number of scores.... which is 150
anonymous
  • anonymous
75
anonymous
  • anonymous
we know that 99.7 % of data lie between 3sd
anonymous
  • anonymous
c) I got 149.55
campbell_st
  • campbell_st
well remember its \[mean \pm \sigma\] which is \[95 \pm 3 \times 12\]
anonymous
  • anonymous
yeah I got that 59------131
campbell_st
  • campbell_st
yep... thats it
campbell_st
  • campbell_st
but you need to find 150 * 99.7% =
anonymous
  • anonymous
I am little bit confused, are those numbers ( 95, 97, 99 etc) days or number of paying customers
anonymous
  • anonymous
150 * 99.7% =149.55
campbell_st
  • campbell_st
so you need to say that 3sd of the mean gives the range 59 - 131... which is 99.7% of all scores.. then 99.7% of 150 = .... to support your answer
anonymous
  • anonymous
should I round it..
campbell_st
  • campbell_st
well I would
anonymous
  • anonymous
are those numbers ( 95, 97, 99 etc) days or number of paying customers?
campbell_st
  • campbell_st
but include the decimal in your calculations... then your rounded... answer
anonymous
  • anonymous
got it..
campbell_st
  • campbell_st
they are the number of paying customers.... during the noon hour... over 150 days... so the set has 150 scores.... of the number of people shopping ...during a set time
anonymous
  • anonymous
so, you mean data could possibly be say, 1st day 20 people pay 2nd day 87, 3->99, 4->150, ......150->3. but mode is 99, most of the these 150 days, 99 people pay.. average # people is 95.. right..

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