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(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.
a) that's what I thought b)what percentage of the data lies between Q1 and Q3= 50% right
thats correct... so find 50% of the number of scores.... which is 150
we know that 99.7 % of data lie between 3sd
c) I got 149.55
well remember its \[mean \pm \sigma\] which is \[95 \pm 3 \times 12\]
yeah I got that 59------131
yep... thats it
but you need to find 150 * 99.7% =
I am little bit confused, are those numbers ( 95, 97, 99 etc) days or number of paying customers
150 * 99.7% =149.55
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
should I round it..
well I would
are those numbers ( 95, 97, 99 etc) days or number of paying customers?
but include the decimal in your calculations... then your rounded... answer
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
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..