Question

Stath Question..

Answer 1

@ⒶArchie☁✪

Answer 2

(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.

Answer 3

a) that's what I thought
b)what percentage of the data lies between Q1 and Q3= 50% right

Answer 4

thats correct... so find 50% of the number of scores.... which is 150

Answer 5

75

Answer 6

we know that 99.7 % of data lie between 3sd

Answer 7

c) I got 149.55

Answer 8

well remember its

\[mean \pm \sigma\]

which is \[95 \pm 3 \times 12\]

Answer 9

yeah I got that

59------131

Answer 10

yep... thats it

Answer 11

but you need to find 150 * 99.7% =

Answer 12

I am little bit confused, are those numbers ( 95, 97, 99 etc) days or number of paying customers

Answer 13

150 * 99.7% =149.55

Answer 14

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

Answer 15

should I round it..

Answer 16

well I would

Answer 17

are those numbers ( 95, 97, 99 etc) days or number of paying customers?

Answer 18

but include the decimal in your calculations... then your rounded... answer

Answer 19

got it..

Answer 20

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

Answer 21

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..