Question

A large number of children were surveyed. The table below shows the distribution of the time they spent watching TV in the week before the survey. Intervals include the left endpoint but not the right. You can assume that the times are uniformly distributed within each interval. The table is used to draw a histogram of the distribution.

time (hours) percent
0-4 28
4-8 36
8-12 12
12-20 12
20-28 12

Estimate the 80th percentile of the distribution of times, in hours; give the best estimate you can, based on the information in the prb

Answer 1

i think the answer is 9.77 but em not sure

Answer 2

if memory serves, we can take the midpoint of a class and multiply it by the percentage to help out

Answer 3

0-4 28(2)
4-8 36(6)
8-12 12(10)
12-20 12(16)
20-28 12(24)

with any luck this gives us 100 number positions

the average of the 80th and 81st should be the 80th percentile

28
36
----
64
12
---
76
12
---
88 .... so between the 76th and 88th positions we have alot of 16

im going to take a gander and say its 16

Answer 4

how you found 16 i cant understand?

Answer 5

my idea, if im correct, is:

there are 28 positions that average 2. (0+4)/2 = 2

there are 36 positions that average 6. (4+8)/2 = 6

etc....

out of 100 possible positions, the 80th percentile would be the average of the 80th and 81st postiions, if the values are placed in order from lowest to highest.

It just so happens, in my mind, that the average of 16 and 16 is ... 16

Answer 6

ok

Answer 7

thats only my idea, i got no idea how correct it is :)