5.1+Estimating+the+Size+of+a+Crowd

Using sample and ratios to estimate the size of a population, which this investigation incude the the size of crowd. I also use the idea of population density to to describe how crowded a space was with a crowd of people. * Essential Question: We can use the amount of something of a sample for the whole thing. So thing can be used like a amount how many peolpe is there in small equal porpotion compare to the whole for a estimation. Like theres 1 to 100 part amount people is 30. If its 2 to 100 it's 60, so it's clearly if it's 100 to 100 its 3000. > Strategies to solve the problem is by estimating like: > * Count a small part or a grid of a crowd and estimate by times it with the total amount of width and length > * Count down on length and width and multi it together. > > > > > > Prob 5.1
 * 16 Feb
 * M.A
 * Big Idea:
 * Notes from Class:

=**Sometimes the size of a crowd is estimated from aerial photographs. Imagine that the illustartion below is arial photograph of a crowd rally. Each dot represents on person. Estimate how many people attended the rally. Explain the moethod you use to arrive at your answer.**=

18 v

By counting the dots on one of the grid will get a small part of the whole crowd. On one of the grid is 18. By finding 18 tells me that there is 18 to the total. What I know on the grid that there is a 9 rows and 14 columns. 9 X 14 is 126. That shows me that I can estimate that there 126 of 18 people. Since it counted 1 grid out of 126. So when I count and estimating by calculating 18 x 126 is equal to 2286. My estimation is that they are 2286 amoun of people in the crowd.

18(amount people in 1 grid) x 9(amount of side on the grid) x 14(amount of people on the length) = 2286 ( Estimated amount of people)

F.U = =

**In your group, discuss ways your method might lead to a poor estimate of the crowd size.**
Since it just a estimation, it is not accurate. It does only count 1 particular grid, not all. If the method included more example migth lead us closer to the real accurate amount since we use more samples. So the method is distributed unevenly. It also can lead on miscounting a dot.