## Estimation Station

## Class Problems

To view the problem, click the title.

## Jelly BellyThis problem definitely brings memories of holiday games where themed prizes are given for the closest guess given to the actual answer. This of course included getting the jar of candy. Helpful information in this problem include the net weight (in ounces) of the jar along with the measurement of the jelly belly itself. What could be more helpful, however, would be the measurements of the jar itself, not how much it can hold. This would be helpful because both measurements given were in different units and difficult to compare to to find a reasonable estimation.
My estimate is 1080 jelly bellies in the jar. The reason I believe this is correct is because I gave a rough estimate as to how many jelly bellies are in an ounce and multiplied that by the amount of ounces in the container (45). I used 24 as my approximation of how many jelly bellies are in an ounce based on my knowledge of an ounce. As well, whether or not this is true, I took the number of question marks on the image as a context clue. ## TreesThis problem is a fairly challenging one to answer because of the enormity and randomness that nature sometimes has. In this case, not only is there a large forest with wildly growing trees in no apparent pattern, but there is also a river. However, the author was kind enough to remind those estimating that there is a river and your guess should exclude that area. Another helpful bit was having a smaller picture showing what one square foot looked like.
I estimated that there were around twenty trees per square foot from the small, additional photo provided. Therefore, my answer to the problem would be about 300 trees since the formal problem asked for how many trees in fifteen square feet. However, since there is a river I removed fifty trees from the final answer making it 250 trees. |
## FishMy initial thoughts to this problem would be "wow there is a lot of fish". Information that was helpful is stating how much of the tank is being viewed in the first picture. As well as giving an image of the tank that they are in. A way to add more information to help would be to give a fuller picture of the tank, as only part of it is seen. Along with the dimensions of said tank.
My estimate is 450 fish. When looking at the picture of the number of fish in 2/3 of the tank, I saw three levels of shadows. If I estimated I would say there is probably around a hundred fish per level. Therefore, there are 300 fish in 2/3 of the tank. This leads me to believe there to be about 150 fish per third of the tank which means there are 450 fish. ## Animal Cracker ConspiracyWhen I first saw this problem, the title just screamed out at me. It was interesting and clever and was one of the main reasons I chose to complete this problem. There were several dimensions included that were also very helpful through the process. I also liked that they were slightly misleading because one set was helpful for volume, and another for area.
My estimate is that there are around 40 animal crackers. I came to this conclusion by first finding the volume of the cylinder by information provided and then dividing it by the approximate volume of one cracker. I found that by using the information provided, but turning it into millimeters since I estimated the additional side needed to find volume in millimeters. Since the volume of the cylinder was in inches, I had to convert this amount back into inches, though. ## Phone Battery LifeThis problem is not completely unlike the example first given, and also includes a flair of modern day teen relevance. Showing the change over time is very helpful in a problem like this, however, the lapse of time throughout the day is not consistent. It goes from two hours, to one, and then back to two. It would be more helpful to see if there is a pattern throughout the day at consistent intervals. Although, not having such makes it more of a fun estimation problem. Helpful information would be the span of time in this problem. It goes from the morning to early evening giving a wider range of information to form an estimation on.
My estimate is that it will die before the time reaches 11:00 pm. My reasoning behind this conclusion is that the average time depletion for two hours (since there was an ability to take an average of two hours), is 15.5% battery power lost. Therefore from 4:58 pm, or basically five, there are three more stretches of two hour periods. That would mean that 46.5% battery power will have been used. However, at 4:58 pm, battery power was already at 38% which is a lower percentage. Therefore, battery power will have been spent. |