swazination

New Member
Joined
Oct 11, 2018
Messages
9
1.1 The average number of deer seen per hour 3.6.
Calculate the probability of observing 0, 1, 2, …. 10+ below
Number of Deer Seen

1
2
3
4
5
6
7
8
9
10+

How would i solve this on excel?

<colgroup><col><col span="6"></colgroup><tbody>
</tbody>
 
A​
B​
1​
mean
3.6​
2​
Seen
Prob
3​
0​
2.73%​
4​
1​
9.84%​
5​
2​
17.71%​
6​
3​
21.25%​
7​
4​
19.12%​
8​
5​
13.77%​
9​
6​
8.26%​
10​
7​
4.25%​
11​
8​
1.91%​
12​
9​
0.76%​
13​
10+​
0.40%​
 
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Take a look at the Poisson probability.
That makes some assumptions which I am not sure would be appropriate here, such as:
  • The occurrence of one event does not affect the probability that a second event will occur. That is, events occur independently.
  • The rate at which events occur is constant. The rate cannot be higher in some intervals and lower in other intervals.
(from: https://en.wikipedia.org/wiki/Poisson_distribution)

As an outdoor enthusiast, I know for a fact that:
- deer seldom travel alone
- deer are much more active at night (especially dusk and dawn) than during the day

So I would argue that this may not be the best fit for a Poisson distribution, but perhaps your teacher doesn't know that much about deer! ;)

Without some assumptions (or the standard deviation), it is tough to say exactly what method they expect you to use. What kind of stuff have they been teaching lately? It probably falls into one of those categories. If they have been teaching Poisson distribution, I would probably go with that (even though it is not really a good fit).
 
Last edited:
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Reckon it depends on whether this is a class for math or wildlife management.
 
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Reckon it depends on whether this is a class for math or wildlife management.
Kind of reminds of a funny saying I came across years ago.
You might be an engineer if you ever have assumed that a horse is a sphere to make the math easier!
 
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