# Statistics

• Dec 20th 2009, 06:41 AM
Romanka
Statistics
Linda found that bluebirds in her area spend 10% of their day at her bird feeder. If 8 bluebirds live the area, how many minutes during a 12 hour period are 3 bluebirds expected to be at the feeder?

Thank you so much!
• Dec 20th 2009, 12:05 PM
ANDS!
The most important thing about these types of problems is understanding what is being asked, and figuring out certain "code words". In your problem for example they tell you that of a birds day, 10% of it is spent doing some activity. Which means 90% is spent doing something else. If we were to take the sample space of All Blue Bird Activity, it would be composed of two parts: one part milling about at some bird feeder and one part doing something else.

Secondly, they are asking what behavior you can expect to see. Well, expectation (as you should know) is just another way of saying "average": On average, how many minutes should three birds spend at her bird feeder?

And thirdly, you should recognize, that the fact that one bird has gone to this persons bird feeder, probably isn't going to affect another bird going in. So if we call X1, X2 and X3 independent events, where X is the number of minutes bird N is spending at this bird feeder, then we can go about figuring out the problem which is simple E[# of minutes of three birds] = X1+X2+X3.

Therefore, over a 12 hour period (or 720 minutes), one blue bird should spend 72 minutes at her bird feeder. How many minutes should three spend?
• Dec 20th 2009, 06:21 PM
Romanka
But how can I get the right result - 24 min ?(Itwasntme)
• Dec 21st 2009, 03:57 AM
awkward
Quote:

Originally Posted by Romanka
Linda found that bluebirds in her area spend 10% of their day at her bird feeder. If 8 bluebirds live the area, how many minutes during a 12 hour period are 3 bluebirds expected to be at the feeder?

Thank you so much!

Hello Romanka,

If the probability that a bird is at the feeder at any one point in time is 0.1, then the probability that exactly 3 birds of the 8 are at the feeder is

$\binom{8}{3} \cdot (0.1)^3 \cdot (0.9)^5 \approx 0.03307$,

so the expected number of minutes in 24 hours that exactly 3 birds will be at the feeder is approximately

$0.03307 \times 12 \times 60 = 23.81$.
• Dec 21st 2009, 05:57 AM
Romanka
Thanks!!!
• Dec 21st 2009, 09:59 AM
ANDS!