Question

Karen is starting a career as a professional wildlife photographer and plans to photograph Canadian Geese at one of the staging grounds during their migration in North Eastern Manitoba. She booked a place in a hide at the edge of a lake and plans to photograph the geese as they land on the water. For the price of a room in the Hilton ($200 per day), she gets a spot on a wooden bench shared by other photographers, a muddy floor, a bracket to mount her telephoto lens, a tent to sleep in and delicious meals of freshly caught fish. Last year, during a stay of 3 days in this hide, she got 2 shots worth $5000 each. She regards this as typical for this time of year although good shots happen at random and each day is independent of the others. To establish her reputation Karen only sells $5000 photographs. This year she has booked 4 days in the hide. What is the standard deviation of her revenue from one day? (The answer is $4082, I just don't know how they got the answer)

Answer 1

Answer:

$5000*0.816 = $4082

Step-by-step explanation:

It's a strange question, but based on the statement and the question it sounds like it's a poisson distribution:

* For 3 days she was able to get 2 good shots (typical of that time of the year)

* Good shots happen randomly

* Each day is independent of another

Let's call 'p' the probability that she makes a good shot per day

Let's call 'n' the number of days Karen is taking shots.

So, if in 3 days he got 2 good shots and that is typical at that time of the year, then the expected value for the number of good shots (X) is:

$E(x)=\frac{2}{3}$

For a Poisson distribution $E (x)=\lambda\\\lambda= np$

So:

$\lambda =\frac{2}{3}$

For a Poisson distribution the standard deviation is:

$\sigma = \sqrt{\lambda}\\\sigma = \sqrt{\frac{2}{3}}$

$\sigma = 0.816$ this is the standard deviation for the number of buentas taken.

So the standard deviation for income is the price of each shot per sigma

$5000*0.816 = $4082, which is the desired response.