It is known (only in my head, this a math question) that pygmy flying squirrels normally have a weight distrib

Question

this is haaaard

here is the whole question:

It is known (only in my head, this a math question) that pygmy flying squirrels normally have a weight distributed with a mean of 40 grams and that the standard deviation from this mean is 3.5g. If 4000 pygmy flying squirrels inhabit one area, how many would weight more than 49grams?

Answer 1

hi, i am in awe of you, you are so clever for even knowing this question, it just made my head hurt, have a nice day

Answer 2

Assuming the weight of the squirrles follow a normal distribution, you are looking for the cumulative normal distribution of the tail of the distribution (49-40)/3.5=2.571 standard deviations higher than the mean. You find this by subtracting the cumulative normal distribution of everything below 2.571 standard deviation about the mean from 1. You can look up the value for cumulative normal distribution at 2.571 standard deviation in a statistics handbook. If you have Excel, you can find it using the NORMDIST function. For your case, type in:

=NORMDIST(49,40,3.5,1)

That gives you the value of the percentage of the population that weighs up to 49 grams. What you want is 1 minus that to get you the percentage of squirrels that are heavier than 49 grams. Multiply that percentage remaining by 4000, and you have your answer. In this case, it's 4000*0.005064=20.25 squirrels. Since squirrels can't be fractional unless you count roadkills and the question does say "living", I'd round that off to 20 little porky squirrels weight more than 49g living in the area.

Alternatively, you can use the fact that normal distribution is symmetric about the mean. The population of squirrels weighing more than 49 grams is the same as the population of squirrels less than 31 grams. So you can find the percentage directly by finding the population of squirrels weighing less than 31 grams instead of doing 1 minus the population of squirrels weighing up to 49 grams.

In this case, you will find:

NORMDIST(31,40,3.5,1) will equal 1-NORMDIST(49,40,3.5,1)

Multiply that by 4000 to get your answer.

Answer 3

I'm fairly certain the answer is .03%, or 1 squirrel. I've had a TINY amount of experience with standard deviation, and by combining that with the information on Wikipedia I came up with the answer. I hope this helps, and if it doesn't I suggest that you post this in a mathematics section; zoology is a science, and you'll probably get more help by switching sections

Answer 4

It's 21. I answered on my Wajas account and got it correct.