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?
A) 18
B) 19
C) 21
D) 0
E)7.8
I have only taken up to geometry so I have no idea how to work this out. I don't want the answer just the formulas needed and some tips on how to work it out, thanks
2007-02-05 04:44:30 · 5 answers · asked by milknhoney4ever 1 in Science & Mathematics ➔ Mathematics
so what does the z stand for?
I get what the Z is for but not the z
2007-02-05 05:10:16 · update #1
OK thanks guys but I am horrible at math. It has been 5 years since I went to school and it just doesn't make any sense to me. This is not for a test or homework. Could someone just give me the answer?
2007-02-05 05:25:13 · update #2
This question has nothing to do with squirrels. It is just a made up math question. I just need the answer thanks.
2007-02-05 06:10:47 · update #3
Hey... how odd... this question is a trivia question written word for word from a virtual pet website known as wajas.com
she wants to get the right answer to get the prize from the question.
2007-02-06 10:17:55 · answer #1 · answered by zipperaward 1 · 0⤊ 0⤋
Statistics is an area where even the simplified mathematics tends to bite back. It is also an area of math that nobody has a "feel" for, so even professionals need to approach it with caution. Your example illustrates some of the problems quite well.
You have a mean of 40gm and standard deviation of 3.5 gm. A 49 gm squirrel is thus 9/3.5 = 2.6 standard deviations above the mean. Reading off the graphical bell curve, there is an expectation of about 1% of the population being this heavy, or 40 squirrels. You can calculate this value with greater precision and also establish confidence limits based on the normal distribution. Most of it will be based on unjustified assumptions.
The first assumption is the normal curve. The amount of data needed to establish a mean and a standard deviation to any degree of precision is surprisingly large, and the ability of those numbers to tell you something useful is limited to the area very close to the mean. Deviating to the extent of 99% of the population being closer to the mean than your study group increases your sample size needs hugely.
Flying or gliding animals tend to have very sharp size cutoffs within a species, primarily to do with lift-to-mass ratios. If your population of squirrels had an absolute cutoff of 47 grams for maximum weight, it could still have the same mean and standard deviation, and you would have to sample most of the population to establish that there probably was an upper weight limit of less than 49 grams.
Grinding through the formulas in statistics does not gain you as much insight as you might think. There are good software statistics packages that can run the numbers for you. You need to think very carefully to be sure how those numbers relate to the question you thought you asked.
2007-02-05 05:49:52 · answer #2 · answered by virtualguy92107 7 · 0⤊ 0⤋
All normal distributions can be collapsed into the standard normal by making this variable change:
z=(x-mu)/sigma
where mu is the mean and sigma is the standard deviation.
Tables then give probabilities P( z< Zc )
So, for your problem:
1. get the standard normal Zc=(49-40)/3.5
2. to get P( z>Zc) use P(z>Zc) = 1-P(z
It's helpful to draw a sketch of the bell curve for these data. It's peak is at the mean, 40.
So, 49 is over to the right. You want the area, probability, under the curve starting at 49 and going off the right tail. The curve is symmetrical and so you can get the same area by looking at the left tail at Zleft=-Zc.
2007-02-05 04:59:31 · answer #3 · answered by modulo_function 7 · 0⤊ 0⤋
When looking at these sort of Stats questions, we need to see which distribution the data follows, the symbol ~ means 'has distribution of ...'. This particular question looks like the Central Limit Theorem (CLT) to me.
The central limit theorem states that given a distribution with a mean μ and variance σ², the sampling distribution of the mean approaches a normal distribution with a mean (μ) and a variance σ²/N as N, the sample size, increases.
You will need to calculate some z values for your inequality and transform them using tables to get the answer.
Check it out in a stats book
Hope this helps Leigh
2007-02-05 05:15:41 · answer #4 · answered by leigh w 1 · 0⤊ 0⤋
A
2007-02-06 01:53:25 · answer #5 · answered by cle_goose 1 · 0⤊ 0⤋