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Scientists are weighing frogs. The mean weight is 5.5oz and the median weight is 4.1oz. Which of the following statements is true regarding the frogs?

A) Frogs are normally distributed

B) Frogs are exponentially distributed

C) There are a few very heavy frogs

D) There are a few very light frogs.

I know it can't be choice A because median=mean=mode for a normal distribution. Can't be choice B either because this is usually modeled for waiting times. Can't be choice C because median is less than 5.5oz. I have to say it's choice D because median value says weights are near 4.1oz (considered lighter).

My question to you all is, are you all in accordance with my reasoning? I'm not sure if I'm correct (haven't seen this material in 10+ years). Thanks in advance.

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This is a poor question. Your reasoning on A is right. For B you would need median = mean$\times ln(2)$, which you don't have. C and D are miss leading. What do we mean by light or heavy. The set 3, 4.1, and 9.4 fit your criterion. I might interpret this set as having the same number of light and heavy frogs. –  mtiano Dec 19 '13 at 0:44
    
Yes thanks at least we're on the same boat on the question. This is a practice question asked on a California Subject Matter examination test. Yes I'm confused at C and D also and my reasoning is probably not helping either. –  User69127 Dec 19 '13 at 0:46

2 Answers 2

up vote 1 down vote accepted

The intended answer is C. Having the mean above the median is (usually) caused by some very large entries. In a case like this, think about 900 frogs at 4.1 oz plus 100 frogs at 18.1 oz. This distribution has the desired properties. It is far from the only one, but the mean gets pulled by things far from the median.

One could argue that a distribution of 501 frogs at 4.1 oz and 499 frogs at 6.9056 oz also satisfies the desired properties and 6.9056 oz does not qualify as a very heavy frog.

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What a terrible question! None of the options is really ruled out by the wording*. (It's clear enough what answer they probably want, but that's really beside the point here)

* though I can rule one out from other considerations ... but in a way that would invalidate the choices they'd no doubt want you to pick on other questions of this type/

Let me show you why:

  • "Scientists are weighing frogs" ... this tells us we're dealing with a sample,

  • "The mean weight is 5.5oz and the median weight is 4.1oz." ... sample mean > sample median

That's the information. Nothing about sample size. No definition of what 'few' or 'many' not 'heavy' or 'light' mean in this context.

I know it can't be choice A because median=mean=mode for a normal distribution.

For a population. This is a sample. You can easily get samples where mean and median differ. Imagine we have a sample of size 3 where the weights are 4.09, 4.10 and 8.31 oz. If the population mean was 5.5 and the population standard deviation was somewhere near 1.5, this wouldn't be a surprising outcome.

So no, that information doesn't rule out option A

Can't be choice B either because this is usually modeled for waiting times.

So? How does what people often choose to a distribution for rule it out as a possibility?

If it were the case that we had the population mean and median, those values would rule out an exponential distribution (since for an exponential the median should be $\mu\ln 2$ and it's bigger than that), but again, these are sample figures.

Can't be choice C because median is less than 5.5oz.

Again, so? What if a proportion of $0.5+\epsilon$ are around 4.1 oz and a proportion of $0.5-\epsilon$ are around 6.9 oz?

The problem is that without some kind of precise definition of 'few' or 'many', 'heavy' or 'light' in this context, you can't rule out C or D.

[We might argue from biology that the exponential is unlikely (since it implies that the most likely sizes are very small). We could use biological arguments to rule out the normal (frogs don't have negative weights, so they can't be exactly normal). We still can't choose between C and D.]

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