Standard Deviation Word Problem

AI Thread Summary
The problem presents a scenario where the average time spent online is 14 hours, with a standard deviation of 17 hours, raising questions about the validity of a normal distribution assumption. Subtracting the standard deviation from the mean results in a negative value, which is not feasible for time spent online. This discrepancy highlights the limitations of assuming normal distribution in real-world scenarios, particularly when dealing with restricted domains like time. The discussion suggests that the problem may be designed to illustrate these shortcomings rather than being an error in the textbook. Ultimately, it emphasizes the need for careful consideration of distribution assumptions in statistical analysis.
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Homework Statement


It has been projected that the average and standard deviation of the amount of time spent online using the Internet are, respectively, 14 and 17 hours per person per year (approximately normally distributed). What value is exactly 1 standard deviation below the mean?

Homework Equations


Emiprical Rule
\mu \pm \sigma contains approximately 68% of the measurements.
\mu \pm 2\sigma contains approximately 95% of the measurements.
\mu \pm 3\sigma contains approximately almost all of the measurements.

The Attempt at a Solution


In similar problems, the mean is the larger number in the problem, so solving the problem is a simple matter of subtracting the standard deviation from the mean to find out the percentage of population.

In this case though, the standard deviation (17) is greater than the mean(14)? If I solve this like I do normal problems, this would leave me with a negative value for time spent on the Internet.

Is this an error in the textbook or is there something I'm missing here?
 
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My guess is that the problem is meant to emphasize the shortcomings of assuming a normal distribution, e.g. the normal distribution spans the entire real line while examples may have a restricted domain, as in this case only positive reals.
 
Thanks
 
It should be used to reinforce the idea that assuming things are normally distributed isn't always justified: the actual question as you describe it shows that the times can't be normal, for the reason you point out.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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