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A statistical analysis of 1,000 long-distance phone calls made from a mobile phone indicates that the length of these calls is normally distributed. It is known that half of these phone calls are less than or equal to 240 seconds duration and the standard deviation is 40 seconds.

a) What percentage of these calls lasted less than 180 seconds?
b) How many calls had a length between 180 and 300 seconds?
c) What is the length of a particular call if only 1% of all calls are shorter?

I'm confused by the bit that says ".... known that half of these phone calls are less than or equal to 240 seconds duration and the standard deviation is 40 seconds. ....".

Half the calls are 240 seconds, so is the mean 240? And, how can the SDeviation be used to solve a, b, c?

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    Answer can be written in terms of the error function http://en.wikipedia.org/wiki/Error_function The limited range $[0,\infty]$ not a serious problem here as zero lies more than a few sigma (standard deviations) away from the mean.2011-07-25
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    Clearly this is a fictional example. Phone call durations do not come from a normal distribution - calculate the probability that a call takes less than 0 seconds, it is not zero!2011-07-25
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    Of course it is a fictional example (from a book). For the sake of solving, any help?2011-07-25
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    Your probability distribution is $\propto e^{-(t-\mu)^2\over 2\sigma^2}$ where $\sigma$ is the standard deviation and $\mu$ is the mean. To find the fraction of calls with t between two values you just need to integrate up the Gaussian distribution. The integral of the Gaussian is an error function, so the answer can be written in terms of error functions.2011-07-25
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    Ok, if I were doing this in Excel with: Norm.Dist(X, MEAN, SD, CUMULATIVE), then for question a, would this be correct? Norm.Dist(180, 240, 40, TRUE)2011-07-25
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    OneQOnly: There are online tables that allow you to match z-scores with probabilities, and , conversely, to find percentiles in a given distribution:http://www.sjsu.edu/faculty/gerstman/EpiInfo/z-table.htm , but I it is pretty likely that you have a table somewhere in your book; maybe the front- or back- jackets.2011-07-25
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    For (a) you need the probability that a standard normal is $\lt (180-240)/40$. So you want probability than standard normal is $\lt -1.5$. For (b) want probability that standard normal is between $-1.5$ and $1.5$. That's $1$ minus twice the tail you computed in (a). For (c) you want to be roughly $(2.33)(40)$ down from the mean.2011-07-25
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    "Half the calls are 240 seconds, so is the mean 240?" Your question is dead on target. The sample median is $240$ which is not necessarily the same as the sample mean. Presumably the claim that the call lengths seem to be "normally distributed" is supposed to be used to infer that _the sample mean is the same as the sample median_ because the mean and the median of a normal distribution happen to be the same. The inference is not necessarily valid. I would say that the statistical analysis claimed to have been done seems to be shoddy or the problem is very poorly posed.2011-10-23

1 Answers 1

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Hints:

If $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma > 0$, then:

1) ${\rm P}(X \leq \mu) = 1/2$ (and ${\rm P}(X \leq x) \neq 1/2$ for any $x \neq \mu$).

2) $\frac{{X - \mu }}{\sigma }$ is a standard normal (i.e. ${\rm N}(0,1)$) random variable.

3) For any $a,b \in [-\infty,\infty]$ with $a < b$, $$ {\rm P}(a < X < b) = {\rm P}\bigg(\frac{{a - \mu }}{\sigma} < \frac{{X - \mu }}{\sigma } < \frac{{b - \mu }}{\sigma }\bigg). $$

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    Regarding the first hint, consider ${\rm P}(X \leq 240) = 1/2$.2011-07-25
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    Another way of putting point (1) is that for the normal distribution, the mean is the same as the median.2011-07-25
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    @Michael: Good point (and I was aware of it).2011-07-25
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    Shai: Is it possible to have ${\rm P}(X \leq \mu) = 1/2$ (and ${\rm P}(X \leq x) \neq 1/2$ for any $x \neq \mu$ for $x\neq \mu$?2011-08-24