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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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    "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

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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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    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