# Help solving: Problem on Normal Distribution of Data

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 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. – Alice Jul 25 '11 at 14:07
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! – PeterR Jul 25 '11 at 14:10
Of course it is a fictional example (from a book). For the sake of solving, any help? – OneQuestionOnly Jul 25 '11 at 14:12
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. – Alice Jul 25 '11 at 14:13
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) – OneQuestionOnly Jul 25 '11 at 14:14

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 ${\rm P}(X \leq 240) = 1/2$. – Shai Covo Jul 25 '11 at 14:46
Another way of putting point (1) is that for the normal distribution, the mean is the same as the median. – Michael Hardy Jul 25 '11 at 22:29
@Michael: Good point (and I was aware of it). – Shai Covo Jul 25 '11 at 22:40
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 $x\neq \mu$? – gary Aug 24 '11 at 17:22