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In general, how do do you calculate the mean and standard deviation of a normal distribution given 2 values on the distribution with their respective probabilities?

For Example:

Suppose that the ages of students in an intro to statistics class are normally distributed. We know that 5% of the students are older than 19.76 years. We also know that 10% of students are younger than 18.3 years.

What are the mean and standard deviation of the ages?

In my attempts to solve a similar problem I can't see how to calculate the mean or standard deviation without first knowing one of the two. I can find the z-score for 95% and 10%, and if I could somehow derive the values for 5% or 90% I could then average the 5% and 95% or 10% and 90% values to then find the mean, but I don't see a way to do so. Is it even possible to solve this problem or is there not enough information?

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3 Answers 3

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Let's take the example in question. Assume that the mean is $\mu$ and that the standard deviation is $\sigma$. If we have two z-values $z_1$ and $z_2$ corresponding to our two observations, 19.76 and 18.3 then we can solve the following equations for $\mu \ \text{and} \ \sigma$. $$\frac{19.76 - \mu}{\sigma} = z_1 \\ \frac{18.3 - \mu }{\sigma} = z_2$$ We have two equations in two unknowns, solving which, we can find $\mu$.

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The accepted answer correctly points out that we have two parameters that we don't know (the mean and the standard deviation), which cannot be calculated from the ages alone. However we have two more important facts: we know the distribution is normal, and we know the probabilities at each age. I was wondering myself, do these help?

To continue, it helps to rephrase our ages in the form of the cumulative probability density function (CDF), i.e. 5% older than 19.76 = 0.95 < 19.76, 10% younger than 18.3 = 0.10 < 18.3. From this point it looks like we might simply have two simultaneous equations to solve. However, the formula for the CDF is unfortunately this (from here):

$CDF(x) = \int_{-\infty}^{x} \frac{e^{-x^{2}/2}} {\sqrt{2\pi}}$

Which cannot be represented as a closed formula. Traditionally you could lookup the values in a "normal distribution table", but to resolve both variables would require an iterative procedure. Fortunately today, computers can perform both the integration and iteration at once.

Wolfram alpha for instance, is able to resolve your simultaneous equations (here):

CDF[NormalDistribution[μ, σ], 19.76] = 0.95
CDF[NormalDistribution[μ, σ], 18.3] = 0.1

Which gives an answer to your example as:

$N(18.9, 0.499)$

However, as far as I have found, and like the accepted answer says, there is no general formula.

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From your z-score table the data at $95\%$ is at about mean +$1.65$ standard deviations. Taking $\mu$ as the mean and $\sigma$ as the standard deviation, this tells us that $\mu+1.65\sigma=19.76$ You should be able to write a similar equation from the other piece of data. That gives two equations in two unknowns.

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