Solved – Normal Distribution – finding mean

normal distributionself-study

I'm studying Statistical Methods and trying to revise for the upcoming exam. Looking at the past exam paper, there is a question:

 X ~ N (µ, 10²), and P ( X > 50) = 0.9, find µ

I've only before done the calculations finding z = (x – mean)/standard deviation
and looking at the values from the Normal distribution function table. I have no idea how to approach this question. Any help would be much appreciated.

Best Answer

Let us suppose that the variance is 100. Then the standard deviation is 10.

P$(X>50)$=P($X-mu>50-mu)$=P($(X-mu)/10 > (50-mu)/10)$=

P$(Z> (50-mu)/10)$=0.9.

So now you can go to the standard normal table and find what value of Z say a gives

P($Z>$a)=0.9.

After you find a you solve for mu by setting $(50-mu)/10$ = a and then solving for $ mu$.

Related Solutions

Solved – How to use standard normal distribution tables

When finding the probability using the table, when should I use the z to find a value

when you have a $z$ and need a probability.

and when should I use the probabilities to find a value?

when you have a probability and need a $z$.

Sometimes when I have this ($z≤0.6$) they say that I should look for the value by using $z$ and sometimes by searching for the nearest two probabilities and which $z$ they are, add them and divide them by 2. So I would like to know when to use each one?

If your probability value is in the body of the table/$z$-value in the margins (or really close to a value), you can use the probability.

If your probability(/$z$ value) is in between, the corresponding $z$ (/probabilities) are in between. The usual thing is to do linear interpolation, but it sounds from your description like they're just asking you to take the midpoint of the two probabilities either side.

[There's some explicit examples of using the tables here and here.]

Maximum Likelihood Estimates – Finding Maximum Likelihood Estimates for Multiple Normal Populations

As hinted in a comment, you have to set up an hypothesis test. Your hypothesis $H_0$ is that the mean is the same. In that case, the likelihood will be $$ \mathcal{L} = \prod_{i=1}^n\frac1{\sqrt{2\pi}\sigma} \exp\left(-\frac{(x_i-\mu)^2}{2\sigma^2}\right) $$ where the product is in all the data points together ($n=\sum_k n_k$). We know that the optimal estimator is $$ \hat{\mu}=\frac1{n}\sum_{i=1}^nx_i $$ and $$ \hat{\sigma}^2=\frac1{n}\sum_{i=1}^nx_i^2-\left(\frac1{n}\sum_{i=1}^nx_i\right)^2 $$ The value of the log-likelihood you get it from replacing in the previous formula is $$ \log\mathcal{L}_0=-n(\log(\hat{\sigma}\sqrt{2\pi})+1/2) $$

Your hypothesis $H_1$ is that the mean is different from each sample. Since the samples are indipendent, the likelihood of that is simply the product of the likelihoods of each sample: $$ \mathcal{L} = \prod_{j=1}^k \prod_{i=1}^{n_j}\frac1{\sqrt{2\pi}\sigma} \exp\left(-\frac{(x_i-\mu_j)^2}{2\sigma^2}\right) $$ The solution for the $\mu_j$ is analogous: $$ \hat{\mu}_j=\frac1{n_j}\sum_{i=1}^{n_j}x_i $$ before replacing it in the likelihood formula, let's do some 'massage' $$ \mathcal{L} = \frac1{(\sqrt{2\pi}\sigma)^n} \prod_{j=1}^k \prod_{i=1}^{n_j} \exp\left(-\frac{(x_i-\hat{\mu}_j)^2}{2\sigma^2}\right) $$ then the product becomes $$ \prod_{j=1}^k \prod_{i=1}^{n_j} \exp\left(-\frac{(x_i-\mu_j)^2}{2\sigma^2}\right) = \exp\left( \sum_{j=1}^k \sum_{i=1}^{n_j} -\frac{(x_i-\hat{\mu}_j)^2}{2\sigma^2} \right) $$ apply logarithm $$ \log\mathcal{L}_1=-n\log(\hat{\sigma}\sqrt{2\pi})-\left( \sum_{j=1}^k \sum_{i=1}^{n_j} \frac{(x_i-\hat{\mu}_j)^2}{2\hat{\sigma}^2} \right) $$ Let's call $\hat{\sigma}^2_j$ the variance of the sample $j$; we have that $$ \log\mathcal{L}_1=-n\log(\hat{\sigma}\sqrt{2\pi})-\left( \sum_{j=1}^k \frac{n_j \hat{\sigma}_j^2}{2\hat{\sigma}^2} \right) $$ by derivating and equating to $0$ you get that $$ \hat{\sigma}^2=\sum_{j=1}^k\frac{n_j}{n}\hat{\sigma}_j^2 $$ and we get $$ \log\mathcal{L}_1=-n(\log(\sum_j n_j/n \hat{\sigma}^2_j \sqrt{2\pi})+1/2) $$ The likelihood ratio, can be computed as the substraction of the two log-likelihoods: $$ \log(\mathcal{L}_1/\mathcal{L}_0)=-n\log\left( \frac{ \frac1n \sum_{i=1}^n (x_i-\mu)^2 }{ \frac1n \sum_{j=1}^k \sum_{i=1}^{n_j}(x_i-\mu_j)^2}\right) $$ You can replace the values of $\mu$, you get eventually to $$ \Lambda = e^n\exp\left(1-\frac{\frac1{n^2}\sum_{j=1}^k\frac{n}{n_j}\left(\sum_{i=1}^{n_j}x_i\right)^2-\frac1{n^2}\left(\sum_{i=1}^nx_i^2\right)}{\frac1n\sum_{i=1}^nx_i^2-\frac1{n^2}\left(\sum_{i=1}^nx_i\right)^2}\right) $$

Best Answer

Related Solutions

Solved – How to use standard normal distribution tables

Maximum Likelihood Estimates – Finding Maximum Likelihood Estimates for Multiple Normal Populations

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