Help with question about Normal distribution

Question

The heights of people at some city is $N(\mu,\sigma^2)$.
25% of the people are above 176 cm. and 20% are below 165 cm.

1. How do I found $\mu,\sigma^2$?
2. We choose randomly a group of 100 people. What is the probability that at least 8 of them will above 180 cm.?
3. How to I found $a,b$ s.t. at probability of 99% the average of those 100 people will be between $a$ and $b$?

I don't have any idea how to solve 1 and 3.
About two I assume that is Negative Binomial distribution, I'm right?
I'd like get any help...

Thank you!

Answer 1

Here's Q1: heights distributed as H, $Z \sim N(0,1)$.

$$H \sim \sigma Z + \mu$$

$$\frac{H-\mu}{\sigma} \sim Z $$

$$\mathbb{P}(H<165)=0.2=\mathbb{P}(Z<\frac{165-\mu}{\sigma})$$

$$\mathbb{P}(Z<-0.8416)=0.2 \therefore -0.8416=\frac{165-\mu}{\sigma}$$

Use $\mathbb{P}(Z>0.6745)=0.25$ to get the other equation and find $\sigma$, $\mu$.

Answer 2

For 1, looking at a standard normal table: http://www.mathsisfun.com/data/standard-normal-distribution-table.html , we see that the 25th percentile matches with (approx.)z=0.68, so that:

$ 0.68 =\frac {176-\mu}{\sigma} $ and the lower 20% matches with (approx.) $z=-0.84$ , so that $\frac{165-\mu}{\sigma}=-0.84$

Now you have two equations with two variables, so you can solve for $\mu, \sigma$

For 3, looking at the table again . By symmetry of the standard normal , the z-value interval (-2.57, 2.57) will contain 99% of the values. Use the value of $\mu$ to find the actual intervals $(\mu-2.57\sigma, \mu+2.57\sigma)$

Answer 3

Here's a sketchtoy: http://sketchtoy.com/61232763

Basically what I did was "plot" the points we have on the normal distribution curve. We got the areas that are to the right or left of each point, and from that using invNorm() we can get the z-scores of these points. The z-scores are, as shown, invNorm(0.2) for the $165\,cm$ point and invNorm(0.75) for the $176 \, cm$ point. The approximate values of the z-scores for each is $-0.8416$ and $0.6745$, respectively. We can now set up two equations that will help us find $\mu$ and $\sigma$:

$$\begin{cases} -0.8416=\frac{165-\mu}{\sigma} \\ 0.6745=\frac{176-\mu}{\sigma} \end{cases}$$

Using systems of equations, we can manipulate this to find what we want. Here's just one way by using subtraction (you can use substitution or what ever else, it doesn't matter):

$$\require{cancel}\begin{align} -0.8416\sigma&=165\cancel{-\mu} \\ -\;\;\;0.6745\sigma &= 176\cancel{-\mu} \\ \hline -1.5161\sigma &=-11 \end{align}$$

From this you get:

$$\begin{align}\therefore \sigma &\approx 7.2554 \\ \sigma^2&\approx 52.6409\end{align}$$

Plugging that back in to whichever equation, you'll get that:

$$\therefore \mu \approx 171.1063$$

From this you can do the rest. Since you have no problem with #2, I'll jump to #3. That question actually just literally asks for a confidence interval. A confidence interval is modeled as: $$\bar{x}\pm z^*\frac{\sigma}{\sqrt{n}}$$

We calculated $\sigma \approx 7.2554$. $n$ is the number of tests, which is given to be $100$. Last piece is $z^*$. Because you want it to be a 99% confidence interval, your z-star is |invNorm(0.005)|. Here's a sketch toy why. That's just a concept you need to know, but it should make sense from the equation.

Anyhow, we can assume that $\bar{x}\approx \mu$. Let's rewrite it as: $$\mu \pm z^*\frac{\sigma}{\sqrt{n}}$$ Point $a$ will be $\mu - z^*\frac{\sigma}{\sqrt{n}}$, and point $b$ is $\mu + z^*\frac{\sigma}{\sqrt{n}}$ Guess what? We know all the values. We can plug them in, and get that our 99% confidence interval is:

$$\therefore [169.2374 \, cm, 172.9751 \, cm]$$