How big should the sample A random variable X has a normal distribution with mean 100 and standard deviation 10.
(a) What is the P( 90< X < 110)?
(b) If Average X is the average of a sample of 16 elements removed this population, calculate P (90≤ Average X ≤ 110).
(c) Represent on a graph, the X and Average X  distributions.
(d) How big should the sample so that P (90≤ Average X ≤ 110) = 0.95.
Guys, could you help me in b) and c). I do boy need the answers, I wanna understand it.
 A: For the example you have a random variable $$X \sim N(100,10^2).$$
In number (b) you have to compute the average over 16 r. v. $X_i,\ i = 1, \dots , 16$. I call this average $Z$. With $\alpha + \beta X \sim N(\alpha + \beta \mu, \beta^2\sigma^2)$ for a normal distributed r. v. with expectation $\mu$ and variance $\sigma^2$ and $X_i \overset{i.i.d}{\sim}X$ you get:
$$Z = \frac{1}{16}\sum_{i=1}^{16} X_i \sim N\left(\frac{1}{16}\sum_{i=1}^{16} \mu, \left(\frac{4\sigma}{16}\right)^2 \right) = N\left(\frac{1}{16}\sum_{i=1}^{16} \mu, \left(\frac{\sigma}{4}\right)^2 \right)$$
With the numbers given you have
$$Z \sim N\left(100, \left(\frac{10}{4}\right)^2 \right)$$
Now you can compute the probability
$$P(Z \in [90,110]) = P(Z \geq 90) - P(Z \geq 110)$$
or more common and easyer to make computations
$$P(Z \in [90,110]) = P(Z \leq 110) - P(Z \leq 90)$$
by standardisation or with a statistic software.
Because the mean from both is equal you just have to compare the standard deviation, which is much lower from the Average X. You can see this in the picture (blue = Average X, green = X)

I hope I don't miss something, now. :)
