How do you compute this normal distribution?

Question

The question is:

Given that X is normally distributed with mean 100 and standard deviation 9, compute the following for n = 16.

(a) Mean (Round your answer to the nearest integer.) and variance of X

(b)P (X) ≤ 105)

(c) P (X > 100)

(d)P (94 ≤ X ≤ 102)

For the record, the X in a-d is actually X-bar (X with the line above it) - which didn't make any sense to me...

For 'a', I know the mean is 100. I thought the variance was 81, but apparently that's wrong...

For 'b', I tried normalcdf(0, 105, 100, 9) = .711 but that isn't correct either.

'c' was easy: .5

For 'd', similar to 'b', I tried normalcdf(94,102,100,9) = .335, but that was wrong as well.

I'm guessing my problem has something to do with the n=16 portion of the question, since I didn't utilize that. Could someone explain how I might go about answering these questions?

Answer 1

(a) We are taking a random sample $X_1,X_2,\dots,X_n$ from a normal distribution with mean $100$ and standard deviation $9$, where $n=16$. The random variable $\bar{X}$ is the sample mean, defined by $$\bar{X}=\frac{1}{n}\left(X_1+X_2+\cdots+X_n\right).\tag{1}$$ Note that the sample mean is a random variable, it depends on the actual values taken on by the $X_i$. The random variable $\bar{X}$ has normal distribution.

For the mean $E(\bar{X})$ of $\bar{X}$, we use the linearity of expectation and from (1) conclude that $$E(\bar{X})=\frac{1}{n}\left(E(X_1)+E(X_2)+\cdots+E(X_n)\right)=\frac{1}{n}(100)(n)=100.$$

For the variance, we use two facts: (i) the variance of $kY$ is $k^2$ times the variance of $Y$ and (ii) the variance of an independent sum is the sum of the variances. Then from (1) we get $$\text{Var}(\bar{X})=\frac{1}{n^2}(81)(n)=\frac{81}{n}.$$ Since $n=16$, the variance of $\bar{X}$ is $\frac{81}{16}$. The standard deviation of $\bar{X}$ is the square root of that, which is $\frac{9}{4}$.

(b) To find the probability that $\bar{X}$ is $\le 105$, just find the probability that a normal with mean $100$ and standard deviation $\frac{9}{4}$ is $\le 105$. You know how to do that.

(c) By symmetry, this is indeed easy.

(d) Much like (b).

Answer 2

Use [(standard deviation)/(n^.5)] as your sigma value in the calculator command because we are dealing with a sample of the population, not the whole population. Change this, and keep everything else the same, and you should get the right answer.