normally distributed

Question

A population is normally distributed with μ = 50, σ = 8, N = 52. Answer the following questions based on this information. (show all work)

 a. The mean of the distribution of means is ______ .

b. The variance of the distribution of means is ______.

c. The shape of the distribution of means is ______ .

Answer 1

Teresa, the form of these questions makes me wonder if you have made an entry error in posting them.  If (a), (b) and (c) relate to what is commonly called the "sample mean" then below I provide the answer.  The point is that the quantity you are trying to gauge the statistics of must be a "random variable" well defined from the population that contains the data (the "observables")

 

So....for example, let us consider the SAMPLE MEAN, which is actually an estimate of the population mean that is obtained by combining the samples.  It works like this.

 

 

Let mhat=sample mean=(1/N)*Sum from 1 to N of x_i where x_i denotes the i^th sample and i ranges from 1 to N, N = number of samples.  In equation form

 

                       mhat = (1/N)∑x_i where the sum is over all values of i = 1, 2, ...., N to account for the N samples.

 

mhat is a random variable composed of the sum of the (assumed independent) samples.  Because the samples are statistically independent and Gaussian, mhat is also Gaussian.  It should be noted that if the samples were NOT independent, the sum of the samples would not necessarily be Gaussian.  However, if the samples were ASSUMED to be JOINTLY Gaussian, then it's still true that the sum of the samples is Gaussian.  Indeed, one of the ways the JOINT distribution of Gaussian samples is DEFINED is by defining the distribution of the SUM of the samples.  When the samples are NOT statistically independent, if they are ASSUMED to be JOINTLY Gaussian, then one need only define the covariance matrix of the samples to completely specify the joint distribution.  This all goes a bit beyond the question you posed but I wanted to make sure you appreciate the subtlety involved in the incomplete statement of your problem at the outset, since you did not state that the samples were statistically independent nor did you state they were jointly Gaussian.  This means some assumptions are needed to proceed at all!  The assumption of statistical independence between the samples is commonly made but not always true in real life situations.

 

To answer some of your questions you must assume (as is usually the case in these problems) that the N samples are all identically distributed and statistically independent (iid).  In this case, the Mean value of mhat is the true mean of the distribution, which is 50 in this example.  This is easily shown by realizing that the mean value is a linear operation and can be brought inside the summation that defines mhat.  So then you are simply adding up N values of μ giving you 52μ and dividing by 52, which yields μ = 50.  This is the answer to (a)

 

 

The answer to (b) is slightly more complicated.  The variance of mhat is computed by taking the average of the SQUARE of mhat and then subtracting from that the square of the mean.  This follows from the DEFINITION of the Variance of a random variable.  Carrying out the algebra:

 

eqn A:    Var(mhat) = Avg[mhat²] - μ²

 

As we have already shown, the mean of mhat is μ=50.

 

When you compute the first term on the RHS of Eqn A above, you must first carry out the expansion of mhat².  The coefficient is 1/N² and this multiplies a double summation: ∑_i∑_jAVG(x_ix_j) = N*AVG[(x_i)¹] +N(N-1)μ²,

 

Here the first term on the RHS follows from using the assumption of independent samples when taking the average of x_ix_j when i is NOT equal to j.  Hint: break up the double sum into the sum of terms where i=j (of which there are N terms) and the rest of the sum over terms where i differs from j (of which there are N[N-1] terms.  The total number of terms inside the double sum is N². 

 

The easiest way to visualize the above computation is to arrange the result of taking the average of x_ix_j into a square MATRIX of elements ρ_i,j = AVG[ x_ix_j ].  Now, as i and j range over their values from 1 to N, all the matrix entries will be obtained.  There are N² total elements in the matrix.  All the DIAGONAL terms are equal to the average of [(x_i)²] and all the off-diagonal terms equal μ² because the samples are independent.  If you subtract the number of diagonal elements (N) from the total of N² matrix elements, there are (N² - N) = N(N-1) off-diagonal elements. 

 

Carrying out the algebra we have:

 

Eqn B:       Avg[(mhat)²] = (1/N²)*[N*Avg{(x_i)²) +(N-1)N*μ²] = (1/N²)*{N(σ² + μ²) + N² μ² - Nμ²}= (1/N)σ² + μ²

 

Here σ² denotes the variance of any of the samples and we've used the fact that  σ² = AVG[(x_i)²] - μ² in Eqn B.

 

Now use Eqn A and B together and you find that the Variance of mhat is equal to (σ²/N) 

This is the answer to question (b), but I leave it to you to plug in the numerical values.  Be careful: The given value for σ must be squared to get the Variance.

 

Question (c) was addressed above.  The shape of the distribution of the sample mean is Gaussian

 

Note: Typing equations into this tool is rather slow and tedious.  I cannot guarantee that they format correctly.  It would be far easier if I could have prepared the answer in Word in attached it.