Question

In a certain population, the random variable x can only assume the values 4 and 8. The probability distribution for x is given below. Construct the sampling distribution for the sample mean x bar if the random samples of size n=3 are drawn from the population.

x|4|8
p(x)|.2|.8

Answer 1

\[E(X)=\sum_{}^{}xp(x)=(4\times 0.2)+(8\times 0.8)=7.2\]
\[Var(X)=E(X ^{2})-[E(X)]^{2}=(4^{2}\times 0.2)+(8^{2}\times 0.8)-7.2^{2}=2.56\]
\[E(X)=\mu=7.2\]
\[\sigma _{x}=\sqrt{Var(X)}=\sqrt{2.56}=1.6\]
By the Central Limit Theorem the sampling distribution for the sample means has a mean 7.2 and standard deviation given by
\[\frac{\sigma _{x}}{\sqrt{n}}=\frac{1.6}{\sqrt{3}}\]

Answer 2

How would I make this into a sampling distribution?

Answer 3

Doesn't it have to be in a table?

Answer 4

When sampled with replacement there are 8 possible outcomes as follows:
Outcome Sum Mean
#1 8 + 8 + 8 24 8
#2 8 + 8 + 4 20 6 2/3
#3 8 + 4 + 8 20 6 2/3
#4 4 + 8 + 8 20 6 2/3
#5 8 + 4 + 4 16 5 1/3
#6 4 + 8 + 4 16 5 1/3
#7 4 + 4 + 8 16 5 1/3
#8 4 + 4 + 4 12 4

Answer 5

Oh okay, I get it.
So to find the mean and standard deviation of the sample distribution, would it be using this data or the data in your first reply? it would be this data right?

Answer 6

My first post gave the calculations for, and values of, the mean and standard deviation of the sampling distribution.
Please disregard my previous third post which I have deleted and replaced with this.
The sampling distribution for the sample means is:
x bar Frequency Probability
8 1 1 * 0.8 * 0.8 * 0.8 = 0.512
6 2/3 3 3 * 0.8 * 0.8 * 0.2 = 0.384
5 1/3 3 3 * 0.8 * 0.2 * 0.2 = 0.096
4 1 1 * 0.2 * 0.2 * 0.2 = 0.008

Answer 7

I actually do have to calculate the mean and standard deviation of the sampling mean. I did not initially mention it in the question because I thought I was able to figure it out. However, I am confused about how to get it since I guess there is more than one way depending on the type of distribution. Would it be like this?:

\[mean= 4(0.008)+5\frac{ 1 }{ 3 } (0.096)+6\frac{ 2 }{ 3 }+8(0.512)=7.20 \]


\[standard deviation=[4^{2}(0.008)+(5\frac{ 1 }{ 3 })^{2}(0.096)+(6\frac{ 2 }{ 3 })^{2}+8^{2}(0.512)]-7.20^{2}=\sqrt{52.69}-7.20^{2}=44.48\]

Answer 8

Wouldn't the data from your first post be the mean and standard deviation of the probability distribution?

Answer 9

The right hand side of your equation for the mean should show the term in 6 2/3 being multiplied by 0.384. The value for the sample mean agrees with the value calculated in by first post.
The right hand side of your equation for the standard deviation should show the term in 6 2/3 squared being multiplied by 0.384. Also the whole of the right hand side should be under a square root radical sign, otherwise it is the variance that is calculated. When these corrections are made the value of the standard deviation is very close to the value in my first post given by
\[\frac{1.6}{\sqrt{3}}=0.92376\]

Answer 10

Thank you for catching those.

For the standard deviation, my result is 52.69-51.84=\[\sqrt{0.85}\]=0.92195

am I still supposed to get \[\sqrt{2.56}\] using the (newly revised) data that I had written in my prior comment?

Answer 11

Using the method for calculating the standard deviation that you posted I get
square root of (52.69296 - 51.84) = square root of (0.852960) = 0.9236
I suggest that you use your method and this result.

Answer 12

so using the centeral limit theorem is not required?

Answer 13

No need to use the Central Limit Theorem, although it does confirm the result.

Answer 14

I see, I think I'll play it safe and still with the 0.9236 result. Thank you so much for your help, I really appreciate it.

Answer 15

You're welcome :)