Peeved about a simple Sample Mean question

[trap101]

Let X(bar)[sample mean] be the average of a sample of 16 independent normal random variables with mean 0 and variance 1. Determine c such that

P(|X(bar)| < c) = 0.5


Ok here's the problem, I know this is SUPPOSED to be a simple computation but for one reason or another little ol' fooliosh me can't figure it out.

Since we are dealing with X(bar) I figure it is defined as 1/n\sum(from 1 to n) X_i

now since we're talking abiout 16 independent Normal random variables, I figured it would be as easy as taking the sum of all of these normals and solving for x. Well according to the solution this isn't the case.

Another thing I tried was taking the product of the iid normals:

1/(2\pi\sigma²)^1/2 e^\sum(X_i2)/2

set this equal to 0.5 and solve for X. Again the wrong answer. Help please

 

[jbunniii]

If
$$Y = \frac{1}{N}\sum_{n=1}^{N} X_n$$
then what can you say about $Y$? What kind of random variable is it? What are its mean and variance?

 

[HallsofIvy]

You seem to be completely misunderstanding the problem! What you should be doing is using the theorem:

"The average of n random variable, each having normal distribution with mean \mu and standard deviation \sigma, has the standard deviation with mean ]\mu and standard deviation \frac{\sigma}{\sqrt{n}}. "

The problem is simply asking you to find c the probability of that "X" is less than c is 0.5 when X has the normal distribution with 0 and standard deviation 4. That can be done by looking up "z" such that P(z)< 0.5 on a table of the standard normal distribution
(A good one online is at http://www.math.unb.ca/~knight/utility/NormTble.htm )
and then solving \frac{x}{4}= z.

(You really shouldn't need a table to find z such that P(z)< 0.5!)
(I say after looking it up in the table myself!)

 

[trap101]

You seem to be completely misunderstanding the problem! What you should be doing is using the theorem:

The average of n random variable, each having normal distribution with mean \mu and standard deviation \sigma, has the standard deviation with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

The problem is simply asking you to find c the probability of that "X" is less than c is 0.5 when X has the normal distribution with 0 and standard deviation 4. That can be done by looking up "z" such that P(z)< 0.5 on a table of the standard normal distribution
(A good one online is at http://www.math.unb.ca/~knight/utility/NormTble.htm )
and then solving \frac{x}{4}= z.

(You really shouldn't need a table to find z such that P(z)< 0.5!)
(I say after looking it up in the table myself!)

Ok, I got the right solution, but I got it extremely flukey...Let me see if I make sense of it.

So looking at the table of the standard normal distribution, for the value of 0.5 I got a probability of 0.6915. Taking this 0.6915 I put it into the expression of \frac{x}{4}= z and got a value of 0.1728 for z. A couple questions: the 0.6915 is the percentage in relation to the z-score of 0.5? and what exactly was the motivation behind dividing 0.6915 by 4? Sure the answer might be right but what was that expression representing?

 

[Ray Vickson]

Ok, I got the right solution, but I got it extremely flukey...Let me see if I make sense of it.

So looking at the table of the standard normal distribution, for the value of 0.5 I got a probability of 0.6915. Taking this 0.6915 I put it into the expression of \frac{x}{4}= z and got a value of 0.1728 for z. A couple questions: the 0.6915 is the percentage in relation to the z-score of 0.5? and what exactly was the motivation behind dividing 0.6915 by 4? Sure the answer might be right but what was that expression representing?

The sample mean $\bar{X}$ is normally distributed, has expected value = 0 and variance $\sigma^2=16$, so has standard deviation $\sigma = 4$. That is, we can write
\bar{X} = \frac{Z}{\sigma} = \frac{Z}{4},
where $Z \sim N(0,1).$ So,
P\{ |\bar{X}| &lt; c \} = P\{ -c &lt; \bar{X} &lt; c \} = P\{-c &lt; Z/4 &lt; c\}.

 

[Ray Vickson]

Ok, I got the right solution, but I got it extremely flukey...Let me see if I make sense of it.

So looking at the table of the standard normal distribution, for the value of 0.5 I got a probability of 0.6915. Taking this 0.6915 I put it into the expression of \frac{x}{4}= z and got a value of 0.1728 for z. A couple questions: the 0.6915 is the percentage in relation to the z-score of 0.5? and what exactly was the motivation behind dividing 0.6915 by 4? Sure the answer might be right but what was that expression representing?

The sample mean $\bar{X}$ is normally distributed, has expected value = 0 and variance $\sigma^2=16$, so has standard deviation $\sigma = 4$. That is, we can write
\bar{X} = \sigma Z = 4 Z,
where $Z \sim N(0,1).$ So,
P\{ |\bar{X}| &lt; c \} = P\{ -c &lt; \bar{X} &lt; c \} = P\{-c &lt; 4 Z &lt; c\}.

 

[trap101]

The sample mean $\bar{X}$ is normally distributed, has expected value = 0 and variance $\sigma^2=16$, so has standard deviation $\sigma = 4$. That is, we can write
\bar{X} = \sigma Z = 4 Z,
where $Z \sim N(0,1).$ So,
P\{ |\bar{X}| &lt; c \} = P\{ -c &lt; \bar{X} &lt; c \} = P\{-c &lt; 4 Z &lt; c\}.

ok, just to make sure this is a legitimate move I'm going to make we have established:
P\{-c &lt; 4 Z &lt; c\}. = 0.5

now looking at the tables, for a Standard Normal distribution, the value of "c" is satisfied by 0.6915.

Therefore: P\{-0.6915 &lt; 4 Z &lt; 0.6915\}. = 0.5, but since this is not fully standardized I have to divide by the 4:

therefore:

P\{-0.6915/4 &lt; Z &lt; 0.6915/4\}.

which gives me 0.1728. Is this the right set of steps?

 

[Ray Vickson]

ok, just to make sure this is a legitimate move I'm going to make we have established:
P\{-c &lt; 4 Z &lt; c\}. = 0.5

now looking at the tables, for a Standard Normal distribution, the value of "c" is satisfied by 0.6915.

Therefore: P\{-0.6915 &lt; 4 Z &lt; 0.6915\}. = 0.5, but since this is not fully standardized I have to divide by the 4:

therefore:

P\{-0.6915/4 &lt; Z &lt; 0.6915/4\}.

which gives me 0.1728. Is this the right set of steps?

Sorry: I meant that variance is 1/16 and $\sigma = 1/4$. We have $\bar{X} = \sigma Z = Z/4,$ so the probability is $P \{ -c < Z/4 < c \} = P \{ -4c < Z < 4c \}$.

 

[Ray Vickson]

ok, just to make sure this is a legitimate move I'm going to make we have established:
P\{-c &lt; 4 Z &lt; c\}. = 0.5

now looking at the tables, for a Standard Normal distribution, the value of "c" is satisfied by 0.6915.

Therefore: P\{-0.6915 &lt; 4 Z &lt; 0.6915\}. = 0.5, but since this is not fully standardized I have to divide by the 4:

therefore:

P\{-0.6915/4 &lt; Z &lt; 0.6915/4\}.

which gives me 0.1728. Is this the right set of steps?

Your answer is 16 times too small.

 

[Ray Vickson]

The sample mean $\bar{X}$ is normally distributed, has expected value = 0 and variance $\sigma^2=16$, so has standard deviation $\sigma = 4$. That is, we can write
\bar{X} = \frac{Z}{\sigma} = \frac{Z}{4},
where $Z \sim N(0,1).$ So,
P\{ |\bar{X}| &lt; c \} = P\{ -c &lt; \bar{X} &lt; c \} = P\{-c &lt; Z/4 &lt; c\}.

Please disregard this message, and instead refer to the others I gave. I wanted to delete this message, but the system won't allow it. Basically, $\bar{X} = 4 Z$, NOT what I wrote immediately above.