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July 13th, 2017, 01:44 PM   #1
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Exclamation Normal Distribution: P for mean of random sample

Hey guys, I have a task that is really giving me a hard time:

You are taking a random sample (n = 25) from a normal distributed Variable X with μ = 50 and σ = 20. What is the probability that the mean from the
randomly taken sample is 58 or higher?

I would really like to understand how to calculate this and especially WHY

Thanks for any tips!
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July 13th, 2017, 02:24 PM   #2
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The average (sample mean) will have a normal distribution with a mean $\displaystyle =\mu =50$ and a standard deviation $\displaystyle =\frac{\sigma}{\sqrt{n}}=4$.
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July 13th, 2017, 11:52 PM   #3
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Thanks! But how do I calculate the probability for the mean of the sample being 58 or higher?
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July 14th, 2017, 12:04 AM   #4
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Quote:
Originally Posted by flowe View Post
Thanks! But how do I calculate the probability for the mean of the sample being 58 or higher?
mathman showed you that the sample mean is distributed as a normal random variable with

$\mu = 50$

$\sigma = 4$

Thus

$\begin {align*}
&P[\bar{X} \geq 58] = \\ \\
&1 - P[\bar{X} < 58] =\\ \\
&1 - \Phi\left(\dfrac{50-58}{4}\right) =\\ \\
&1 - \Phi(-2) \approx \\ \\
&1 - 0.977 = 0.023
\end{align*}$

where $\Phi(x)$ is the CDF of the standard normal distribution
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July 14th, 2017, 12:55 AM   #5
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Question

Thank you for the clarification!

We are using the z-formula (z = X - μ / σ). The formula gives us the probability of any X value in the standard normal distribution - why exactly can we use this to find the probability for the mean of the sample?

If you want to find P for any X value it makes sense to me to use this formula - but I do not understand why the X¯ is represented by this, since it is the mean. Can we handle the P of the mean of the sample like the P of any other X value?

This is how I visualised it for myself: (image attached)
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File Type: jpg Bildschirmfoto 2017-07-14 um 09.55.37.jpg (22.2 KB, 1 views)
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July 14th, 2017, 01:03 AM   #6
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Quote:
Originally Posted by flowe View Post
Thank you for the clarification!

We are using the z-formula (z = X - μ / σ). The formula gives us the probability of any X value in the standard normal distribution - why exactly can we use this to find the probability for the mean of the sample?

If you want to find P for any X value it makes sense to me to use this formula - but I do not understand why the X¯ is represented by this, since it is the mean. Can we handle the P of the mean of the sample like the P of any other X value?

This is how I visualised it for myself: (image attached)
suppose $X$ and $Y$ are iid normal rvs with mean $\mu$ and standard deviation $\sigma$

Then $S=X + Y$ is also a normal rv with mean $\mu_S=2\mu$ and $\sigma_S = \sqrt{2 \sigma^2} = \sqrt{2}\sigma$

If you continue this with $n$ iid normal rvs $X_i$ you find that

$\bar{X} = \dfrac 1 n \sum \limits_{k=1}^n X_i$ is a normal rv with mean $\mu_{\bar{X}} = \mu$ and $\sigma_{\bar{X}} = \dfrac{\sigma}{\sqrt{n}}$

just as mathman showed you.
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July 14th, 2017, 02:20 PM   #7
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Quote:
Originally Posted by flowe View Post
Thanks! But how do I calculate the probability for the mean of the sample being 58 or higher?
I think it would be clearer if you used "mean" for the true mean (50) and "average" for the sample mean, as observed, which has a normal distribution.
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