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 July 13th, 2017, 12:44 PM #1 Newbie   Joined: Jul 2017 From: Germany Posts: 3 Thanks: 0 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!
 July 13th, 2017, 01:24 PM #2 Global Moderator   Joined: May 2007 Posts: 6,259 Thanks: 510 The average (sample mean) will have a normal distribution with a mean $\displaystyle =\mu =50$ and a standard deviation $\displaystyle =\frac{\sigma}{\sqrt{n}}=4$. Thanks from flowe
 July 13th, 2017, 10:52 PM #3 Newbie   Joined: Jul 2017 From: Germany Posts: 3 Thanks: 0 Thanks! But how do I calculate the probability for the mean of the sample being 58 or higher?
July 13th, 2017, 11:04 PM   #4
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Quote:
 Originally Posted by flowe 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

July 13th, 2017, 11:55 PM   #5
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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)
Attached Images
 Bildschirmfoto 2017-07-14 um 09.55.37.jpg (22.2 KB, 1 views)

July 14th, 2017, 12:03 AM   #6
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Quote:
 Originally Posted by flowe 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.

July 14th, 2017, 01:20 PM   #7
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Quote:
 Originally Posted by flowe 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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