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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: 6,805 Thanks: 716 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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Thanks: 0 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. Tags distribution, normal, random, sample, statistics or probability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post IndependentThinker Advanced Statistics 1 February 27th, 2015 02:01 PM nsvcivil Advanced Statistics 2 January 9th, 2012 03:18 PM Freddy7 Algebra 0 October 20th, 2011 08:52 AM oahz Advanced Statistics 1 July 29th, 2010 04:00 PM callkalpa Advanced Statistics 0 May 17th, 2010 06:47 AM

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