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July 13th, 2017, 01:44 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, 02:24 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,628 Thanks: 622 
The average (sample mean) will have a normal distribution with a mean $\displaystyle =\mu =50$ and a standard deviation $\displaystyle =\frac{\sigma}{\sqrt{n}}=4$.

July 13th, 2017, 11:52 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 14th, 2017, 12:04 AM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,174 Thanks: 1143  Quote:
$\mu = 50$ $\sigma = 4$ Thus $\begin {align*} &P[\bar{X} \geq 58] = \\ \\ &1  P[\bar{X} < 58] =\\ \\ &1  \Phi\left(\dfrac{5058}{4}\right) =\\ \\ &1  \Phi(2) \approx \\ \\ &1  0.977 = 0.023 \end{align*}$ where $\Phi(x)$ is the CDF of the standard normal distribution  
July 14th, 2017, 12:55 AM  #5 
Newbie Joined: Jul 2017 From: Germany Posts: 3 Thanks: 0 
Thank you for the clarification! We are using the zformula (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) 
July 14th, 2017, 01:03 AM  #6  
Senior Member Joined: Sep 2015 From: USA Posts: 2,174 Thanks: 1143  Quote:
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, 02:20 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,628 Thanks: 622  

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distribution, normal, random, sample, statistics or probability 
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