MHB Standard Deviation: Get 10 Cents on Probability Problem

AI Thread Summary
The discussion revolves around calculating the probability using standard deviation and z-scores, with one participant initially arriving at a probability of 2.14%. It is confirmed that this percentage is correct for the probability between 2 and 3 standard deviations from the mean. However, the participant faced issues with an online system not recognizing 2.14 as the correct answer. Another user suggests that the problem may stem from entering a percentage instead of the decimal value of the probability, which should be 0.0214. Ultimately, a glitch in the system was identified as the source of the confusion.
Coder74
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Hey everyone!

I'm learning about some statistics and its past office hours with my teacher but I'm stuck on this problem.. I came up with 2.14% as the probabillity\final answer..
Could you guys give me your 10 cents on this?

Thanks again!

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You need to convert the two raw data (called $x$) into $z$-scores:

$$z=\frac{x-\mu}{\sigma}$$

Then, use a table to find the area under th standard normal curve between the two $x$-scores to find the requested probability.

Can you proceed?
 
Coder74 said:
Hey everyone!

I'm learning about some statistics and its past office hours with my teacher but I'm stuck on this problem.. I came up with 2.14% as the probabillity\final answer..
Could you guys give me your 10 cents on this?

Thanks again!

Hi Coder74,

2.14% is the correct answer.
It's the probability between 2 and 3 standard deviations from the mean.
 
Thanks for both replies you guys I appreciate it!
However since this is an online schooling when I entered 2.14 as an answer it wouldn't register as correct..
 
Coder74 said:
Thanks for both replies you guys I appreciate it!
However since this is an online schooling when I entered 2.14 as an answer it wouldn't register as correct..

Consulting the table in my old stats textbook, I find:

$$P(X)\approx0.4987-0.4772=0.0215$$

When I use a numeric scheme to approximate the integral I get:

$$P(X)=\frac{1}{\sqrt{2\pi}}\int_2^3 e^{-\frac{x^2}{2}}\,dx\approx0.0214002339165491$$

Perhaps this issue is you are entering a percentage, and the app is expecting the value of probability, i.e. 0.0214. :D
 
Thanks, Mark! As it turns out there was a glitch in the system after all.. Haha, thanks for the help everyone!
 
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