Standard Deviation: Get 10 Cents on Probability Problem

[Coder74]

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!

View attachment 6508

 

[MarkFL]

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

\(\displaystyle z=\frac{x-\mu}{\sigma}\)

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?

 

[I like Serena]

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.

 

[Coder74]

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..

 

[MarkFL]

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:

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

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

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

Perhaps this issue is you are entering a percentage, and the app is expecting the value of probability, i.e. 0.0214. :D

 

[Coder74]

Thanks, Mark! As it turns out there was a glitch in the system after all.. Haha, thanks for the help everyone!