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i have a small statistics problem:

The time spent by students to solve an exam, following a normal distribution, has a mean of 80 minutes, and a standard deviation of 20 minutes.

.. and asks:

What's the minimum time that 30% of the students spent on it?

Can anyone help me finding the way to solve this?

Thanks

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This must be a homework. If it is, please read on how to ask a homework question on this site. –  Sasha Nov 13 '11 at 17:24
    
This should be one of the "general" types of questions asked for in the meta. I can't think of how to phrase it in a general way though that would be recognizable to someone who'd ask it... –  David Mitra Nov 13 '11 at 17:32
2  
This question (and its variants) has been hashed and rehashed any number of times on stats.SE. I would vote to close, if I could. –  cardinal Nov 13 '11 at 17:33
    
It's not homework, i'm just practicing on solving statistics problems for my exam. –  Inside Nov 13 '11 at 18:40
    
@Inside: For future reference, Sasha's comment (and associated link) still apply. The more effort you put into providing context for your question, the more likely people will make the effort to provide a nice response. In this case, David has already provided essentially the complete answer. –  cardinal Nov 13 '11 at 18:45

1 Answer 1

up vote 0 down vote accepted

Let $X=$the time spent by a student on the test. $X$ has normal distribution with mean 80 and standard deviation 20. You need to solve $$P[X>a]=.3$$ Passing to the standard normal distribution, need to solve $$P[{X-80\over 20}>{a-80\over20} ]=.3,$$ or, $$ [Z>{a-80\over20} ]=.3 $$ where $Z$ is the standard normal variable. You need to do a "reverse lookup": go to a table/calculator for the CDF of the standard normal and find the value $z$ with $P[Z>z]=.3$. Call this $z_3$. Then set $$ z_3= {a-80\over20} $$ and solve for $a$.

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