# Need help with 2 probabilty questions:

• Oct 22nd 2008, 09:48 PM
jjoaquin
Need help with 2 probabilty questions:
1. If a population has a mean income of $50,000 with a standard deviation of$4000, what is the probability of selecting a random same (n=400) from this population that has a mean $50,200 or greater? 2. On a true/false exam with 10 questions, what is the probability of getting 90% or more just by chance (you didn't study!). any help will do!! • Oct 22nd 2008, 09:59 PM mr fantastic Quote: Originally Posted by jjoaquin 1. If a population has a mean income of$50,000 with a standard deviation of $4000, what is the probability of selecting a random same (n=400) from this population that has a mean$50,200 or greater?

[snip]

1. You should know that $\bar{X}$ ~ Normal $\left( \mu = 50,000, \, \sigma = \frac{4000}{\sqrt{400}}\right)$.

Now calculate $\Pr(\bar{X} > 50, 200)$.
• Oct 22nd 2008, 10:01 PM
mr fantastic
Quote:

Originally Posted by jjoaquin
[snip]
2. On a true/false exam with 10 questions, what is the probability of getting 90% or more just by chance (you didn't study!).

any help will do!!

has what you need.
• Oct 26th 2008, 08:32 PM
jjoaquin
thank you for the link to the probability thread. I got alot of help from it. As for my first question, i'm still stuck on it. Can you post the solution? thanks
• Oct 26th 2008, 09:18 PM
mr fantastic
Quote:

Originally Posted by jjoaquin
thank you for the link to the probability thread. I got alot of help from it. As for my first question, i'm still stuck on it. Can you post the solution? thanks

Where are you stuck? How have you been taught to calculate a probability using a normal distribution?
• Oct 26th 2008, 10:04 PM
jjoaquin
i have the answer, but the problem I have is with the z-score formula. the formula that i use is the x minus the mean divided by the standard deviation. the solution calls for this z-score formula: x minus the mean divided by the standard deviation by the square root of N. i'm confused as to why the solution changes the standard z-score formula. maybe because my prof didn't teach us that yet... basically, i didn't know there was more than one z-score formula
• Oct 26th 2008, 10:16 PM
mr fantastic
Quote:

Originally Posted by jjoaquin
i have the answer, but the problem I have is with the z-score formula. the formula that i use is the x minus the mean divided by the standard deviation. the solution calls for this z-score formula: x minus the mean divided by the standard deviation by the square root of N. i'm confused as to why the solution changes the standard z-score formula. maybe because my prof didn't teach us that yet... basically, i didn't know there was more than one z-score formula

There's not.

If you read my first post you'll see that the sd of the sample mean is 4000/sqrt{400} = 200.

You should know that the standard deviation for the sampling mean is (population mean)/sqrt{sample size} (and I said as much in my first post). So have you been taught about the sampling distribution of the mean or not?

So 200 is the value for sd that you substitute into your z-score formula.
• Oct 26th 2008, 10:20 PM
jjoaquin
we've learned the sampling distrubution of means but he never explained the last part clearly (sqrt of sampling size), so thanks for explaining it to me