Probability questions

• Nov 21st 2006, 10:39 AM
GoneFishing
Probability questions
If standard deviation is 80 and mean is 10:

P ( X < 100)

How do you do that?

...

And, confidence intervals are very hard. Can someone break how to do a 95% interval down. Say 1000 people were interviewed about smoking and 600 said they did. How would you do that interval? Can do you do it on a calculator?
• Nov 21st 2006, 11:13 AM
galactus
Quote:

And, confidence intervals are very hard. Can someone break how to do a 95% interval down. Say 1000 people were interviewed about smoking and 600 said they did. How would you do that interval? Can do you do it on a calculator?
Confidence intervals aren't difficult. Since you have a proportion, Just use the formula $E=z\sqrt{\frac{pq}{n}}$. This is the margin of error.

p=0.60, q=1-p, n=1000. z that corresponds to 95% CI is 1.96

$E=1.96\sqrt{\frac{(.60)(.40)}{1000}}=0.0304$

0.60+0.0304=0.6304; 0.60-0.0304=0.5696

So, you can say with 95% confidence that the proportion of people who say they smoke is between 56.96% and 63.04%.
• Nov 21st 2006, 11:51 AM
GoneFishing
how do you determine the margin of error for a dataset? is it fixed?
• Nov 21st 2006, 12:22 PM
CaptainBlack
Quote:

Originally Posted by GoneFishing
If standard deviation is 80 and mean is 10:

P ( X < 100)

How do you do that?

let z=(x-10)/80, then if x is normally distributed with mean 10 and sd 80,
then z has a standard normal distribution and:

p(X<100)=p(z<1.125)

and this can be looked up in a table of the standard normal distribution,
and is ~0.87.

RonL
• Nov 21st 2006, 12:26 PM
CaptainBlack
Quote:

Originally Posted by galactus
Confidence intervals aren't difficult. Since you have a proportion, Just use the formula $E=z\sqrt{\frac{pq}{n}}$. This is the margin of error.

p=0.60, q=1-p, n=1000. z that corresponds to 95% CI is 1.96

$E=1.96\sqrt{\frac{(.60)(.40)}{1000}}=0.0304$

0.60+0.0304=0.6304; 0.60-0.0304=0.5696

So, you can say with 95% confidence that the proportion of people who say they smoke is between 56.96% and 63.04%.

You need to say that you are using the normal approximation to the
binomial distribution here. The 1.96 only applies to the normal distribution
which is justified here as we have a large population and we are not
skulking in the far skirts of the distribution.

RonL