# Lost and need HELP!

#### missycl

##### New Member
I need assistance on the following problem (I'll state it and then state my thoughts on it so far):

Scores made on a certain aptitude test by nursing students are approximately normally distributed with a mean of 500 points and a variance of 10,000. A random sample of the students is to be taken. How many students must be selected in the sample so that there is a 90% chance or better of at least one of them having a test score outside the range 304 to 696?

Now, I know the range of scores constitutes 97.5% of the area under the curve (the z value is 1.96). And that the probability that someone scores outside this range is 2.5%. But I am not sure how to find a sample size with a 90% chance of someone being outside the range HELP!

#### JohnM

##### TS Contributor
I think you can solve this by looking at it a different way - how many scores must be sampled before the chance of 0 scores outside the range is 10% or less?

On the first sample, the probability of the score being within the range is .975. The probability of the first two scores being outside the range is .975*.975. For 3 scores, .975*.975*.975. And so on.....

Now, just determine how many times you need to multiply .975 by itself so that it gets to 0.10 or lower. In other words, to what power do you need to raise 0.975 in order to get an answer of 0.10 or less?

0.975^n = 0.10 --> solve for n

Now, using a little bit of algebra, take the natural log of both sides:

n * ln(0.975) = ln(0.10)

n = ln(0.10)/ln(0.975)

n = -2.302/-0.025

n = 90.95

so round up to 91

#### missycl

##### New Member
Yes, I knew I could reverse the problem to use the 10% rather than 90%, but the way to incorporate that percentage at all was throwing me.

Thanks for the response!