Given a random variable X with distribution N(mu,225) how can I find the sample size n required for a 95% confidence interval for the mean mu with an error of at most epsilon=0.01?

The only formula I can find is n=(z^2*s^2)/.25 Can I somehow use the population variance (take the square root of 225) and substitute it into an equation? I hope someone can help me. My search through websites has only made it worse. THANKYOUUUU!

Since variance is given to you, you can apply the following formula , where and . Can you find and take it from here?

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thankyou! I knew it was something like that. For the 0.025, I would get than answer from the table, but rather than the answer, it is the answer (2.240)? I'm just so amazed they even have that on the table.

?

Did I mess up somewhere? Than number seems way too huge. Thankyou

since the n formula divides by 2, for 95% would it be 0.025/2 for a z value of 2.240? or is it 0.05/2 for a z value of 1.96? Thanks

Since the you're looking at a 95% CI, it follows . So when calculating the z value, you use

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Thank you! That does change the answer quite a bit, from 11289600 to 8643600. Still big, but this time hopefully its right.

Don't be scared of the size of the result...

As you make , (i.e. the smaller you make the error, the larger the sample size will be)

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