Why is my answer wrong? (statistics)

[Artusartos]

Let $\bar{X}$ be the mean of a random sample of size n from a distribution that is $N(\mu,9)$. Find n such that $P(\bar{X}-1 < \mu < \bar{X}+1)=.90$, approximately.

My answer:

$$-Z^* < \frac{\bar{X}-\mu}{9/\sqrt(n)} < Z^*$$ where $-Z^*$ and $Z^*$ are the critical values.

So...

For the confidence interval, we have $$\bar{x} \pm z^*(\frac{9}{\sqrt(n)})$$

When I looked up the normal table for $z^*$, I found that it was approximately equal to 1.29Since the question tells us that the confidence interval is $$\bar{X} \pm 1$$, so I just solved $$1 = (1.29)(\frac{9}{\sqrt(n)})$$...but my answer was wrong...it was supposed to be 24 or 25. Can anybody please help?

Thanks in advance

 

[mfb]

A probability of 0.9 to be inside gives a probability of 0.05 for both sides. You have to look for 0.95 and not 0.90. In addition, this value should be at the other side of your =.

 

[Ray Vickson]

Let $\bar{X}$ be the mean of a random sample of size n from a distribution that is $N(\mu,9)$. Find n such that $P(\bar{X}-1 < \mu < \bar{X}+1)=.90$, approximately.

My answer:

$$-Z^* < \frac{\bar{X}-\mu}{9/\sqrt(n)} < Z^*$$ where $-Z^*$ and $Z^*$ are the critical values.

So...

For the confidence interval, we have $$\bar{x} \pm z^*(\frac{9}{\sqrt(n)})$$

When I looked up the normal table for $z^*$, I found that it was approximately equal to 1.29Since the question tells us that the confidence interval is $$\bar{X} \pm 1$$, so I just solved $$1 = (1.29)(\frac{9}{\sqrt(n)})$$...but my answer was wrong...it was supposed to be 24 or 25. Can anybody please help?

Thanks in advance

Besides what 'mfb' has told you, there is always the issue of what the '9' means in N(μ,9). Most commonly, the notation is $N(\mu, \sigma^2),$ so that $\sigma = 3.$ I have seen the other notation $N(\mu,\sigma)$ used occasionally, but it is rarer. You need to check the convention used by your textbook and/or course notes. (I won't spoil your fun by telling you the answer.)