If the mean= 9, Standard deviation= 1.5, then a 95% confidence interval is ?? must show work!

the material given in my online class does not explain well how to answer! please help

A 95% confidence interval covers 95% of the normal curve -- the probability of observing a value outside of this area is less than 0.05. Because the normal curve is symmetric, half of the area is in the left tail of the curve, and the other half of the area is in the right tail of the curve.

A 95% confidence interval for the standard normal distribution, then, is the interval (-1.96, 1.96), since 95% of the area under the curve falls within this interval.

Upper Limit = 9 - (1.96) (0.05) = 8.902
Lower Limit = 9 + (1.96) (0.05) = 9.089

These information is not enough.
I will also need the sample size to decide which test statistic to use and thus to find the confidence interval for u.
Suppose the sample size is n.
Assume that the distribution is normal.

95% confidence interval
= mean +/- Z(0.025)(standard deviation / sqrt(sample size))
= 9 +/- (1.96)(1.5 / sqrt(n))
= 9 +/- (2.94 / sqrt(n))