statistics finding the confidence level of the interval estimate medical expenses A confidence interval for the true mean of the annual medical expenses of a middle-class American family is given as ($ 738, $ 777). If this interval is based on interviews with 110 families and a standard deviation of $ 120 is assumed. Suppose all annual medical expenses of middle-class American families follow an approximately normal distribution.
(a) What is the sample mean of annual medical expenses? 
sample mean = 777 + 738 / 2
            = 757.5

(b) What is the confidence level of the interval estimate (as a decimal)
CI = sample mean + z alpha/2 * (standard deviation / square root n)
Isolate for z alpha/2
777 = 757.5 + z alpha/2 * (120/square root 110)
Rearrange
(777 * square root 110) / (757.5 * 120) = z alpha/2
                        0.089650657 = z alpha/2

What are the correct solutions and answers?
 A: Foremost, you are told that the standard deviation of the expenses is assumed to be $\sigma = 120$.  This is your first hint that the confidence interval was derived using a normal distribution, not a student's $t$ distribution, because the latter uses the sample variance $s$ that is calculated from the data.
Thus, you have correctly inferred that the interval has the form $$[738, 777] = [L, U] = \left[\bar x - z_{\alpha/2} \frac{\sigma}{\sqrt{n}},\; \bar x + z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\right].$$  It follows that $$\bar x = \frac{L + U}{2} = \frac{738+777}{2}$$ as you again correctly compute.  It also follows that $$z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = \frac{U - L}{2},$$ which is a quantity usually given the name "margin of error."  Hence the critical value $$z_{\alpha/2} = \frac{\sqrt{n}}{\sigma} \times \frac{U - L}{2} \approx 1.70431,$$ and if you recall, this means $$\Pr[Z > z_{\alpha/2}] = \Pr[Z > 1.70431] = \alpha/2,$$ where $Z$ is the standard normal distribution.  So you would either use a calculator or a normal distribution table to find that $\alpha/2 \approx 0.0441612$, which means that the provided confidence interval corresponds to a significance level of about $\alpha = 0.088$, or $100(1-\alpha)\% = 91.17\%$ confidence.
A: There is an algebra error while isolating the variable

777 = 757.5 + z alpha/2 * (120/square root 110)

In the next step, you seem to be multiplying "square root 110" through out. May I ask why ?
