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1. A physician wanted to estimate the mean length of time that a patient had to wait to see him after arriving at the office. A random sample of 36 patients showed a mean waiting time of 23.4 minutes and a standard deviation of 7.2 minutes. Find a 96% confidence interval for .
sd/sqrt(n) sounds familiar
\[z=\frac{x-mean}{sd/\sqrt{n}}\]

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Other answers:

where z is the value at Za = .02 maybe?
|dw:1327284026014:dw|
i know im missing something :/
we want the interval of Lx to Rx right?
z = 2.06 if i see the table right
\[z=\frac{x-mean}{sd/\sqrt{n}}\] \[z(sd/\sqrt{n})={x-mean}\] \[z(sd/\sqrt{n})+mean={x}\] \[2.06(7.2/\sqrt{36})+23.4={x}\text{ to the right}\] \[-2.06(7.2/\sqrt{36})+23.4={x}\text{ to the left}\]
with any luck that is :) im sure im forgetting something important tho

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