need some help (Random selection, mean, SD...)

[oryxncrake]

hi, i posted this on a different math help forum a couple weeks ago but am still confused.

here is the question:

"A survey was conducted for a hospital's emergency room to find out the average waiting time for patients to see a doctor. The data collected in this survey showed that the standard deviation was 30 minutes and the mean was 55 minutes. What is the probability that a patient who is randomly selected has to wait in the emergency room MORE than two hours?

So, since the mean = 55 and the SD = 30, will x = 120? If so, where do I go from here? aren't i supposed to find the z-score prior to doing any of this first?

 

[mmm4444bot]

… will x = 120 ? Yes

If so, where do I go from here? Use the formula to calculate the z-score, then look up the corresponding probability in a table or database.

aren't i supposed to find the z-score prior to doing any of this first? Of what were you thinking, when you wrote "any of this" ?

 

[oryxncrake]

so i should have:

z= (120-55)/30

= 2.1666....

then use 2.1666 to look up the z-score which is 0.9850 ... is this correct?


thanks for the reply, btw

 

[chrisr]

\(\displaystyle With\ a\ mean\ of\ 55\ and\ a\ standard\ deviation\ of\ 30,\)
\(\displaystyle if\ you\ are\ evaluating\ the\ probability\ of\ waiting\ more\ than\ 120\ minutes,\)
\(\displaystyle assuming\ the\ distribution\ of\ waiting\ times\ follows\ the\ Normal\ curve,\)
\(\displaystyle then\ the\ curve\ is\ skewed\ as\ waiting\ times\ cannot\ be\ <0.\)
\(\displaystyle You\ might\ consider\ a\ different\ distribution.\)