The general manager of an amusement park wants to estimate the length of time people spend in the park. From the data collected over several months it was found that the average time an individual spends in the park is 3.4 hrs with a standard deviation of 1.1 hrs. It was not known whether the distribution is normal but it appears to be mound-shaped. Thirty five people just got off the city bus and went into the amusement park. Find the probability that the sample mean time these people spend in the park is more than 3 hrs.

Ok...so I solved it like this...

z= x - m / stand.dev

z = 3 - 3.4 / 1.1 = -.09 = .4641; 1 - .4641 = .5359