SOLUTION: Hi! I'm stumped on the second part of this question. Could someone explain the steps for me please?
The mean travel time to work in the US is 25.1 minutes with a standard deviat
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-> SOLUTION: Hi! I'm stumped on the second part of this question. Could someone explain the steps for me please?
The mean travel time to work in the US is 25.1 minutes with a standard deviat
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Question 1138417: Hi! I'm stumped on the second part of this question. Could someone explain the steps for me please?
The mean travel time to work in the US is 25.1 minutes with a standard deviation of 6.4 minutes.
a. Find the probability that a random sample of 36 people will have a mean travel time greater than 23 minutes. (I got 0.9755)
b. Find the 90th percentile of the sample mean for those 36 people. (The answer key says it's 26.467, but I cannot figure out the steps!)
Thanks in advance. :) Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! z>(x bar-mean)/sigma/sqrt (n)
z>(23-25.1)*6/6.4
z>-1.96875
0.9755 is correct
z for the 90th percentile is 1.28
1.28=(x bar-mean)/sigma/sqrt (n) or (x bar-mean) divided by 6.4/6, which is 1.067
8.19=6(x bar-mean)
1.367=x bar-mean
x bar=26.467 min
