Homework SourseThe mean time to download the homepage of a popular hotel in
The mean time to download the homepage of a popular hotel in town is 1.2 second. Suppose that the download time is normally distributed with a standard deviation of 0.6 second. Suppose that you select a random sample of 36 download times:
1) What is the probability that the sample mean is more than 1 second?
2) The probability is 0.6 that the sample mean is between what two symmetrical values?
3) The probability is 0.2 that the sample mean is different than the population mean by at least how many seconds?
Solution
1.
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 1
u = mean = 1.2
n = sample size = 36
s = standard deviation = 0.6
Thus,
z = (x - u) * sqrt(n) / s = -2
Thus, using a table/technology, the right tailed area of this is
P(z > -2 ) = 0.977249868 [ANSWER]
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2.
As the middle area is
Middle Area = P(x1<x<x2) = 0.6
Then the left tailed area of the left endpoint is
P(x<x1) = (1-P(x1<x<x2))/2 = 0.2
Thus, the z score corresponding to the left endpoint, by table/technology, is
z1 = -0.841621234
By symmetry,
z2 = 0.841621234
As
u = mean = 1.2
s = standard deviation = s/sqrt(n) = 0.6/sqrt(36) = 0.1
Then
x1 = u + z1*s = 1.115837877
x2 = u + z2*s = 1.284162123 [ANSWERS]
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3.
This is represented by the distance of the 10th and 90th percentiles from the mean.
First, we get the z score from the given left tailed area. As
Left tailed area = 0.9
Then, using table or technology,
z = 1.281551566
As
E = z * s,
where
z = the critical z score = 1.281551566
s = standard deviation = 0.1
Then
x = critical value = 0.128155157 [ANSWER]
