Suppose a geyser has a mean time between eruptions of 65 min. Let the interval o
ID: 3125754 • Letter: S
Question
Suppose a geyser has a mean time between eruptions of 65 min. Let the interval of time between the eruptions be normally distributed with standard deviation 19 min. Complete parts (a) through (e) below.
a. What is the probability that a randomly selected time interval between eruptions is longer than 73 min.?
Answer: The probability that a randomly selected time interval is longer than 73 minutes is approximately _______.
b. What is the probability that a random sample of 15 time intervals between eruptions has a mean longer than 73 min.?
Answer: The probability that mean of a random sample of 15 time intervals is more than 73 minutes is approximately _______.
c. What is the probability that a random sample of 33 time intervals between eruptions has a mean longer than 73 min?
Answer: The probability that the mean of a random sample of 33 time intervals is more than 73 minutes is approximately _________.
d. What effect does increasing the sample size have on the probability? Provide an explanation for this result. Fill in the blanks below.
Answer: If the population mean is less than 73 minutes, then the probability that the sample mean of the time between eruption is greater than 73 minutes (increases/decreases) because the variability in the sample mean (increases/decreases) as the sample size (increases/decreases).
e. What might you conclude if a random sample of 33 time intervals between eruptions has a mean longer than 73 minutes? Select all that apply.
__ The population mean may be greater than 65.
__ The population mean is 65, and this is just a rare sampling
__ The population mean must be less than 65, since the probability is so low
__ The population mean may be less than 65
__ The population mean must be more than 65, since the probability is so low
__ The population mean is 65, and this is an example of a typical sampling result
__ The population mean cannot be 65, since the probability is so low
Explanation / Answer
A)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 73
u = mean = 65
s = standard deviation = 19
Thus,
z = (x - u) / s = 0.421052632
Thus, using a table/technology, the right tailed area of this is
P(z > 0.421052632 ) = 0.336858325 [ANSWER]
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b)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 73
u = mean = 65
n = sample size = 15
s = standard deviation = 19
Thus,
z = (x - u) * sqrt(n) / s = 1.63072983
Thus, using a table/technology, the right tailed area of this is
P(z > 1.63072983 ) = 0.05147367 [ANSWER]
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c)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 73
u = mean = 65
n = sample size = 33
s = standard deviation = 19
Thus,
z = (x - u) * sqrt(n) / s = 2.41876322
Thus, using a table/technology, the right tailed area of this is
P(z > 2.41876322 ) = 0.007786687 [ANSWER]
*****************
d)
If the population mean is less than 73 minutes, then the probability that the sample mean of the time between eruption is greater than 73 minutes DECREASES because the variability in the sample mean DECREASES as the sample size INCREASES. [ANSWER]
*****************
e)
The population mean may be greater than 65. [ANSWER]
The population mean must be more than 65, since the probability is so low. [ANSWER]
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