Suppose a geyser has a mean time between eruptions of 67 minutes Let the interva

Question

Suppose a geyser has a mean time between eruptions of 67 minutes Let the interval of time between the eruptions be normally distributed with standard deviation 20 minutes Complete through (c) below. The probability that the mean of a random sample of 11 time intervals is more than 75 minutes is approximately Witt] (Round to four decimal places as needed) What is the probability that a random sample of 42 time intervals between eruptions has a mean longer than 75 minutes? The probability that the mean of a random sample of 42 time intervals is more than 75 minutes is approximately 0048 (Round to four decimal places as needed.) What effect docs increasing the sample size have on the probability0 Provide an explanation for this result. Fill in the blanks below. If the population mean is less than 75 minutes, then the probability that the sample mean of the time between eruptions is greater than 75 minutes decreases because the variability m sample mean' decreases as the sample size increases. What might you conclude if a random sample of 42 time intervals between eruptions has a mean longer than 75 minutes? Select all that apply. The population mean may be less than 67 The population mean is 67, and this is an example of a typical sampling result The population mean is 67. and this is just a rare sampling The population mean must be more than 67, since the probability is so low The population mean cannot be 67, since the probability is so low The population mean must he less than 67, since the probability is so low The population mean may be greater than 67

Explanation / Answer

b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    75      
u = mean =    67      
n = sample size =    11      
s = standard deviation =    20      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.326649916      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.326649916   ) =    0.092312258 [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 =    75      
u = mean =    67      
n = sample size =    42      
s = standard deviation =    20      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.592296279      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.592296279   ) =    0.004766881 [ANSWER]

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d)

i. DECREASES
ii. DECREASES
iii. INCREASES

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e)

As this is a very low probability from part c), we are likely wrong that the true mean is 67, it is likely higher.

Hence,

OPTION G: u may be greater than 67. [ANSWER]

[Other options such as options D and E could also be correct, but they seem too confident that the mean is not 67. I would just say "maybe", but these options seem so sure. It's really up to you and what you usually do in class. Thanks!]

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