Suppose a geyser has a mean lime between eruption of 71 minutes. Let the interva

Question

Suppose a geyser has a mean lime between eruption of 71 minutes. Let the interval of Une between the eruptions be noonaty distributed with standard deviation 24 minules. Compete parts (a) through (e) below What Is the probabtvty that a randomly selected time interval between erupbans is longer than 82 minutes' The probability that a randomly selected ume intervalislonger than 82 minutes is approximately What is the probaNity that a random sample ol 13 time intervals between eruptions has a mean longer than 82 nMMt The probability that the mean of a random sampfe of 13 time intervals is more than 82 minutes is approximately What Is the probabiWy mat a random sample ol 27 tune tervals between erupbons has a mean longer than 82 minutes? The probability that the mean of a random sample of 27 time intervals is more than 82 minutes is approximately What effect does increasing the sample size have on the probability' Provide an explanation or ttvs result FM m the blanks betow If the population mean is less than 82 mwiutes. then the probaMity that the sample mean of the time between eruptons is greater than 82 minutes What might you conclude of a random sample of 27 fcme intervals between eruptions has a mean longer than 82 mmules? Select at that apply The population mean must be less than 71. since the probabMy is so low. The population mean is 71, and this is an example of a typical sampung result The population mean may be greater than 71. The population mean must be more than 71. since the probabilty is so low. The population mean may be less than 71. The population mean is 71. and ttvs is just a rare sampling The population mean cannot be 71. 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 =    82      
u = mean =    71      
          
s = standard deviation =    24      
          
Thus,          
          
z = (x - u) / s =    0.458333333      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.458333333   ) =    0.32335649 [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 =    82      
u = mean =    71      
n = sample size =    13      
s = standard deviation =    24      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.652544335      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.652544335   ) =    0.049211818 [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 =    82      
u = mean =    71      
n = sample size =    27      
s = standard deviation =    24      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.38156986      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.38156986   ) =    0.00861951 [ANSWER]

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d)

First blank: DECREASES
2nd: DECREASES
3rd: INCREASES

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e)

As it is highly unlikely to get an average greater than 82 if the true mean is really 71, then it is likely that the true mean is greater than 71 (must be closer to 82).

OPTION C. [ANSWER]
OPTION G. [ANSWER]

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