(a) What is the probability that a randomly selected time interval between erupt

Question

(a) What is the probability that a randomly selected time interval between eruptions is longer than 105 minutes? The probability that a randomly selected time interval is longer than 105 minutes is approximately(Round to four decimal places as needed ) (b) What is the probability that a random sample of 9 time intervals between eruptions has a mean longer than 105 minutes? The probability that the mean of a random sample of 9 time intervals is more than 105 minutes is approximately(Round to four decimal places as needed ) (c) What is the probability that a random sample of 23 time intervals between eruptions has a mean longer than 105 miniutles? The probability that the mean of a random sample of 23 time intervals is more than 105 minutes is approxmatelyRound to four decima easing the sample size have on the probability? Provide an explanation for this result Choose the correct answer below O A. The probability decreases because the variability in the sample mean increases as the sample size inc O B. The probability increases because the variability in the sample mean increases as the sample size increases. O C. The probability decreases because the variability in the sample mean decreases as the sample size increases O D. The probability increases because the variability in the sample mean decreases as the sample size increases (e) What might you conclude if a random sample of 23 time intervals between eruptions has a mean longer than 105 minutes? Choose the best answer below O A. The population mean may be greater than 96 O B. The population mean cannot be 96, since the probability is so low O C. The population mean must be more than 96, since the probability is so low 0 D. The population mean is 96 minutes, and this is an example of a typical sampling Click to select your answer(s) 0 Type here to search

Explanation / Answer

a) Since =96 and =20 we have:

P ( X>105 )=P ( X>10596 )=P ((X)/>(10596)/20)

Since Z=x/ and (10596)/20=0.45 we have:

P ( X>105 )=P ( Z>0.45 )

Use the standard normal table to conclude that:

P (Z>0.45)=0.3264

b) n= 9

Mean= 96

Standard deviation= 20/sqrt(9)= 20/3= 6.67

Since =96 and =6.67 we have:

P ( X>105 )=P ( X>10596 )=P ((X)/>(10596)/6.67)

Since Z=(x)/ and (10596)/6.67=1.35 we have:

P ( X>105 )=P ( Z>1.35 )

Use the standard normal table to conclude that:

P (Z>1.35)=0.0885

c) n= 23

Mean = 96

And Standard deviation = 20/sqrt (23)= 4.17

Since =96 and =4.17 we have:

P ( X>105 )=P ( X>10596 )=P ((X)/>(10596)/4.17)

Since Z=(x)/ and (10596)/4.17=2.16 we have:

P ( X>105 )=P ( Z>2.16 )

Use the standard normal table to conclude that:

P (Z>2.16)=0.0154

d) Probability decrease as variability in the sample mean decreases with increasing the sample size.

e) The population mean can not be greater than 96 Since probability is too low.

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