Suppose a geyser has a mean time between eruptions of 94 minutes. Let the interv

Question

Suppose a geyser has a mean time between eruptions of 94 minutes. Let the interval of time between the eruptions be normally distributed with standard deviation 23 minutes. (a) What is 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 11 time intervals between eruptions has mean longer than 105 minutes? (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 minutes? The probability that the mean of a random sample of 23 time intervals is more than 105 minutes is approximately....(Round to four decimal places as needed.) (d) If the population mean is less than 105 minutes, then the probability that the sample mean of the time between eruptions is greater than 105 minutes increases,decreases because the variability of the sample mean increases,decreases as the sample size decreases, increases. (e) What might you conclude if a random sample of 23 time intervals between eruptions has a mean longer than 105 minutes? Select all that apply. (a)The population mean may be greater than 94. (b) The population mean must be less than 94, since the probability is so low. (c) The population mean is 94, and this is just a rare sampling. (d) The population mean must be more than 94, since the probability is so low. (e) The population mean cannot be 94, since the probability is so low. (f) The population mean is 94, and this is an example of a typical sampling result.

Explanation / Answer

a) From informaation given Xbar=94, s=23, Xi=105. Compute Z score.

Z=(Xi-Xbar)/s=(105-94)/23=0.48

P(X>105)=P(Z>0.48)=0.3156

b) Xbar=94, s=23/sqrt 11=6.93, Xi=105

P(X>105)=P[Z>{(105-94)/6.93}]=P(Z>1.59)=0.0559

c)

Xbar=94, s=23/sqrt 23=4.80, Xi=105

P(X>105)=P[Z>{(105-94)/4.80}]=P(Z>2.29)=0.0110

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