Suppose a geyser has a mean time between eruptions of 90 minutes. Let the interv

Question

Suppose a geyser has a mean time between eruptions of 90 minutes. Let the interval of time between the eruptions be normally distributed with standard deviation 30 minutes. Complete parts (a) through (e) below. (a) What is the probability that a randomly selected time interval between eruptions is longer than 103 minutes? The probability that a randomly selected time interval is longer than 103 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 a mean longer than 103 minutes? The probability that the mean of a random sample of 11 time intervals is more than 103 minutes is approximately (Round to four decimal places as needed) (c) What is the probability that a random sample of 30 time intervals between eruptions has a mean longer than 103 minutes? The probability that the mean of a random sample of 30 time intervals is more than 103 minutes is approximately Round to four decimal places as needed.) (d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. Fill in the blanks below. If the population mean is less than 103 minutes, then the probability that the sample mean of the time between eruptions is greater than 103 minutes sample size V as the Y because the variability in the sample mean

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 90
standard Deviation ( sd )= 30
a.
P(X > 103) = (103-90)/30
= 13/30 = 0.4333
= P ( Z >0.4333) From Standard Normal Table
= 0.3324
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b.
mean of the sampling distribution ( x ) = 90
standard Deviation ( sd )= 30/ Sqrt ( 11 ) =9.0453
P(X > 103) = (103-90)/30/ Sqrt ( 11 )
= 13/9.045= 1.4372
= P ( Z >1.4372) From Standard Normal Table
= 0.0753
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P(X > 103) = (103-90)/30/ Sqrt ( 30 )
= 13/5.477= 2.3735
= P ( Z >2.3735) From Standard Normal Table
= 0.0088
d.
the missing words are decrease, decrease and increase respectively.
e.
The population mean must be less than 103, since the probability low.

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