Question

Suppose a geyser has 23 minutes. Complete parts (a) through (e) below. mean time between eruptions of 80 minutes. Let the interval of time between the eruptions be normally distributed with standard deviation (a) What is the probability that a randomly selected time interval between eruptions is longer than 91 minutes? The probability that a randomly selected time interval is longer than 91 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 91 minutes? The probability that the mean of a random sample of 9 time intervals is more than 91 minutes is approximately (Round to four decimal places as needed.) (c) What is the probability that a random sample of 29 time intervals between eruptions has a mean longer than 91 minutes? The probability that the mean of a random sample of 29 time intervals is more than 91 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 91 minutes, then the probability that the sample mean of the time between eruptions is greater than 91 minutes hecause the variahility in the sample mean as the samnle size. Click to select your answer(s).

Answer 1

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^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 ) = 80
standard deviation ( sd )= 23
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(a)
P(X > 91) = (91-80)/23
= 11/23 = 0.4783
= P ( Z >0.4783) From Standard Normal Table
= 0.3162
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(b)
sample size (n) = 9
mean of the sampling distribution ( x ) = 80
standard deviation ( sd )= 23/ Sqrt ( 9 ) =7.6667
P(X > 91) = (91-80)/23/ Sqrt ( 9 )
= 11/7.667= 1.4348
= P ( Z >1.4348) From Standard Normal Table
= 0.0757
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(c)
sample size (n) = 29
mean of the sampling distribution ( x ) = 80
standard deviation ( sd )= 23/ Sqrt ( 29 ) =4.271
P(X > 91) = (91-80)/23/ Sqrt ( 29 )
= 11/4.271= 2.5755
= P ( Z >2.5755) From Standard Normal Table
= 0.005
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(d)
From the normal distribution concept, observed that the mean is less
than 87 minutes, then the probability that the sample mean of the time
between eruptions is greater than 87 minute decrease, since the
variability in the sample mean decrease as the sample size increases.