Question

Suppose a geyser has a mean time between eruptions of 82 minutes. Let the interval of time between the eruptions be normally distributed with standard deviation 19 minutes. Complete parts (a) through (e) below. (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 10 time intervals between eruptions has a mean longer than 91 minutes? The probability that the mean of a random sample of 10 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 25 time intervals between eruptions has a mean longer than 91 minutes? The probability that the mean of a random sample of 25 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 because the variability in the sample mean as the sample size (e) What might you conclude if a random sample of 25 time intervals between eruptions has a mean longer than 91 minutes? Select all that apply. A. The population mean must be more than 82, since the probability is so low. B. The population mean may be less than 82. C. The population mean is 82, and this is an example of a typical sampling result OD. The population mean may be greater than 82 E. The populat mean must be less than 82, since the probability is so low. F. The population mean cannot be 82, since the probability is so low. G. The population mean 82, and this is just a rare sampling.

Answer 1

Answer:

Given,

Mean = 82

Standard deviation = 19

a)

P(X > 91) = P(z > (91 - 82)/19)

= P(z > 0.47)

= 0.3191775 [since from z table]

= 0.3192

b)

P(X > 91) = P(z > (91 - 82)/(19/sqrt(10)))

= P(z > 1.50)

= 0.0668072 [since from z table]

= 0.0668

c)

P(X > 91) = P(z > (91 - 82)/(19/sqrt(25)))

= P(z > 2.37)

= 0.008894  [since from z table]

= 0.0089

d)

Decreases due to the variability decreases as sample size increases.

e)

A, D , G