Question

question 5, 8.1.22 part 6 of 6 hw score: 50%, 4 of 8 points o points: 0 of 1 the most famous geyser in the world, old faithful in yellowstone national park, has a mean time between eruptions of 85 minutes. if the interval of time between the eruptions is normally distributed with standard deviation 21.25 minutes, complete parts (a) through (f). click here to view the standard normal distribution table (page 1). click here to view the standard normal distribution table (page 2). if the sample size increases, the probability decreases because the variability in the sample mean decreases. (e) what might you conclude if a random sample of 34 time intervals between eruptions has a mean longer than 94 minutes? select all that apply. a. the population mean is probably greater than 85 minutes. b. the population mean is probably less than 85 minutes. c. the population mean is 85 minutes, and this is just a rare sampling result. d. the population mean must be less than 85 minutes, since the probability is so low. e. the population mean is 85 minutes, and this is an example of a typical sampling result. f. the population mean cannot be 85 minutes, since the probability is so low. g. the population mean must be more than 85 minutes, since the probability is so low. (f) on the start of a certain day, suppose there are 23 time intervals for old faithful. treating these 23 intervals as a random sample, there is a 0.20 likelihood that the mean length of time between eruptions will exceed what value? the likelihood the mean length of time between eruptions exceeds minutes is 0.20. (round to one decimal place as needed.)

Explanation:

For part (e), when a random sample has a mean different from the population mean (here, sample mean > 85, population mean is 85? Wait, no—wait, the population mean for Old Faithful is 85 minutes? Wait, no, the interval between eruptions is normally distributed with mean 85 minutes? Wait, the sample of 34 has a mean longer than 94? Wait, no, the question is: if a random sample of 34 time intervals has a mean longer than 94 minutes, what can we conclude?

Wait, the options:

A: The population mean is probably greater than 85 minutes.

B: The population mean is probably less than 85 minutes.

C: The population mean is 85 minutes, and this is an example of a typical sampling result.

D: The population mean must be less than 85 minutes, since the probability is so low.

E: The population mean is 85 minutes, and this is just a rare sampling.

F: The population mean cannot be 85 minutes, since the probability is so low.

G: The population mean must be more than 85 minutes, since the probability is so low.

Wait, if the sample mean is much higher than the supposed population mean (85), a rare sample mean (like mean >94) would suggest that maybe the population mean is actually higher, because if the population mean were 85, getting a sample mean that high would be very unlikely. So option A makes sense: the population mean is probably greater than 85 minutes. Also, options F and G are wrong because "must" is too strong. E: if it's rare, then it's not typical. C: sample mean much higher than 85 isn't typical. B and D: sample mean is higher, so population mean being less is unlikely. So correct options are A (and maybe others? Wait the green check is on A. So the conclusion is that the population mean is probably greater than 85 minutes, because a sample mean that high is unlikely if the population mean were 85.

For part (f): We need to find the value \( x \) such that \( P(\bar{X} > x) = 0.20 \), where \( \bar{X} \) is the sample mean of \( n = 23 \) intervals. The population mean \( \mu = 85 \), standard deviation \( \sigma = 21.25 \). The sampling distribution of \( \bar{X} \) is normal with mean \( \mu_{\bar{X}} = \mu = 85 \) and standard deviation \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{21.25}{\sqrt{23}} \).

First, calculate \( \sigma_{\bar{X}} \): \( \sqrt{23} \approx 4.7958 \), so \( \sigma_{\bar{X}} \approx \frac{21.25}{4.7958} \approx 4.431 \).

We need \( P(\bar{X} > x) = 0.20 \), so \( P(\bar{X} \leq x) = 0.80 \). The z-score corresponding to the 80th percentile is \( z = 0.84 \) (from standard normal table: \( \Phi(0.84) \approx 0.8000 \)).

Then, \( x = \mu_{\bar{X}} + z \cdot \sigma_{\bar{X}} = 85 + 0.84 \cdot 4.431 \approx 85 + 3.722 = 88.722 \), rounded to one decimal place: 88.7.

Wait, let's check: \( z = 0.84 \) (since \( P(Z \leq 0.84) \approx 0.80 \)). So \( x = 85 + 0.84*(21.25/sqrt(23)) \). Let's recalculate \( 21.25 / sqrt(23) \):

\( sqrt(23) ≈ 4.79583 \)

\( 21.25 / 4.79583 ≈ 4.431 \)

\( 0.84 * 4.431 ≈ 3.722 \)

\( 85 + 3.722 ≈ 88.722 \), so 88.7 when rounded to one decimal.

Answer:

(e) The correct conclusion is: A. The population mean is probably greater than 85 minutes.
(f) The value is \(\boxed{88.7}\) (rounded to one decimal place).