Question

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). (a) what is the probability that a randomly selected time interval between eruptions is longer than 94 minutes? the probability that a randomly selected time interval is longer than 94 minutes is approximately 0.3372. (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 94 minutes? the probability that the mean of a random sample of 10 time intervals is more than 94 minutes is approximately 0.0901. (round to four decimal places as needed.) (c) what is the probability that a random sample of 34 time intervals between eruptions has a mean longer than 94 minutes? the probability that the mean of a random sample of 34 time intervals is more than 94 minutes is approximately \boxed{} (round to four decimal places as needed.)

Explanation:

Step1: Identify the distribution of the sample mean

The population mean \(\mu = 85\) minutes, population standard deviation \(\sigma = 21.25\) minutes, and sample size \(n = 34\). The sampling distribution of the sample mean \(\bar{X}\) is normal (by the Central Limit Theorem, since \(n\) is reasonably large, or because the population is normal) with mean \(\mu_{\bar{X}}=\mu = 85\) and standard deviation \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{21.25}{\sqrt{34}}\).

First, calculate \(\sigma_{\bar{X}}\):
\(\sqrt{34}\approx5.83095\), so \(\sigma_{\bar{X}}=\frac{21.25}{5.83095}\approx3.644\)

Step2: Calculate the z-score

We want to find \(P(\bar{X}> 94)\). The z-score is calculated as \(z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{94 - 85}{3.644}=\frac{9}{3.644}\approx2.47\)

Step3: Find the probability using the standard normal table

We need to find \(P(Z > 2.47)\). Since \(P(Z > z)=1 - P(Z\leq z)\), we look up \(P(Z\leq2.47)\) in the standard normal table. From the table, \(P(Z\leq2.47) = 0.9932\) (approximate value, more precise value can be obtained from the table or calculator). Then \(P(Z > 2.47)=1 - 0.9932 = 0.0068\) (rounded to four decimal places). Wait, let's check the z-score calculation again. Wait, maybe I made a mistake in the z-score. Wait, \(\mu = 85\), \(\bar{x}=94\), so \(94 - 85 = 9\). \(\sigma_{\bar{X}}=\frac{21.25}{\sqrt{34}}\approx\frac{21.25}{5.83095}\approx3.644\). So \(z=\frac{9}{3.644}\approx2.47\). Now, looking up \(z = 2.47\) in the standard normal table: the value for \(z = 2.47\) is \(0.9932\) (from standard normal table: for \(z = 2.4\), the area is \(0.9918\), for \(z = 2.47\), it's \(0.9932\) (more accurately, using a calculator, \(P(Z\leq2.47)=0.9932\)). So \(P(Z > 2.47)=1 - 0.9932 = 0.0068\)? Wait, but maybe I should use more precise calculation. Wait, let's recalculate \(\sigma_{\bar{X}}\): \(\sqrt{34}\approx5.8309518948\), \(21.25\div5.8309518948\approx3.64435\). Then \(9\div3.64435\approx2.47\). Now, using a more precise standard normal table or calculator, \(P(Z\leq2.47)\): the exact value from the standard normal distribution is \(P(Z\leq2.47)=0.9932\) (from the table: z = 2.47, the cumulative probability is 0.9932). So \(1 - 0.9932 = 0.0068\). Wait, but let's check with a calculator. Alternatively, maybe the initial approach is correct. Wait, but let's re-express the steps.

Wait, the problem is about the sampling distribution of the sample mean. The population is normal (since it's stated that the time between eruptions is normally distributed), so the sampling distribution of the sample mean is normal with mean \(\mu = 85\) and standard deviation \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{21.25}{\sqrt{34}}\). Then, to find \(P(\bar{X}>94)\), we standardize:

\(z=\frac{94 - 85}{\frac{21.25}{\sqrt{34}}}=\frac{9\sqrt{34}}{21.25}\approx\frac{9\times5.83095}{21.25}\approx\frac{52.47855}{21.25}\approx2.47\)

Then, \(P(Z > 2.47)=1 - P(Z\leq2.47)\). From the standard normal table, \(P(Z\leq2.47)=0.9932\), so \(P(Z > 2.47)=1 - 0.9932 = 0.0068\). Rounded to four decimal places, it's approximately \(0.0068\). Wait, but maybe the answer is different. Wait, let's check with a calculator. Using a calculator, the z-score is \(z=\frac{94 - 85}{21.25/\sqrt{34}}=\frac{9}{21.25/\sqrt{34}}\approx\frac{9\times\sqrt{34}}{21.25}\approx\frac{9\times5.83095}{21.25}\approx\frac{52.47855}{21.25}\approx2.47\). Then, using the standard normal distribution, \(P(Z > 2.47)=1 - \Phi(2.47)\), where \(\Phi\) is the cumulative distribution function. \(\Phi(2.47)\) is approximately 0.9932, so \(1 - 0.9932 = 0.0068\). So the probability is approximately \(0.0068\) (rounded to four decimal places).

Answer:

0.0068 (The answer may vary slightly depending on the precision of the z - score and the standard normal table used, but the approximate value is around 0.0068)