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 eruptions is normally distributed with standard deviation 21.25 minutes, answer the following questions: (a) What is the probability that a randomly selected time interval between eruptions is longer than 95 minutes? (b) What is the probability that a random sample of 20 time intervals between eruptions has a mean longer than 95 minutes? (c) What is the probability that a random sample of 30 time intervals between eruptions has a mean longer than 95 minutes? (d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. (e) What might you conclude if a random sample of 30 time intervals between eruptions has a mean longer than 95 minutes? (f) On a certain day, suppose there are 22 time intervals for Old Faithful. Treating these 22 eruptions as a random sample, the likelihood the mean length of time between eruptions exceeds ____ minutes is 0.20.

d) Increasing the sample size narrows the sampling distribution. Larger samples will have a mean which has more probability of being closer to the population mean. This is reflected by the decrease in the standard deviation of the sample mean when the sample size increases.

e) If a random sample of 30 time intervals between eruptions has a mean longer than 95 minutes, i can conclude that the mean time between eruptions may not be 85 minutes. This sample result is such an unlikely result that it can serve as evidence to question the validity of the mean time between eruptions.

f) Treating these 22 eruptions as a random sample, the likelihood the mean length of time between eruptions exceeds 88.8 minutes is 0.20.

Step-by-step explanation:

We have the the time between eruptions as normally distributed with mean = 85 min. and standard deviation = 25.25.

a) We have to calculate P(x>95)

[tex]z=(X-\mu)/\sigma=(95-85)/21.25=10/21.25=0.47\\\\P(X>95)=P(z>0.47)=0.32[/tex]

b) The sampling distribution of samples of size n=20 is

[tex]\mu_s=\mu=85\\\\\sigma_s=\sigma/\sqrt{n}=21.25/\sqrt{20}=4.75[/tex]

The probability that the sample mean is longer than 95 is

[tex]z=(95-85)/4.75=10/4.75=2.11\\\\P(M_{20}>95)=P(z>2.11)=0.02[/tex]

c) Now, with n=30, the mean and standard deviation becomes:

[tex]\mu_s=\mu=85\\\\\sigma_s=\sigma/\sqrt{n}=21.25/\sqrt{30}=3.88[/tex]

[tex]z=(95-85)/3.88=10/3.88=2.58\\\\P(M_{30}>95)=P(z>2.58)=0.005[/tex]

f) The sample size is n=22

[tex]\mu_s=\mu=85\\\\\sigma_s=\sigma/\sqrt{n}=21.25/\sqrt{22}=4.53[/tex]

We have to calculate x so that P(X>x)=0.2

This probability happens for a z=0.84.

Then, we can calculate x as:

[tex]x=\mu+z*\sigma=85+0.84*4.53=85+3.8=88.8[/tex]

Treating these 22 eruptions as a random sample, the likelihood the mean length of time between eruptions exceeds 88.8 minutes is 0.20.

ogorwyne ogorwyne · Answer 2 · 2022-04-08T19:53:37+02:00

The probability that the randomly selected time interval is longer than 95 minutes while solving for the mean time of the eruptions is 0.3192.

How to solve for the probability of a ramdom sample

mean of x = 85 minutes

σ = 21.25 minutes

x ~ N(σ², μ)

p(x < 95)

1-p(x≤95)

[tex]1-z(\frac{95-85}{2.25} )[/tex]

1 - p(z≤0.47)

Using the z table here

1-0.6808 = 0.3192

Probability of x > 95 = 0.3192

b. n = 20

[tex]1-p(z < \frac{95-85}{21.25/\sqrt{20} } )[/tex]

from central limit theorem

1 - p (z < 2.1045)

1-0.99826 = 0.0174

c. n = 30

[tex]1-p(z < \frac{95-85}{21.25/\sqrt{30} } )[/tex]

1 - p (z < 2.58)

1-0.9951

= 0.0049

d. What we can observe from these results is that as the sample increase is increasing from 1, 20 to 30, the probability keeps decreasing.

e. The probability is very small so one has to be doubtful if they are a really the population mean or if it has to be more than that.

Read more on probability here: https://brainly.com/question/24756209