Mathematics: List below how times interval between eruption of a geyser.

List below how times interval between eruption, №12918091, 14.09.2021 16:30

Mathematics

Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally distributed with standard deviation 23 minutes. Complete parts (a) through (e) below. (a) What is the probability that a randomly selected time interval between eruptions is longer than 82minutes? The probability that a randomly selected time interval is longer than 82 minutes is approximately nothing. (Round to four decimal places as needed.) (b) What is the probability that a random sample of 13 time intervals between eruptions

Step-by-step answer

P Answered by PhD

The probability that a randomly selected time interval between eruptions is longer than 82minutes = 0.3336

The probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 minutes = 0.0594

The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes = 0.0057

Step-by-step explanation:

From the given data

mean, u = 72

Standard deviation $\rho$ = 23

A) Probability that a randomly selected time interval between eruptions is longer than 82minutes

$P (x 82) = P[\frac{x-u}{\rho} \frac{82-72}{23}]\\\\P (x 82) = P[z 0.43]\\\\P (x 82) = 0.3336$

B)

$P (x 82) = P[\frac{x-u}{\frac{\rho}{\sqrtn}} \frac{82-72}{\frac{23}{\sqrt{13}}}]\\\\P (x 82) = P[z 1.5676]\\\\P (x 82) = 0.0594$

C)

$P (x 82) = P[\frac{x-u}{\frac{\rho}{\sqrtn}} \frac{82-72}{\frac{23}{\sqrt{34}}}]\\\\P (x 82) = P[z 2.5351]\\\\P (x 82) = 0.0057\\\\$

D) If the mean is less than 82minutes, then the probability that the sample mean of the time between eruptions is greater than 83 minutes decrease because the variability in the sample mean decrease as the sample size increases

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Mathematics

Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally distributed with standard deviation 23 minutes. Complete parts (a) through (e) below. (a) What is the probability that a randomly selected time interval between eruptions is longer than 82minutes? The probability that a randomly selected time interval is longer than 82 minutes is approximately nothing. (Round to four decimal places as needed.) (b) What is the probability that a random sample of 13 time intervals between eruptions

Step-by-step answer

P Answered by PhD

The probability that a randomly selected time interval between eruptions is longer than 82minutes = 0.3336

The probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 minutes = 0.0594

The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes = 0.0057

Step-by-step explanation:

From the given data

mean, u = 72

Standard deviation $\rho$ = 23

A) Probability that a randomly selected time interval between eruptions is longer than 82minutes

$P (x 82) = P[\frac{x-u}{\rho} \frac{82-72}{23}]\\\\P (x 82) = P[z 0.43]\\\\P (x 82) = 0.3336$

B)

$P (x 82) = P[\frac{x-u}{\frac{\rho}{\sqrtn}} \frac{82-72}{\frac{23}{\sqrt{13}}}]\\\\P (x 82) = P[z 1.5676]\\\\P (x 82) = 0.0594$

C)

$P (x 82) = P[\frac{x-u}{\frac{\rho}{\sqrtn}} \frac{82-72}{\frac{23}{\sqrt{34}}}]\\\\P (x 82) = P[z 2.5351]\\\\P (x 82) = 0.0057\\\\$

D) If the mean is less than 82minutes, then the probability that the sample mean of the time between eruptions is greater than 83 minutes decrease because the variability in the sample mean decrease as the sample size increases

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Mathematics

Suppose a geyser has a mean time between eruptions of 65 min. Let the interval of time between the eruptions be normally distributed with standard deviation 19 min. Complete parts (a) through (e) below. a. What is the probability that a randomly selected time interval between eruptions is longer than 73 min.? b. What is the probability that a random sample of 15 time intervals between eruptions has a mean longer than 73 min.? c. What is the probability that a random sample of 33 time intervals between eruptions has a mean longer than 73 min? d.

Step-by-step answer

P Answered by Master

The answer is provided in the attachment

Suppose a geyser has a mean time between eruptions of 65 min. Let the interval of time between the e

Mathematics

Suppose a geyser has a mean time between eruptions of 65 min. Let the interval of time between the eruptions be normally distributed with standard deviation 19 min. Complete parts (a) through (e) below. a. What is the probability that a randomly selected time interval between eruptions is longer than 73 min.? b. What is the probability that a random sample of 15 time intervals between eruptions has a mean longer than 73 min.? c. What is the probability that a random sample of 33 time intervals between eruptions has a mean longer than 73 min? d.

Step-by-step answer

P Answered by Master

The answer is provided in the attachment

Mathematics

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally distributed with a standard deviation 20 minutes, answer the following questions. (A) What is the probability that a randomly selected Time interval between irruption’s is longer than 84 minutes? (B) what is the probability that a random sample of 13 time intervals between irruption‘s has a mean longer than 84 minutes? (C) what is the probability that a random sample of 20 time intervals between irruption‘s has a mean longer

Step-by-step answer

P Answered by PhD

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

Mathematics

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally distributed with a standard deviation 20 minutes, answer the following questions. (A) What is the probability that a randomly selected Time interval between irruption’s is longer than 84 minutes? (B) what is the probability that a random sample of 13 time intervals between irruption‘s has a mean longer than 84 minutes? (C) what is the probability that a random sample of 20 time intervals between irruption‘s has a mean longer

Step-by-step answer

P Answered by PhD

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

Social Studies

Can we predict the next eruption of the old faithful geyser by knowing the duration of the previous eruption? the observations and least squares regression line appear in the scatterplot. the scatterplot and the least-squares regression line for predicting the interval (minutes) between eruptions and the duration (minutes) of the previous eruption is given below. which of the following is the correct interpretation of the slope of the least squares regression line for the data about old faithful?

Step-by-step answer

P Answered by PhD

1

The correct answer is We can predict the time between eruptions will be 10.73 minutes and this can be stated with 0.901 probability.

Eruptions do not always produce symmetrical and imposing cones. Volcanic relief has varied shapes and we must analyze the properties of the lava and the conditions of spill occurrence to understand.

Social Studies

Can we predict the next eruption of the old faithful geyser by knowing the duration of the previous eruption? the observations and least squares regression line appear in the scatterplot. the scatterplot and the least-squares regression line for predicting the interval (minutes) between eruptions and the duration (minutes) of the previous eruption is given below. which of the following is the correct interpretation of the slope of the least squares regression line for the data about old faithful?

Step-by-step answer

P Answered by PhD

1

The correct answer is We can predict the time between eruptions will be 10.73 minutes and this can be stated with 0.901 probability.

Eruptions do not always produce symmetrical and imposing cones. Volcanic relief has varied shapes and we must analyze the properties of the lava and the conditions of spill occurrence to understand.

Mathematics

Listed below are time intervals (min) between eruptions of a geyser. Assume that the recent times are within the past few years, the past times are from around 20 years ago, and that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Does it appear that the mean time interval has changed? Is the conclusion affected by whether the significance level is 0.10 or 0.01? Recent 78 91 89 78 58 99 62 87 70 89 82 84 56 81 75 101 61 Past 90

Step-by-step answer

P Answered by Master

p value = 0.039

t = - 2.169

Step-by-step explanation:

Applying the null and alternate hypothesis

$H_{0} : U1 = U2$

$H_{\alpha } : U1 \neq U2$

using excel worksheet to calculate for ( t and p )

t = -2.169

p = 0.039

from the results obtained

The conclusion is affected by the significance level because : 0.1 < p > 0.01

so when the significance level is = 0.1 the Null hypothesis is rejected and we can say the mean time interval will change while

if the significance level = 0.01 the Null hypothesis is accepted and we can not say the mean time interval has changed because the p -value is greater than 0.01

attached is the excel solution