Math / Data & Statistics

QuestionHigh-rent district: The mean monthly rent for a one-bedroom apartment without a doorman in Manhattan is $\$ 2557$ . Assume the standard deviation is $\$ 486$ . A real estate firm samples 86 apartments. Use Excel.
Part: $0 / 5$ $\square$
Part 1 of 5 (a) What is the probability that the sample mean rent is greater than $\$ 2627$ ? Round the answer to at least four decimal places.
The probability that the sample mean rent is greater than $\$ 2627$ is $\square$ .

Studdy Solution

STEP 1

1. The sample size is 86 apartments.
2. The population mean ( $\mu$ ) is \$2557.
3. The population standard deviation ($\sigma$) is \$486.
4. The sample mean ($\bar{x}$) we are comparing to is \$2627.
5. The distribution of sample means is approximately normal due to the Central Limit Theorem.

STEP 2

1. Calculate the standard error of the mean.
2. Calculate the z-score for the sample mean of \$2627.
3. Use Excel to find the probability corresponding to the z-score.
4. Interpret the result to find the probability that the sample mean is greater than \$2627.

STEP 3

Calculate the standard error of the mean (SEM) using the formula:
$\text{SEM} = \frac{\sigma}{\sqrt{n}}$
where $\sigma = 486$ and $n = 86$ .
$\text{SEM} = \frac{486}{\sqrt{86}} \approx 52.42$

STEP 4

Calculate the z-score using the formula:
$z = \frac{\bar{x} - \mu}{\text{SEM}}$
where $\bar{x} = 2627$ , $\mu = 2557$ , and $\text{SEM} \approx 52.42$ .
$z = \frac{2627 - 2557}{52.42} \approx 1.336$

STEP 5

Use Excel to find the probability corresponding to the z-score. In Excel, use the formula:
$\text{=1 - NORM.S.DIST(1.336, TRUE)}$
This gives the probability that the sample mean is greater than \$2627.

STEP 6

Interpret the Excel result. The probability that the sample mean rent is greater than \$2627 is approximately:
$\boxed{0.0909}$

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Studdy Solution

STEP 1

1. The sample size is 86 apartments.
2. The population mean ( $\mu$ ) is \$2557.
3. The population standard deviation (\(\sigma\)) is \$486.
4. The sample mean (\(\bar{x}\)) we are comparing to is \$2627.
5. The distribution of sample means is approximately normal due to the Central Limit Theorem.

STEP 2

1. Calculate the standard error of the mean.
2. Calculate the z-score for the sample mean of \$2627.
3. Use Excel to find the probability corresponding to the z-score.
4. Interpret the result to find the probability that the sample mean is greater than \$2627.

STEP 3

Calculate the standard error of the mean (SEM) using the formula:
$\text{SEM} = \frac{\sigma}{\sqrt{n}}$
where $\sigma = 486$ and $n = 86$ .
$\text{SEM} = \frac{486}{\sqrt{86}} \approx 52.42$

STEP 4

Calculate the z-score using the formula:
$z = \frac{\bar{x} - \mu}{\text{SEM}}$
where $\bar{x} = 2627$ , $\mu = 2557$ , and $\text{SEM} \approx 52.42$ .
$z = \frac{2627 - 2557}{52.42} \approx 1.336$

STEP 5

Use Excel to find the probability corresponding to the z-score. In Excel, use the formula:
$\text{=1 - NORM.S.DIST(1.336, TRUE)}$
This gives the probability that the sample mean is greater than \$2627.

STEP 6

Interpret the Excel result. The probability that the sample mean rent is greater than \$2627 is approximately:
$\boxed{0.0909}$

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Studdy Solution

STEP 1

1. The sample size is 86 apartments.2. The population mean (μ\muμ) is \$2557.3. The population standard deviation (\(\sigma\)) is \$486.4. The sample mean (\(\bar{x}\)) we are comparing to is \$2627.5. The distribution of sample means is approximately normal due to the Central Limit Theorem.

STEP 2

1. Calculate the standard error of the mean.2. Calculate the z-score for the sample mean of \$2627.3. Use Excel to find the probability corresponding to the z-score.4. Interpret the result to find the probability that the sample mean is greater than \$2627.

STEP 3

Calculate the standard error of the mean (SEM) using the formula:SEM=σn \text{SEM} = \frac{\sigma}{\sqrt{n}} SEM=n​σ​where σ=486\sigma = 486σ=486 and n=86n = 86n=86.SEM=48686≈52.42 \text{SEM} = \frac{486}{\sqrt{86}} \approx 52.42 SEM=86​486​≈52.42

STEP 4

STEP 5

Use Excel to find the probability corresponding to the z-score. In Excel, use the formula:=1 - NORM.S.DIST(1.336, TRUE) \text{=1 - NORM.S.DIST(1.336, TRUE)} =1 - NORM.S.DIST(1.336, TRUE)This gives the probability that the sample mean is greater than \$2627.

STEP 6

Interpret the Excel result. The probability that the sample mean rent is greater than \$2627 is approximately:0.0909 \boxed{0.0909} 0.0909​

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The sample size is 86 apartments.
2. The population mean ( $\mu$ ) is \$2557.
3. The population standard deviation (\(\sigma\)) is \$486.
4. The sample mean (\(\bar{x}\)) we are comparing to is \$2627.
5. The distribution of sample means is approximately normal due to the Central Limit Theorem.

1. Calculate the standard error of the mean.
2. Calculate the z-score for the sample mean of \$2627.
3. Use Excel to find the probability corresponding to the z-score.
4. Interpret the result to find the probability that the sample mean is greater than \$2627.

Calculate the standard error of the mean (SEM) using the formula:
$\text{SEM} = \frac{\sigma}{\sqrt{n}}$
where $\sigma = 486$ and $n = 86$ .
$\text{SEM} = \frac{486}{\sqrt{86}} \approx 52.42$

Use Excel to find the probability corresponding to the z-score. In Excel, use the formula:
$\text{=1 - NORM.S.DIST(1.336, TRUE)}$
This gives the probability that the sample mean is greater than \$2627.

Interpret the Excel result. The probability that the sample mean rent is greater than \$2627 is approximately:
$\boxed{0.0909}$