Math / Data & Statistics

Question18. A population has a mean of 200 and a standard deviation of 50 . A sample of size 1 will be taken and the sample mean $\bar{x}$ will be used to estimate the population mean. a. What is the expected value of $\bar{x}$ ? b. What is the standard deviation of $\bar{x}$ ? c. Show the sampling distribution of $\bar{x}$ . d. What does the sampling distribution of $\bar{x}$ show?

Studdy Solution

STEP 1

1. The population mean is $\mu = 200$ .
2. The population standard deviation is $\sigma = 50$ .
3. A sample of size $n = 1$ is taken.

STEP 2

1. Determine the expected value of $\bar{x}$ .
2. Calculate the standard deviation of $\bar{x}$ .
3. Describe the sampling distribution of $\bar{x}$ .
4. Interpret the sampling distribution of $\bar{x}$ .

STEP 3

The expected value of the sample mean $\bar{x}$ is equal to the population mean $\mu$ .
$E(\bar{x}) = \mu = 200$

STEP 4

The standard deviation of the sample mean $\bar{x}$ is given by the formula:
$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$
Since $n = 1$ , the formula simplifies to:
$\sigma_{\bar{x}} = \frac{50}{\sqrt{1}} = 50$

STEP 5

The sampling distribution of $\bar{x}$ when $n = 1$ is the same as the population distribution. Therefore, it is normally distributed with:
- Mean: $\mu = 200$ - Standard deviation: $\sigma = 50$

STEP 6

The sampling distribution of $\bar{x}$ shows that:
- The sample mean $\bar{x}$ is an unbiased estimator of the population mean $\mu$ . - The variability of $\bar{x}$ is the same as the population variability when $n = 1$ .
The expected value of $\bar{x}$ is $200$ , the standard deviation of $\bar{x}$ is $50$ , and the sampling distribution of $\bar{x}$ is normally distributed with mean $200$ and standard deviation $50$ .

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Studdy Solution

STEP 1

1. The population mean is $\mu = 200$ .
2. The population standard deviation is $\sigma = 50$ .
3. A sample of size $n = 1$ is taken.

STEP 2

1. Determine the expected value of $\bar{x}$ .
2. Calculate the standard deviation of $\bar{x}$ .
3. Describe the sampling distribution of $\bar{x}$ .
4. Interpret the sampling distribution of $\bar{x}$ .

STEP 3

The expected value of the sample mean $\bar{x}$ is equal to the population mean $\mu$ .
$E(\bar{x}) = \mu = 200$

STEP 4

The standard deviation of the sample mean $\bar{x}$ is given by the formula:
$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$
Since $n = 1$ , the formula simplifies to:
$\sigma_{\bar{x}} = \frac{50}{\sqrt{1}} = 50$

STEP 5

The sampling distribution of $\bar{x}$ when $n = 1$ is the same as the population distribution. Therefore, it is normally distributed with:
- Mean: $\mu = 200$ - Standard deviation: $\sigma = 50$

STEP 6

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Studdy Solution

STEP 1

1. The population mean is μ=200\mu = 200μ=200.2. The population standard deviation is σ=50\sigma = 50σ=50.3. A sample of size n=1n = 1n=1 is taken.

STEP 2

1. Determine the expected value of xˉ\bar{x}xˉ.2. Calculate the standard deviation of xˉ\bar{x}xˉ.3. Describe the sampling distribution of xˉ\bar{x}xˉ.4. Interpret the sampling distribution of xˉ\bar{x}xˉ.

STEP 3

The expected value of the sample mean xˉ\bar{x}xˉ is equal to the population mean μ\muμ.E(xˉ)=μ=200 E(\bar{x}) = \mu = 200 E(xˉ)=μ=200

STEP 4

The standard deviation of the sample mean xˉ\bar{x}xˉ is given by the formula:σxˉ=σn \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} σxˉ​=n​σ​Since n=1n = 1n=1, the formula simplifies to:σxˉ=501=50 \sigma_{\bar{x}} = \frac{50}{\sqrt{1}} = 50 σxˉ​=1​50​=50

STEP 5

The sampling distribution of xˉ\bar{x}xˉ when n=1n = 1n=1 is the same as the population distribution. Therefore, it is normally distributed with:- Mean: μ=200 \mu = 200 μ=200 - Standard deviation: σ=50 \sigma = 50 σ=50

STEP 6

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1. The population mean is $\mu = 200$ .
2. The population standard deviation is $\sigma = 50$ .
3. A sample of size $n = 1$ is taken.

1. Determine the expected value of $\bar{x}$ .
2. Calculate the standard deviation of $\bar{x}$ .
3. Describe the sampling distribution of $\bar{x}$ .
4. Interpret the sampling distribution of $\bar{x}$ .

The expected value of the sample mean $\bar{x}$ is equal to the population mean $\mu$ .
$E(\bar{x}) = \mu = 200$

The standard deviation of the sample mean $\bar{x}$ is given by the formula:
$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$
Since $n = 1$ , the formula simplifies to:
$\sigma_{\bar{x}} = \frac{50}{\sqrt{1}} = 50$

The sampling distribution of $\bar{x}$ when $n = 1$ is the same as the population distribution. Therefore, it is normally distributed with:
- Mean: $\mu = 200$ - Standard deviation: $\sigma = 50$