Questions: The annual salary for one particular occupation is normally distributed, with a mean of about 134,000 and a standard deviation of about 15,000. Random samples of 44 are drawn from this population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of these sample means. Then, sketch a graph of the sampling distribution. The mean is μbarx= , and the standard deviation is σbarx= . (Round to the nearest integer as needed. Do not include the symbol in your answers.)

The annual salary for one particular occupation is normally distributed, with a mean of about 134,000 and a standard deviation of about 15,000. Random samples of 44 are drawn from this population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of these sample means. Then, sketch a graph of the sampling distribution.

The mean is μbarx= , and the standard deviation is σbarx= .
(Round to the nearest integer as needed. Do not include the symbol in your answers.)

Transcript text: The annual salary for one particular occupation is normally distributed, with a mean of about $\$ 134,000$ and a standard deviation of about $\$ 15,000$. Random samples of 44 are drawn from this population, and the mean of each sample is determined. Find the mean and standard deviation of the sampling distribution of these sample means. Then, sketch a graph of the sampling distribution. The mean is $\mu_{\bar{x}}=$ $\square$ , and the standard deviation is $\sigma_{\bar{x}}=$ $\square$ . (Round to the nearest integer as needed. Do not include the $\$$ symbol in your answers.)

Solution

Solution Steps

Step 1: Determine the Mean of the Sampling Distribution

The mean of the sampling distribution of the sample means, denoted as $\mu_{\bar{x}}$, is equal to the mean of the population. Therefore, \[ \mu_{\bar{x}} = 134,000 \]

Step 2: Determine the Standard Deviation of the Sampling Distribution

The standard deviation of the sampling distribution of the sample means, denoted as $\sigma_{\bar{x}}$, is calculated using the formula: \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \] where $\sigma$ is the standard deviation of the population, and $n$ is the sample size. Substituting the given values: \[ \sigma_{\bar{x}} = \frac{15,000}{\sqrt{44}} \approx 2,262 \]

Final Answer

The mean is $\mu_{\bar{x}} = 134,000$, and the standard deviation is $\sigma_{\bar{x}} = 2,262$.

{"axisType": 3, "coordSystem": {"xmin": 120000, "xmax": 150000, "ymin": 0, "ymax": 0.0002}, "commands": ["y = (1/(2262_sqrt(2_pi)))_exp(-0.5_((x-134000)/2262)**2)"], "latex_expressions": ["$y = \\frac{1}{2262\\sqrt{2\\pi}}e^{-0.5\\left(\\frac{x-134000}{2262}\\right)^2}$"]}

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