Question

the governor of a state wanted to assess the public education system of the state. she instructed a group of researchers to create a test to assess the knowledge of every high school senior in the state. the researchers administered the tests and found that the average score was 77 points out of 100, with a standard deviation of 17 points. complete parts (a) through (c) below.
choose the correct variable below.
a. the variable is the number of high school seniors.
b. the variable is the number of high school students.
c. the variable is the test length.
d. the variable is the test score.
b. for samples of size 200, find the mean \\( \mu_{x} \\) and standard deviation \\( \sigma_{x} \\) of all possible sample mean test scores.
\\( \mu_{x}=77 \\) (type an integer or a decimal.)
\\( \sigma_{x}=1.2 \\) (round to one decimal place as needed.)
c. repeat part (b) for samples of size 400.
\\( \mu_{x}=77 \\) (type an integer or a decimal.)
\\( \sigma_{x}=\square \\) (round to one decimal place as needed.)

Explanation:

Step1: Recall the formula for the mean of the sampling distribution

The mean of the sampling distribution of the sample mean \(\mu_{\bar{x}}\) is equal to the population mean \(\mu\). Since the population mean (average score) is given as \(77\), \(\mu_{\bar{x}} = 77\).

Step2: Recall the formula for the standard deviation of the sampling distribution

The formula for the standard deviation of the sampling distribution of the sample mean (also known as the standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. Given \(\sigma = 17\) and \(n = 200\), we substitute these values into the formula: \(\sigma_{\bar{x}}=\frac{17}{\sqrt{200}}\).
Calculate \(\sqrt{200}\approx14.142\), then \(\sigma_{\bar{x}}=\frac{17}{14.142}\approx1.2\).

Step3: Repeat for \(n = 400\)

Using the formula \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\) again, with \(\sigma = 17\) and \(n = 400\). Calculate \(\sqrt{400}=20\), then \(\sigma_{\bar{x}}=\frac{17}{20}=0.85\approx0.9\) (but wait, no, let's re - check. Wait, original formula: \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), \(\sigma = 17\), \(n = 400\), \(\sigma_{\bar{x}}=\frac{17}{\sqrt{400}}=\frac{17}{20}=0.85\). But if we consider the problem's requirement of one decimal place. Wait, no, wait the formula is correct. Wait, for part b: \(n = 200\), \(\sigma_{\bar{x}}=\frac{17}{\sqrt{200}}\approx\frac{17}{14.14}\approx1.2\). For part c: \(n = 400\), \(\sigma_{\bar{x}}=\frac{17}{\sqrt{400}}=\frac{17}{20}=0.85\approx0.9\) (but wait, no, wait the problem says "Round to one decimal place as needed"). Wait, \(\frac{17}{20}=0.85\), which rounds to \(0.9\). But let's check the arithmetic again. \(\sqrt{200}\approx14.1421\), \(17\div14.1421\approx1.2\) (since \(14.1421\times1.2 = 16.9705\approx17\)), \(\sqrt{400} = 20\), \(17\div20=0.85\approx0.9\)

Answer:

b. \(\mu_{\bar{x}} = 77\), \(\sigma_{\bar{x}}\approx1.2\)
c. \(\mu_{\bar{x}} = 77\), \(\sigma_{\bar{x}}\approx0.9\)