Question

the unacceptable level for insect filth in a certain food item is 3 insect fragments (larvae, body parts, and so on) per 10 grams. a simple random sample of 50 ten-gram portions of the food item is obtained and results in a sample mean of \\( \bar{x} = 3.2 \\) insect fragments per ten-gram portion. complete parts (a) through (c) below.\
click here to view the standard normal distribution table (page 1).\
click here to view the standard normal distribution table (page 2).\
(a) why is the sampling distribution of \\( \bar{x} \\) approximately normal?\
\\( \bigcirc \\) a. the sampling distribution is approximately normal because the sample size is large enough.\
\\( \bigcirc \\) b. the sampling distribution is assumed to be approximately normal.\
\\( \bigcirc \\) c. the sampling distribution is approximately normal because the population is normally distributed.\
\\( \bigcirc \\) d. the sampling distribution is approximately normal because the population is normally distributed and the sample size is large enough.\
(b) what is the mean and standard deviation of the sampling distribution of \\( \bar{x} \\) assuming \\( \mu = 3 \\) and \\( \sigma = \sqrt{3} \\)?\
\\( \mu_{\bar{x}} = \square \\) (round to three decimal places as needed.)

Explanation:

Part (a)

To determine why the sampling distribution of \(\bar{x}\) is approximately normal, we use the Central Limit Theorem (CLT) and properties of sampling distributions. The CLT states that for a sample of size \(n\), the sampling distribution of the sample mean \(\bar{x}\) is approximately normal if either:

1. The population is normally distributed (regardless of sample size), or
2. The sample size \(n\) is large enough (typically \(n \geq 30\), the "large sample" condition).

Looking at the options:

• Option A: "The sampling distribution is approximately normal because the sample size is large enough." This aligns with the CLT's large - sample condition. If the population distribution is unknown, a large sample size (\(n = 50\) in this case, which is greater than 30) ensures the sampling distribution of the sample mean is approximately normal.
• Option B: "The sampling distribution is assumed to be approximately normal." There is no need to assume it; we can justify it with the CLT.
• Option C: "The sampling distribution is approximately normal because the population is normally distributed." We are not given that the population is normally distributed, so we can't assume this.
• Option D: "The sampling distribution is approximately normal because the population is normally distributed and the sample size is large enough." Since we don't know the population distribution, we rely on the sample size being large.

Step 1: Find the mean of the sampling distribution of \(\bar{x}\)

The mean of the sampling distribution of the sample mean \(\bar{x}\) (denoted as \(\mu_{\bar{x}}\)) is equal to the population mean \(\mu\). We are given that \(\mu=3\). So, by the property of the sampling distribution of the sample mean, \(\mu_{\bar{x}}=\mu\).
\(\mu_{\bar{x}} = 3\)

Step 2: Find the standard deviation of the sampling distribution of \(\bar{x}\)

The standard deviation of the sampling distribution of the sample mean (also known as the standard error) is given by the formula \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size.
We are given that \(\sigma = \sqrt{3}\) and \(n = 50\).
First, calculate \(\sqrt{n}=\sqrt{50}\approx7.071\)
Then, \(\sigma_{\bar{x}}=\frac{\sqrt{3}}{\sqrt{50}}\)
We know that \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\), so \(\frac{\sqrt{3}}{\sqrt{50}}=\sqrt{\frac{3}{50}}\approx\sqrt{0.06}\approx0.245\) (rounded to three decimal places)

Answer:

A. The sampling distribution is approximately normal because the sample size is large enough.

Part (b)