Question

the acceptable level for insect filth in a certain food item is 5 insect fragments (larvae, eggs, body parts, and so on) per 10 grams. a simple random sample of 40 ten - gram portions of the food item is obtained and results in a sample mean of $\bar{x}=5.4$ insect fragments per ten - gram portion. complete parts (a) through (c) below
(a) why is the sampling distribution of $\bar{x}$ approximately normal?
a. the sampling distribution of $\bar{x}$ is approximately normal because the sample size is large enough
b. the sampling distribution of $\bar{x}$ is approximately normal because the population is normally distributed.
c. the sampling distribution of $\bar{x}$ is assumed to be approximately normal.
d. the sampling distribution of $\bar{x}$ 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 $\mu = 5$ and $\sigma=\sqrt{5}$?
$\mu_{\bar{x}}=\square$ (round to three decimal places as needed.)

Explanation:

Step1: Find the mean of the sampling distribution

According to the Central Limit Theorem, the mean of the sampling distribution of the sample mean \(\bar{x}\), \(\mu_{\bar{x}}\), is equal to the population mean \(\mu\). Given \(\mu = 5\), so \(\mu_{\bar{x}}=\mu\).

Step2: Calculate the standard deviation of the sampling distribution

The standard deviation of the sampling distribution of the sample mean (also known as the standard error) is given by the formula \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). Here, \(\sigma = \sqrt{5}\approx2.236\) and \(n = 40\). Then \(\sigma_{\bar{x}}=\frac{\sqrt{5}}{\sqrt{40}}=\frac{\sqrt{5}}{2\sqrt{10}}=\frac{\sqrt{50}}{20}=\frac{5\sqrt{2}}{20}=\frac{\sqrt{2}}{4}\approx0.354\)

Answer:

\(\mu_{\bar{x}} = 5.000\) (rounded to three decimal places)