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 ( overline{x}=5.4 ) insect fragments per ten - gram portion. complete parts (a) through (c) below
(a) why is the sampling distribution of ( overline{x} ) approximately normal?
a. the sampling distribution of ( overline{x} ) is approximately normal because the sample size is large enough.
b. the sampling distribution of ( overline{x} ) is approximately normal because the population is normally distributed
c. the sampling distribution of ( overline{x} ) is assumed to be approximately normal.
d. the sampling distribution of ( overline{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 ( overline{x} ) assuming ( mu = 5 ) and ( sigma=sqrt{5} )?
( mu_{overline{x}}=5.000 ) (round to three decimal places as needed.)
( sigma_{overline{x}}=0.354 ) (round to three decimal places as needed.)
(c) what is the probability a simple random sample of 40 ten - gram portions of the food item results in a mean of at least 5.4 insect fragments?
( p(overline{x}geq5.4)=square ) (round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\).
Given \(\bar{x} = 5.4\), \(\mu_{\bar{x}}=5\), and \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\) (where \(\sigma = \sqrt{5}\approx2.236\) and \(n = 40\)).
First, \(\sigma_{\bar{x}}=\frac{\sqrt{5}}{\sqrt{40}}=\frac{\sqrt{5}}{2\sqrt{10}}=\frac{\sqrt{50}}{20}=\frac{5\sqrt{2}}{20}=\frac{\sqrt{2}}{4}\approx0.354\) (already given).
Then \(z=\frac{5.4 - 5}{0.354}=\frac{0.4}{0.354}\approx1.13\)

Step2: Find the probability

We want to find \(P(\bar{X}\geq5.4)\), which is equivalent to \(P(Z\geq1.13)\)
Since \(P(Z\geq z)=1 - P(Z < z)\)
From the standard normal table, \(P(Z < 1.13)=0.8708\)
So \(P(Z\geq1.13)=1 - 0.8708 = 0.1292\)

Answer:

\(0.1292\)