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 60 ten-gram portions of the food item is obtained and results in a sample mean of \(\bar{x}=5.7\) insect fragments per ten-gram portion. Complete parts (a) through (c) below.

(a) Why is the sampling distribution of \(\bar{x}\) approximately normal?

\(\mathrm{A}\). The sampling distribution of \(\bar{x}\) is assumed to be approximately normal.

B. The sampling distribution of \(\bar{x}\) is approximately normal because the population is normally distributed and the sample size is large enough

\(\bigcirc\) C. The sampling distribution of \(\bar{x}\) is approximately normal because the population is normally distributed.

\(\bigcirc\) D. The sampling distribution of \(\bar{x}\) is approximately normal because 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.)

\(\sigma_{\bar{x}}=\square\) (Round to three decimal places as needed.)

(c) What is the probability a simple random sample of 60 ten-gram portions of the food item results in a mean of at least \(5.7\) insect fragments?

\(\mathrm{P}(\bar{x} \geq 5.7)=\square\) (Round to four decimal places as needed.)

Answer 1

Solution - Giren X= 5.7 n=60 of the sampling distribution of X is approximately normal because the population is normally distributed and the sample size is large enough Girent u=5 6=85 Sampling distibution of mean is ex = "=5 पर-51 & standard deviation of mean is 6x= n 6x=_02886 Answer ex = [5] 6x=0.289]

PCX 25.7) = L-PX157) = 1- P(Ky 5.7-5 =l-P(Z<2.9248) = 1-0.9923 P(1254) = 0: 0011 (0.0077 is the probability of a sundom sample of 60 ten-grum portions of the food item result ina mean of at least s7