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more: 0.17 of 1 pt 50 8.1.27-T 10 of 13 (13 completo) HW Score: 56.82%. 7.39 Question Help gran ortions the food item is obtain The acceptable level for insect fit in a certain food item is 5 insect fragments (larvae, eggs, body parts, and so on) per 10 grams. A simple random sample of 50 and results in a sample mean of x = 5.7 insect fragments per ton gram portion Complete parts (a) through (c) below. The sampling distribution of x is assumed to be approximately normal C. The sampling distribution of x is approximately normal because the population is normally distributed D. The sampling distribution of 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 assuming and V5? - 5 (Round to three decimal places as needed) ; - 0.316 (Round to three decimal places as needed.) (c) What is the probably a simple random sample of 50 en-gram portions of the food item results in a mean of at least 5.7 insect fragments? P625.7)- (Round to four decimal places as needed.) e More Enter your answer in the answer box and then click Check Answer 2 ета MacBook Air

Answer 1

Given : Sample size=n=50

a) Therefore , the sampling distribution of $\bar{X}$ is approximately normal because the population is normally distributed and the sample size is large enough.

b) The mean and standard deviation of the sampling distribution of $\bar{X}$ is ,

g=il = 871

$\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{\sqrt{5}}{\sqrt{50}}=0316$

c) Now ,

K-PX Z OX 5.7-5. P(X > 5.7) = PC 0.316

; From normal probability integral table

= P(Z > 2.22)

= 0.0132

Therefore , the probability a simple random sample of 50 ten-gram portion of the food item results in a mean of at least 5.7 insect fragments is 0.0132