The acceptable level for insect filth in a certain food item is 2 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 x overbarequals2.5 insect fragments per​ ten-gram portion. Complete parts​ (a) through​ (c) below.
a. Why is the sampling distribution of x approximately normal?
b. What is the mean and standard deviation of the sampling distribution of assuming t Sand a Mu x = 5 (Round to three decimal places as needed.) Sigma x = 289 (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 5.8 insect fragments? P(x > = 5.8)= (Round to four decimal places as needed.)

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Answer:

Step-by-step explanation:

Given that a simple random sample of 40 ​ten-gram portions of the food item is obtained and results in a sample mean of x overbarequals2.5 insect fragments per​ ten-gram portion.

a) By central limit theorem for samples randomly drawn with large sample sizes (atleast 30) the sample mean follows a normal distribution irrespective of the original distribution it belonged to.

b) Mean of X bar = Sample mean = 5

Std error of x bar = s/sqrt n = [tex]\frac{289}{\sqrt{40} } \\=45.695[/tex]

c) P( x bar >5.8) = P(Z>[tex]\frac{5.8-5}{45.695} \\=0.0176[/tex])

=0.4930

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