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.)

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 =

c) P( x bar >5.8) = P(Z>)

=0.4930

second option

step-by-step explanation:

the measures of a triangle equal 180. use that to find x.

2x+3x+4x=180

9x=180

9x/9=180/9

x=20

now you know x, you know angle a is 40°, angle b is 60°, angle c is 80°.

because angle a is 40°, the first option doesn't work.

because angle b is 60°, the second option does work.

when you add angle a and b, you get 100°. complementary means it equals 90°, so the third option doesn't work.

when you add angle a and c, you get 120°, which is not 100.