Question

Applying the Central Limit Theorem:

The amount of contaminants that are allowed in food products is determined by the FDA (Food and Drug Administration). Common contaminants in cow milk include feces, blood, hormones, and antibiotics. Suppose you work for the FDA and are told that the current amount of somatic cells (common name "pus") in 1 cc of cow milk is currently 750,000 (note: this is the actual allowed amount in the US!). You are also told the standard deviation is 124000 cells. The FDA then tasks you with checking to see if this is accurate.

You collect a random sample of 55 specimens (1 cc each) which results in a sample mean of 776573 pus cells. Use this sample data to create a sampling distribution. Assume that the population mean is equal to the FDA's legal limit and see what the probability is for getting your random sample.

a. Why is the sampling distribution approximately normal?

b. What is the mean of the sampling distribution?


c. What is the standard deviation of the sampling distribution?


d. Assuming that the population mean is 750,000, what is the probability that a simple random sample of 55 1 cc specimens has a mean of at least 776573 pus cells?


e. Is this unusual? Use the rule of thumb that events with probability less than 5% are considered unusual.

Answer 1

a)as sample size is greater than 30 therefore from central limit theorum we can assume normal approximation

b)

mean of the sampling distribution =750,000

c)

standard deviation of the sampling distribution=std deviation/sqrt(n)=124000/sqrt(55)=16720.1566

d)

probability that a simple random sample of 55 1 cc specimens has a mean of at least 776573 pus cells:

probability =

P(X>776573)

=

P(Z>1.59)=

1-P(Z<1.59)=

1-0.9441=

0.0559

e)

as probability is greater than 0.05 ; therefore it is not unusual