# Thread: Any Statistics experts here....

1. ## Any Statistics experts here....

i have no clue what the answer is.....any hint/help will be greatly appreicated.

1. Suppose you are now working at a diagnostic laboratory and are interested in setting up a new test for a recently discovered antibody, Ab X. You test a large number of healthy individuals and obtain a mean of 120 units and a standard deviation of 30 units. Determine the normal range for this test if you want to include 90% of all “normals”.

2.In your never ending search for the truth about the cholesterol levels of College students, you decide to take a random sample of 64 students and find x-bar = 180mg%. You know from your experience in doing such research on the other University of California campuses, that other UC Students have been shown to have an average cholesterol level of 190 and a standard deviation of 40. From your study, what is the probability that a value like your sample mean could have resulted from a population whose mean is 190? (Show Work)

2. Originally Posted by simash4
i have no clue what the answer is.....any hint/help will be greatly appreicated.

1. Suppose you are now working at a diagnostic laboratory and are interested in setting up a new test for a recently discovered antibody, Ab X. You test a large number of healthy individuals and obtain a mean of 120 units and a standard deviation of 30 units. Determine the normal range for this test if you want to include 90% of all “normals”.
For a normal distribution 90% of all observations fall within +/-1.645 standard
deviations of the mean. So the interval containing 90% of norms is:

[120-1.645*30,120+1.645*30]=[70.65,168.35]

2.In your never ending search for the truth about the cholesterol levels of College students, you decide to take a random sample of 64 students and find x-bar = 180mg%. You know from your experience in doing such research on the other University of California campuses, that other UC Students have been shown to have an average cholesterol level of 190 and a standard deviation of 40. From your study, what is the probability that a value like your sample mean could have resulted from a population whose mean is 190? (Show Work)
If the mean is 190, and sd 40, then the z score corresponding to 180 is:

z=(180-190)/40=-0.24,

so we obtain from a table of the cumulative standard normal distribution
the probability of an observation this low or lower is:

p(z<-0.25)=0.4013

RonL