1. ## sampling distribution

Hi,

PROBLEM 1:
Assume the octane rating for a particular type of gasoline is normally distributed ramdom variable with a mean=87.0 and standard deviation=1.5.
-What is the probability the mean of a random sample mean of 5 measurements of the octane rating for this type of gasoline will be between 86.5 and 87.5?

Here I wanted to use the central limit theorem but the sample is not large enough. So I don't kow how to do it.

PROBLEM 2
Suppose the probability that a communication is a signal, X=1, is 0.9, and the probability that a communication is noise, X=0, is 0.1. A random sample of 3 communications is to be taken, the value X, 0, 1, recorded and mean(X) computed. Find the sampling distribution of mean(X).
??

Thank you
B

Hi,

PROBLEM 1:
Assume the octane rating for a particular type of gasoline is normally distributed ramdom variable with a mean=87.0 and standard deviation=1.5.
-What is the probability the mean of a random sample mean of 5 measurements of the octane rating for this type of gasoline will be between 86.5 and 87.5?

Here I wanted to use the central limit theorem but the sample is not large enough. So I don't kow how to do it.
You don't need the central limit theorem, what you need is the fact that
the mean of a sample of size N from a normal distribution N(mu, sigma^2)
is itself normal distributed ~N(mu, sigma^2/N).

So in this case the mean of the sample means is m5 = 87.0, and the standard deviation of the sample means is s5 =1.5/sqrt(5).

Now the interval 86.5 to 87.5 is the interval m5 - 0.75 s5 to m5 - 0.75 s5
or a +/-0.75 sd interval about the mean. The probability of this we look up
in a table of the standard normal distribution and is ~=0.4532

RonL