Download 9.53 Sampling Distributions for Self study Suppose that we have two

9.53 Sampling Distributions for Self study
Suppose that we have two normal populations with the means and standard deviations
listed here. If random samples of size 25 are drawn from each population, what is the
probability that the mean of sample 1 is greater than the mean of sample 2?
Population 1: mean = 40 and SD = 6
Population 2: mean = 38 and SD = 8
The key to solving this problem is to first find the distribution of the difference of two sample
means.
First, using the central limit theorem, we know the sampling distribution of the sample mean
from population 1:

N  x1  40,  x21  6
x1
2
25

Again, from the central limit theorem, we know the sampling distribution of the sample mean
from population 2:

N  x2  38,  x22  8
x2
2
25

Both sample means have normal distributions with a known mean and known variances.
Next, we know that the difference of Normal random variables is also normally distributed with
the following characteristics:
1. The mean of the difference equals the difference of the means.
2. The variance of the difference equals the sum of the variances.
In this case we are looking for the mean and variance of x1  x2 . Knowing the above properties
we can simply write down:
 x  x   x   x  40  38  2
1
2
1
2
 x2  x   x2   x2 
1
2
1
2
62 82

4
25 25
The standard deviation of the difference is just the square root of the variance. So now we know
the difference of sample means is a normal random variable with mean 2 and standard deviation
2.
Our question is to find the probability that this variable is greater than 0. For the sake of
simplicity, we can call the difference between means “X”, just as we might any other normal
random variable, and solve it in the usual manner.
X ~ N (   2.0,   2.0 )
02
 X 
P ( X  0.00 )  P 

 = P( Z  -1.00 )  1  P( Z  -1.00)
2 
 
 1  0.1587 = 0.8413
The probability that the mean of sample 1 is greater than the mean of sample 2 is .8413.