Statistics Homework

Question

The population mean annual salary for environmental compliance specialist is about $60,000. A Random sample of 36 specialist is drawn from this population. What is the probability that the main salary of dissemble is less than $57,500? Assume Standard deviation = $6400

Answer 1

For questions like these, what we do is utilize the central limit theorem.

The central limit theorem states the following:

for a collection of n random variables x_i that are independent and identically distributed (they have the same parameters), then for a large number of them

∑x_i ∼ N( nE[ x_i ] , nVar(x_i ) )

We can divide by n to get the average as opposed to the sum. Then the sample mean will have the following distribution:

(1/n)∑x_i ~N( E[ x_i ] , Var(x_i ) )

36 is a sufficiently large sample (ideally, n>30). Using the numbers from the question:

(1/n)∑x_i ~N( $60,000 , $6400 ).

Now we may ask: P( (1/n)∑x ≤ $57,500 ) "What's the probability of the sample mean being less than $57,500?"

There are several ways to proceed. In a TI-84, you could do normalcdf(60,000, 6400, -100, 57500) which would split out the probability of the sample mean being less than 57500.

We can also take the z-score of 57500 by (57500 - 60,000)/6400 ≈ -0.3906

Looking at a z-score table for -0.39 you will see the P(Z<-0.39) ≈ 0.34287

Hope this helps! If you have any questions let me know.