Intro to Statistics Homework Help

Question

Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 130000 dollars. Assume the standard deviation is 42000 dollars and that the population is normally distributed. 



Find the probability that a random sample of size n=51 has a mean value between 108239.6 and 148819.8 dollars.

P(108239.6 < x-bar < 148819.8) =  (Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

How you solve this will depend on what technology you have available.

The Old School method is to convert both values to an equivalent z-score: [z = (Value - Mean)/StdDev.] and then look up the probability/proportion in a table. Assuming a Simple Random Sample of size 51, the StdDev of the sample means is (Population StdDev)/sqrt(Sample Size) = 5881.18. The lower value has z = -3.70 and left-sided proportion of effectively .0000. Do the same for the upper value and then subtract the proportions.

It's much easier (and will be slightly more accurate) to use technology; the TI-84 has the required functionality, Geogebra's free online calculator has this as well.

I'm happy to schedule a lesson if you need more specific help with this or similar problems. However, if you're looking for free homework help, this isn't the place for that.

Answer 2

The sampling distribution is normal and has mean 130,000 (the same as the original random variable) and stdev ~5881 (equal to the original stdev divided by sqrt(51)). You don't need to invoke the central limit theorem for this, as the different values of the random sample are independent and identically N(130000,42000)-distributed. It is very important to realize that you are doing the sample mean of a random sample here; this will always have a smaller standard deviation than the original population, as long as the sample is truly random (that is, as long as the samples are independent of one another).

You use this to convert the boundary values 108239.6 and 148819.8 to z-scores: (108239.6-130000)/5881 = -3.7 and (148819.8-130000)/5881 = 3.2. These are very large z-scores, so we expect the probability to be close to 1. Precisely it is P(-3.7 ≤ Z ≤ 3.2) = P(Z ≤ 3.2) - P(Z ≤ -3.7) = 0.9992, using a z-score table to look these up or a calculator.

Cheers!