Sup Jose, thanks for the warning that these are the first of many.

First, you need to know the distribution of the sample mean. If your lecturer/class text are any good, you should be able to look this up. The answer is that the sample mean is normally distributed, mean=330 (i.e. same as the population mean, found by the IRS study), and stdev=80/sqrt(40) (NOT the same as the population stdev)

a) You find P[X > 320] = P[Z > (320-330) / (80/sqrt(40))] = 1 - Phi(-0.79) = Phi(0.79) = .7852

If you don't know how that Z and Phi got involved, you're in trouble with this problem.

b) Find P[ 320 < X < 350 ]

c) Find P[X > 350]

a) "the above problem" is concerned with the number of mispelled words in a student essay. Since the number of mispelled words is either 0, 1, 2, or some wholediscretenumber, this should narrow down the choices for distribution. It turns out, that a lot of counts of number of errors/workplace accidents follow the Poisson distribution.

b,c and d) I'm not sure if that sample stdev=2.44 is supposed to play a part in the calculations...I'm not sure how to reconcile it with the sample mean=6.05. Note that the stdev of a Poisson distribution is equal to the square root of the mean (and 2.44 is nearly the sqrt(6.05)).

If I had to write an answer down, I'd say sample mean = 6.05, sample stdev = sqrt(6.05) = 2.46, then use the formula for confidence interval:

(sample mean) +/- z[alpha] * (sample stdev) / sqrt(40)