What you need to do for these is figure out how many standard deviations from the mean the value is. the farther away from the mean, the less likely it is for that value to occur. For a graduate to have a salary of 80,000, that is 10,000 above the mean, which is 2 standard deviations above the mean, since (80,000 - 70,000)/5,000 = 2. It's value minus mean, all divided by standard deviation. If you check on a z chart for normal probability distribution, you will see that .9772 will be less than a salary of 80,000 and only 2.228 percent have more than 80,000. Suppose we test that the mean salaries are indeed 70,000, the null hypothesis, against a hypothesis that the mean salaries are not 70,000 at alpha = .05. The rejection region for this would be when z > 1.96 or z <-1.96. So any value of z in between the -1.96 and 1.96 would be considered a "reasonable" value for z, which would be a reasonable outcome. Since 80,000 gives a z-value of 2, it falls just outside of 1.96 which is not a reasonable outcome.
The way to figure out the highest reasonable outcome, take the 1.96(which is 1.96 standard deviations above the mean) and multiply by the standard deviation and add to the mean.
70,000 + 1.96(10,000) = highest maximum reasonable value.
For part c, we need to take (80,000 - 70,000)/(5,000/sqrt(100)) to get the z value. The formula is (value - mean)/(standard deviation/square root of n)
If this falls between -1.96 and 1.96, it is a reasonable value. It does not, as you will see, so it's not reasonable.
d) For this, we know the z-value that will give the maximum reasonable mean salary is 1.96
so use the formula
1.96 = (x - 70,000)/(5,000/sqrt(100))
The x is the value we are solving for, which is the mean salary for a random sample.
If you do this correctly, you will get 70,980