Question

Suppose college faculty members with the rank of professor at two-year institutions earn
an average of $52,500 per year with a standard deviation of $4,000. In an attempt to
verify this salary level, a random sample of 60 professors was selected from a personnel
database for all two-year institutions in the United States.
n=60 mean=52,500 sd=4,000/squareroot60=516.3977795
C. Calculate the probability the sample mean x-bar is greater than $55,000.
D. If you drew a random sample with a mean of $55,000, would you consider this
sample unusual? What conclusions might you draw? Need help!

Answer 1

Here you have a sample drawn from what could safely be assumed to be a standard normal distribution.

You'll need to find the SAMPLE std. dev., which is smaller than the population std. dev. What is the appropriate formula for that?

Answer 2

sd/the square root of the population right? So 4000/squareroot of 60=516.3977795? @mathmale

Answer 3

I see you already have the sample std. dev. calculated.

C. Calculate the probability the sample mean x-bar is greater than $55,000.

To do this, y ou need to find the z score corresponding to $55,000. That z score is given by \[z=\frac{ 55000-52500 }{ 516 }\]\[z=\frac{ \frac{ 55000-52500 }{ 516 } }{ }\]

Using a table of z scores, please find the area under the std. normal curve which lies to the left of this z score.

Answer 4

Think. Is the resulting fraction the answer you wanted? If not, how would you go about finding that answer from this result (area to the left of this z-score)?

Answer 5

after calculating that i got 4.841227116 but there is no z score for that @mathmale

Answer 6

could you maybe subtract that z score from the zscore of what 55,00 would be or not? @mathmale

Answer 7

Let's check our arith.:

2500
What's z = -------- = ?
516

Indeed it's true you won't find a z-score as large as 4 or 5 in the z-score tables.
If you're looking for the area under the std. normal curve to the left of 55000, that will be very close to 1.00.

Is this your answer? No. You have to subtract that 1.00 from 1.00 to obtain the area under the std.; normal curve to the RIGHT of 55000.

Answer 8

how would you subtract 1.00 from 1.00 im a little confused. @mathmale

Answer 9

So, you are essentially finding that the sample's sd is over 4 std dev from the mean. Is that likely? What would the 68 95 99.7 rule tell you about this?

Answer 10

it can't be done or at least a z score of over 4 would mean that all the sample mean x-bar's are greater than 55,000? @fibonaccichick666

Answer 11

Remember that area under that curve represents probability. Look at the Empirical Rule: 68% of all data lie within 1 std. dev. of the mean,
95% within 2 std. dev.,
99.7% within 3 std. dev., and
just about 100% within 4 std. dev of the mean.

Answer 12

Here's what I was doing:

1. Find the area under the normal curve to the left of 55000 by using the z-score and z-score tables.
2. Subtract this result from 1 to obtain the area under the normal curve to the RIGHT of 55000 (which means "greater than 55000")

Answer 13

how would you find the z score to the left I totally forgot all this stuff you can't do it off the empirical rule right? @mathmale