Question 747311
For this we have to understand that average and "mean" are the same thing so:

LESS THAN 42:
We must first find he z-score using the formula: x-mean/Std.Dev. **or**

(42-48)/3=-2

Now we find the corresponding z score to -2.0 using the standard normal distribution table which = .0228 so:

P(x<42)=.0228

If you are allowed to use a calculator: using a TI-83 or TI-84 the steps are:

2nd DISTR

scroll down to normalcdf

lowerbound=-1,000,000 *this represents a number large enough to approximate - infinity as we are looking for EVERYTHING under 42 (make sure to use a neg sign and not the minus sign)

upperbound=42

mean=48

Std. Dev.=3

Press enter twice and you get .02275, or .0228; the same as if you had used the standard normal distribution table.

MORE THAN 48:
Again we find the z-score using the formula x-mean/Std.Dev. **or**

(48-48)/3 = 0

Now we find the corresponding z score to 0 using the standard normal distribution table which = .5000 so:

P(x>48)=.5000

You can use the same calculator steps replacing upperbound with 48 to find your answer or check your work.

Hope this helps!!