The probabilities you're getting for the z-scores don't match what I'm finding on my table--are you sure you're using the standard normal distribution table?

I don't see why you would subtract .5 from anything. Let me explain what is going on with a z-score--this must be in your book too so refer to that also.

Let's just think about the first investment. Imagine a normal distribution curve (the familiar bell shape) that represents the returns of this investment. You know that the mean is $2,000,000. So the bell is centered at $2,000,000. Draw this on a piece of paper so you really get it. You also know something about how spread out the bell is (that's what the standard deviation is telling you.) We'll use this for the calculations--don't worry about it for the picture right now.

What does the bell represent? The probability that one observation of a random variable with that distribution falls in a given range is the **area under the density curve for that range**. If we know the equation for the curve (the probability density function) we can find the area underneath it using integrals, but we don't want to have to do that every time. Fortunately ,we don't have to--read on.

You're asked to find the probability that if you pick just one value from that distribution, it will be less than $1,900,000. Draw a vertical line on your graph at $1,900,000. What's the probability that one observation is less than $1,900,000? It's the area under the curve to the left of that line, all the way to negative infinity. It would suck to have to integrate the function over that range, so we'd like to have a table of values that someone else has done the work for. The problem is that we'd need a totally separate table for every different mean and every different standard deviation. That's an infinite number of tables, and it's tough to fit an infinite number of tables into the back of a textbook.

Fortunately, there's a way for us to transform *any* normally distributed variable into a normal variable with mean=0 and standard deviation=1. This is called standardizing the variable, and it's done just as you described it: if we call the original variable x and the new standardized variable z, z= (x-μ)/σ. So now what we are interested in in this case is not P(x<1,900,000), but P(z<(1,900,000-2,000,000)/(125,000)).

Why does this help? Well, draw a new bell curve for z. We know that the mean is 0, so it's centered at 0, and we know the standard deviation is 1. Draw your vertical line again, this time at -.8, as you calculated. The probability that z<.8 is exactly the same as the probability that x<1,900,000, but it's a whole lot easier to find, because we have a table for z. Just look up z= -.8 and find the associated probability. That's your answer.

Notice that z<0 corresponds to x<2,000,000, since those are the means of the two distributions. So you'll have a negative z when x is less than the mean, and a positive z when it's more.

I hope this helps. The key points to remember are:

1. The probability that an observation is less than a certain value is the area under the distribution curve to the left of that value.

2. We can standardize any normal random variable by subtracting its mean and dividing by the standard deviation.

3. The new standardized variable will ALWAYS have mean=0 and standard deviation=1, and there's a convenient table for the areas to the left of any value.