Help with a statistics problem.

Question

An investor invested in 9 different investments. The dollar returns on the different investments are probabilistically independent, and each return follows a normative distribution with mean $50,000 and standard deviation $10,000.

There is a 1% chance that the total return on the 9 investments is less than what value?

What is the probability that the investor’s total return is between $400,000 and $520,000?

Answer 1

If you find this problem is still tricky I can explain more.

First we will need to sum the 9 normal distributions. The totals mean with be the sum of the 9 means, and its variance being the sum of the 9 variances. Keep in mind variance is the square of the standard deviation.

Total Mean = 9*50,000

Total Mean = 450,000

Total SD = sqrt(9*100000000)

Total SD = 30,000

Now if we look up the Z-score for 1% (or 0.01) we will find it is between -2.32 and -2.33, so we will just use -2.325.

Next, we back out the value of x from the standard score formula:

Z = (x-mean)/SD

-2.325 = (x-450,000)/30,000

x = 380,250

Therefore, there is a 1% chance the investment total will be below $380,250.

To calculate the probability between 400,000 and 520,000 we first calculate the z-score for each using the same formula.

Z1 = (x-mean)/SD

Z1 = (520,000-450,000)/30,000

Z1 = 2.33

Z2 = (x-mean)/SD

Z2 = (400,000-450,000)/30,000

Z2 = -1.67

Then subtract percentage values correlated to Z2 from Z1, to remove the left tail (below 400,000) from the percentage.

Percentage between two values = 0.9901 - 0.0475 = 94.26%