Question

Fluctuation in the prices of precious metals such as gold have been empirically shown to be well approximated by a normal distribution when observed over short interval of time. In May 1995, the daily price of gold (1 troy ounce) was believed to have a mean of $383 and a standard deviation of $12. A broker, working under these assumptions, wanted to find the probability that the price of gold the next day would be between $394 and $399 per troy ounce. In this eventuality, the broker had an order from a client to sell the gold in the client's portfolio. What is the probability that the client's

Answer 1

please show work

Answer 2

Find the z-scores of the two numbers. Use a z-table to find the probabilities of less than each z-score. Subtract the probability of the 394 score from the 399 one. There's your answer.

Answer 3

I know the answers in 9% just need to see how one gets to that answer.

Answer 4

What do you mean?

Answer 5

I need to see the numbers used in the z tables to show me how you get to the answer. I'm new to all this.

Answer 6

http://allianthawk.org/images/z_table_positive_bbb.png
Use the z-formula (z=x-mean/stddev) to find the two z-values you need.

Answer 7

For example, z of 394 = 394-383/12 = 0.917 = approx 0.92.
The z-value for 0.92 is .8212.

Answer 8

Do you know how to read a z-table?

Answer 9

I see on the chart where the .92= .8212

Answer 10

Use that formula to find the z-score of 399. (z= (x-mean)/stddev)
What value do you get from that z-score?

Answer 11

z(399) = 399-383 / 12
= 1.3333

Answer 12

Ok, now what percentage do you see that corresponds to 1.33 in the table?

Answer 13

.9066

Answer 14

So the probability of the price being between 394 and 399 is .9066-.8212 = .0854

Answer 15

I see it now... THANK YOU!!!