View Full Version : Please Help! I am really stuck on some problems....


Fashion7
11-11-2005, 10:47 PM
During the past forty years, the monthly rate of return of stocks has been approximately normally distrubuted, with u=0.75 percent and sigma=4.2 percent, according to data from *****finance.

a. what is the probability that a randomly selected month has a rate of return of at least 1 percent?

b. what is the probability that a randomly selected month has a rate of return of less than 2 percent?

c. what proportion of months have a positive rate of return? (ie. a rate of return>0%)

d. in march 2000, the month rate of return in stocks was 9.7 percent. Is this rate of return unusual? Why or why not?

Designers in a new contest classifies nominees as brillant, talent, proficient or competent. If a nominee is randomly selected, is the probability of being rated talented equal to 25%?

The median grade for quiz 1 was 79%. What is the probability that a randomly selected student has a grade greater than 79%

A single dice is rolled twice. Find the probability of getting a 2 the first time and a 3 the second time.

quark
11-12-2005, 12:13 AM
Fashion7,

Let's work with one problem at a time. For 1a how far did you get? Show some work and we'll be happy to help.

JohnM
11-12-2005, 01:28 AM
Fashion7,

You may want to check out our Examples section. A lot of your questions deal with the normal distribution and binomial distribution, and we have example problems that help with those topics.

JohnM

Fashion7
11-12-2005, 03:11 AM
OK sounds good to me. I am really confused on how and when to use each formula.

For 1a: This is binomial distrubution right? what does sigma mean? I think if i knew the formula it could get me started.

JohnM
11-13-2005, 04:26 PM
Problems a through d use the normal distribution.

The designers question is true, assuming an equal number of nominees in each category.

The answer to the median question should be obvious (what does the median represent?)

The last question involves probability - the probability of getting a particular value on any roll of a single die is 1/6, so.....

Fashion7
11-13-2005, 05:59 PM
Okay this is what I have gotten...

the median question is 50% correct?

the dice is 1/6 for it to be a two the first time, but for it to be a three the second time it is still a 1/6 chance right?

JohnM
11-13-2005, 06:02 PM
the median question is 50% correct?
yes

the dice is 1/6 for it to be a two the first time, but for it to be a three the second time it is still a 1/6 chance right?
1/6 * 1/6

Fashion7
11-14-2005, 09:21 AM
okay for a-d I am still lost. I know that the standard distribution is z=x-u/o. I am still not sure where sigma comes in the the equation?

quark
11-14-2005, 09:27 AM
sigma is o in (x-u)/o

Fashion7
11-14-2005, 11:17 AM
So then would x be the percentage I am looking for? Or would X be the forty years?

JohnM
11-14-2005, 11:34 AM
For problem a, you are given:

mu = 0.75
sigma = 4.2

mu is the average, and sigma is the standard deviation. In other words, the average rate of return is 0.75%, with a standard deviation of 4.2%.

What is the probability that a randomly selected rate of return is at least 1%? Another way to ask this question is "what proportion of the time is the rate of return at least 1%."

Assuming that the annual rates of return are normally distributed, we want to find out how often we would get a return of at least 1%. We need to figure out where the 1% falls in the distribution - is it near the middle, implying that it happens pretty often, or is it near one of the tails, implying that it is pretty unusual?

First find the z-score for the 1% return rate, which will tel us how far away it is from the middle of the distribution.

z = (x - mu)/s
= (1 - 0.75)/4.2 = .0595

Since z is pretty close to 0, that means it falls near the middle of the standard normal distribution, which is centered at 0 with a standard deviation of 1.

So, looking up z=.0595 in the normal distribution table, we find that the area under the curve between negative infinity and .0595 is 0.5237.

Now, since the total area under the curve is =1, the area beyond z=.0595 is 1 - .5237, which = 0.4763.

So there is a 47.6% chance that a randomly selected annual rate of return is at least 1%.

Use the same logic to figure out problems b,c, and d.