any stats gurus in here?

I have some homework thay i need help with. maybe someone knows?

1. SAT scores are distributed normally with a mean of 505 and standard deviation of 11

a. what percent of the SAT scores are less than 600?

b. what is the probablity of selecting someone whos score was between 500 and 600?

c. What is the probablity of slecting somene who SAT score is greater than 550?

2. if p(a) = .45 and event b is the complement of A what does the P(b) equal?

i think its 99.55?

3. if 6 students line up random order, what is the probability that they will be in alphabetical order?

4. find the probabilty of getting exactly 2 successes in a binomial experiment for which n = 5 and p = 2/3

thanks

^to the top

[howste]

Are you sure that standard deviation value is correct? I think it's way too small. But if it's correct:

1a) 100%
z = (600-505)/11 = 47.5 Normal table says 100% below

1b) 67.53%
z = (505-500)/11 = .4545 Normal table says 67.53% above 500 (100% below 600)

1c) 0%
z = (550-505)/11 = 4.09 Normal table says 0% above 550

If they really wanted you to do some reasonable calculations, they should have given you a larger standard deviation.

[howste]

Question 2: complement of .45 = 1 - .45 = .55

[howste]

Question 3: probability of any given order is 6! = 6x5x4x3x2x1 = 720

[howste]

Question 4: Doh! I don't have a binomial table with me. Use a table and make sure it gives you exactly 2, not 2 or less. If your table gives you n or less, take the value of (2 or less) - (1 or less).

howste! thank and your right! the first problem the standard deviation is 111!

mind if i ask you more thru private messages?

here are some more.

if P(A) = 2/5, finds the odds against event A

and adding on to number 1

SAT scores are distributed normally with a mean of 505 and standard deviation of 111

find the stats score that separates the top 20 percent

[howste]

I found a site that does the Standard Normal table lookups for you here. 111 for a standard deviation makes much more sense.

1a) z = (600-505)/111 = .8559
Plug that into the website and you get .804 (80.4%)

[howste]

1b) z = (600 - 505)/111 = .8559
z = (500 - 505)/111 = -.045

Plug those numbers in to get the difference between the z value and the mean. Then add the two numbers together. .304 + .018 = .322 (32.2%)