Statistics Questions

#1
Ten percent of Americans are left-handed. A statistics class has 25 students in attendance. Find the mean and standard deviation for the number of left-handed students in such classes of 25 students.

When a customer places an order with Rudy’s Online Office Supplies a computerized account information system automatically checks to see if the customer has exceeded his or her credit limit. Past records indicate that the probability of customers exceeding their credit limit is 0.05. Suppose that, on a given day, 15 customers place orders. What is the probability that exactly two customers will exceed their limits?

If you give an answer i would like to know how you got it, thanks!

i havent done much of anything besides stare at them, i am not sure where to start or the formulas to use. (so even if you could just leave me with some formulas i would appreciate it.
 
Last edited:

obh

Active Member
#2
i havent done much of anything besides stare at them, i am not sure where to start or the formulas to use. (so even if you could just leave me with some formulas i would appreciate it.
Hi Premsanity,

Staring will not help :)
1. You need to ask yourself what distribution should I use.
2. As a result of the distribution, you should find the formula for the question.
3. what parameters of the distribution do I have?

Hint for questions.1 please look at the binomial distribution (https://en.wikipedia.org/wiki/Binomial_distribution)
Hint for questions.2 please look at the binomial distribution (http://www.statskingdom.com/1_binomial_distribution.html)
In the discrete binomial distribution, the PDF (also called pmf for discrete) is p(X=x)

Please try to solve now and show the output.

cheers,
 

obh

Active Member
#6
I don’t know if this is right but I got For
1.) Mean = 2.5 s = 15
2.) p = 1.5%
1. 2.5 students, what is the 15?
2. When you use probability you should not use percentages so p=0.1 and the result is without the percentage. writing p=1.5% is like writing p=0.015 which is not correct. you may of course use percentages (only multiplying by 100 and dividing by 100, not really change the result) but the answer will be 150%