Question 624406: Please show me specific detailed answers to these question i got wrong on my test. Thankyou.
An internal survey of a company showed that their staff needs to spend about 55 minutes from home to work every morning on average. It was also reported that 3% of the staff needs over two hours for the travel.
(i) If 50 staff is surveyed on a particular day, what is the probability that 5 will report the travel of over two hours?
(ii) If the company has 3000 staff, what is the expected number of staff who has the travel of over two hours?
(iii) If the company has 3000 staff, what are the variance and standard deviation of the number of staff who has the travel of over two hours?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 55 minutes is the average.
120 minutes is equal to 2 hours.
3% take over 120 minutes.
this looks like a binomial distribution type of problem.
binomial distribution formula is:
p(x) out of n = C(n,x) * p^x * q^(n-x)
p(x) = probability of number of events that are successful
n = total number
p = probability that commute time will be greater than 120 minutes (success)
q = probability that commute time will be less than or equal to 120 minutes (failure)
c(n,x) is the combination formula for x events out of n = n! / (x! * (n-x)!)
based on that:
i) If 50 staff are surveyed on a particular day, what is the probability that 5 will report the travel of over two hours?
p(5) out of 50 = C(50,5) * (.03)^5 * (.97)^45 = .013074229
(ii) If the company has 3000 staff, what is the expected number of staff who has the travel of over two hours?
expected number would be equal to n*p = 50 * .03 = 90
(iii) If the company has 3000 staff, what are the variance and standard deviation of the number of staff who has the travel of over two hours?
based on the binomial distribution formula:
variance is equal to n*p*q = 3000 * .03 * .97 = 87.3
standard deviation is equal to square root of variance is equal to 9.343446901
a reference is here:
http://www.stat.yale.edu/Courses/1997-98/101/binom.htm
i think these are the answers you are looking for but i'm not totally sure.
check them out and see if they agree with the answers you were supposed to have.
|
|
|
