Homework help

I have two homework problems I don't really know where to start or how to do them. If some one could help I would be greatly appreciative.

1) The baggage weights for passengers using a particular airline are normally distributed with a mean of 20lbs and a standard deviation of 4lbs. If the limit on total luggage weight is 2125 lbs, what is the probability that the limit will be exceeded for 100 passengers?





2) A unhappy post office customer is disturbed with the waiting time to buy stamps. Upon registering his complaint he was told the average waiting time in the pas has been about 4 min with a standard deviation of 2 min. The customer collected a sample of n=45 customers and found the mean wait time was 5.3 min. Find the 95% confidence interval for the mean waiting time. Assume the standard deviation is a known value.

Start with the z-scores.

Ok for the post office one the z scores are -1.96 and 1.96. Do I have to figure standard deviation? or do i set 1.96= 5.3 - mu/Standard deviation / square root of 45?

Ok, tell me if I am going in the right direction. For the post office. Since I don't have a standard deviation C= (mean - 1.96, mean + 1.96) which would be C=(5.3-1.96, 5.3+1.96) so the 95% level fits in between 3.34 and 7.26?

Jd2002 said:

1) The baggage weights for passengers using a particular airline are normally distributed with a mean of 20lbs and a standard deviation of 4lbs. If the limit on total luggage weight is 2125 lbs, what is the probability that the limit will be exceeded for 100 passengers?

Click to expand...

Someone tell me if I am in the ball park here.

Mean = n * mu
Standard Deviation = sqrt(n) * Standard Deviation
mean 2000 = 20 * 100
SD 40 = sqrt(100) * 4

(2000-2125) / 40= -3.125

.5-.4991 = .0009

So the answer is less the 1% .0009?

Can you lend me $2.99 ?