# Normal distr

1. Sep 16, 2009

### nicholasch

1. The problem statement, all variables and given/known data

Lifts usually have signs indicating their maximum capacity. Consider a sign in a lift that reads “maximum capacity 1120kg or 16 persons”. Suppose that the weights of lift-users are normally distributed with a mean of 68kg and a standard deviation of 8kg.

(a) What is the probability that a lift-user will weigh more than 70kg?
(b) What is the probability that a lift-user will weigh between 65 and 75kg?
(c) What is the probability that 16 people will exceed the weight limit of 1120kg?
(d) What is the probability that 18 people will not exceed the weight limit?

[Hint for parts (c) and (d): Convert the probability statement about total weight into a probability statement about average weight.]

2. The attempt at a solution
(a) pr(x>70)= pr((70-68)/8))= pr(z>0.25) =0.401294
(b) pr(65<x<75) = pr((65-68)/8<z<(75-68)/8))=pr(-3/8<z<7/8) = 0.455383
(c) not too sure how to approach this but... 1120/16=70. so need to calculate pr(x>70) 16 times? 0.401296^16?
(d) 1120/18= 62.2222222222, that means 18 people cant be above 62.22kgs?

2. Sep 16, 2009

### Staff: Mentor

The answers for a and b look reasonable (I didn't check your math closely), but here's a tip on what you're writing, since you are losing some information about the inequalities in your probabilities.

a)Pr(x > 70) = Pr( (x - mu)/sigma > (70 - 68)/8 ) = Pr(z > 0.25) = ...

Do you see how what I wrote is different from what you have? If you don't keep track of the inequalities inside Pr( ... ), you will probably come to grief sooner rather than later.

b) Similar comment

3. Sep 16, 2009

### nicholasch

cool thanks for the prompt reply. i was trying to speed up the process with the typing, but i guess i best avoid it before it becomes a habit.

how about c+d? am i on the right track?

4. Sep 16, 2009

### Staff: Mentor

For c and d, I don't think you're on the right track. You need to use a different statistic, one that deals with sample collections rather than individual items. If I'm remembering correctly, the statistic is called the standard error of the mean, and looks a lot like the Z statistic, but sigma is different.

SE = sigma/sqrt(n), and
z = (M - mu)/SE

where sigma is the population s.d., mu is the population mean, M is the sample mean, and n is the number in the sample. You can read more about this here: http://en.wikipedia.org/wiki/Z-test, in the section titled Example. Set up your inequalities as before, but use this form of z.