Question

The capacity of a lift is 10 people or 1680 pounds. The capacity will be exceeded if 10 people have weights with a greater mean the 1680/10 pounds. suppose the people have wights that are normally distributed with a mean of 191 pounds and a strandard deviation of 33 lbs.
Find Probability that 10 randomly selected people will have a mean that is greater than 168 pounds.

Answer 1

@CliffSedge okay so i think i figured out another step for changing the standard deviation.

Answer 2

Yeah?

Answer 3

i put 191 AGAIN and it's supposed to be 171. oops
so it would be \[\frac{ 168-171 }{ \frac{ 33}{ \sqrt{10} } }\]
right?

Answer 4

so i get -.2874797873 and i look on the zscore table for that right?

Answer 5

I think that is an equivalent way to do it, yes.

Answer 6

what the heck.. i got .3897 and the answer is .6131
okay so the other way to do it is the normalcdf(lower limit, higher limit, mean, sd)
normalcdf(168,1000,168,(33/sqrt10))

Answer 7

i still dont get the right answer.. :( on my calculator i got .4999999995

Answer 8

I first did it with \[\large Z=\frac{1680-1710}{\sqrt{10*33^2}}\]
It comes out the same.

Answer 9

wait why did you use 1680? and 1710?

Answer 10

You got the correct z-score = -.2874797873
That should get you the right probability. Did you look it up in a table or use the normalcdf function?

Answer 11

i looked that up on the table and i got rounding up to .4847

Answer 12

I did a change of variables, since it said the mean of ten people was 168, then the sum of their weights would be 1680.

Answer 13

hmm well im not sure why i am not getting the right number..

Answer 14

Hmm, let me check a table . . .
I used http://www.math.unb.ca/~knight/utility/NormTble.htm
for 0.28 < z < 0.29, and got
0.6103 < P < 0.6141

Answer 15

Using normalcdf(1680, 100000, 1710, sqrt(10*33^2)), I get the same as using
normalcdf(168, 100000, 171, 33/sqrt(10)), which is P≈0.613127...

Answer 16

Which is the same as using normalcdf(-0.28748,100000).

Answer 17

I see above that this is what you entered in your calculator?
"normalcdf(168,1000,168,(33/sqrt10))"

That second 168 should be the mean, 171.

Answer 18

okay i see what you mean now. let me write all this down! thanks by the way lol

Answer 19

also, in the book, they use the infinite limits as -999999 and 999999 lol so next time i'll use those because i was sooooooooo confused from wondering how you were getting these numbers. the result is the same either way

Answer 20

My pleasure. It's a lot easier since we both have the same calculator.

Answer 21

Yeah, you can use any large number to approximate infinity, I usually just hit 1 then 0 a bunch of times, but theoretically, all you need is a number that is 5-10 times the standard deviation more than the mean to get an accuracy better than 1 in 100,000 or so.

Answer 22

That's why I used 1000 at first because the mean +5σ ≈ 200-400, so any number larger than 500 is close enough to infinity for the calculator to not tell the difference.

Answer 23

whats strange is that im not getting that zscore as a result still.. LOL what the heck! when i use the normalcdf function, isnt the result going to be the zscore? thats not a number i look up right? because it's not the answer either way..

Answer 24

No, no, no, the normalcdf function operates on a z-score to output a p-value.

Answer 25

wait nvm! i got it

Answer 26

forgot to put DIVIDE when i had 33 and sqrt 10. -__-

Answer 27

The inverse-normal function operates on a p-value to output a z-score.

Answer 28

whats a p value..

Answer 29

p-value can mean 'proportion' or 'probability' depending on the context.

Answer 30

You'll see it more when you do hypothesis testing.