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.
a. find probability that if a person is randomly selected, his weight will be GREATER THAN 168 pounds.

Answer 1

@CliffSedge check this one out

Answer 2

Ok, normal distribution calculation.
You are finding P(X>168) with N(191,33)
You can find the z-score for x=168 and find P(z>z*).
Do you know how to do that on the TI-84?

Answer 3

you might have to remind me..

Answer 4

the way you start is that you use the equation \[\frac{ given - mean }{ Standard Devation } \]
which makes which makes \[\frac{ 168 - 171 }{ 33 } \]

Answer 5

that equals -.0909090909
and then im supposed to look that up on the z score table which is .4641 and then i subtract that from 1 which equals .5359 but it says the answer is .5362

Answer 6

171? I thought the mean was 191?

Answer 7

oops! it's supposed to be 191!!!! sorry

Answer 8

Well, in your case of using 171, the table value, .5359 that you got is near enough to the more precise value of .5362.

You can get the more precise value using the normalcdf function on our calculator.

Answer 9

it wont work already tried

Answer 10

Here's a reminder how to do that:
Press 2ND-VARS for DISTR menu.
Do normalcdf(168,1000,171,33)

Answer 11

ooooooooh!!!!!!! this whole time i was entering normalcdf(-.0909090909) lol -__-

Answer 12

why did you use 100 though?

Answer 13

1000*

Answer 14

That's an approximation to infinity.

Answer 15

so what about part b.
says find probability that 10 random people selected with have a mean that is greater that 169 pounds?

Answer 16

oops 168*

Answer 17

Gah, sorry for the delay; browser crashed . . .
Ok, for this one, you will use 1680 as X and 1710 as the mean (since if the mean of ten people is 1680, then the sum of all their weights is 1680)

Answer 18

You'll need to adjust the standard deviation appropriately as well. Do you know how to do that @hmdp4 ?

Answer 19

no im not sure how to adjust the sd

Answer 20

You have to do it in terms of the variance.
If the SD is 33, what is the Var?

Answer 21

i dont remember how to find the variance in this case

Answer 22

i know it's something like the opposite of the standard devation or something of that matter

Answer 23

It's simple. Standard deviation = σ; variance = σ^2.
Since you're multiplying everything by ten, you multiply the variance by ten, then find the new standard deviation.