Question

HELP! PLZ!

Answer 1

@jim_thompson5910 care to help with this one? lol

Answer 2

first you need to find the standard error of the mean

Answer 3

so you use the formula

SE = sigma/sqrt(n)

Answer 4

tell me what you get

Answer 5

what does SE = sigma/sqrt(n) mean again?

Answer 6

it's the standard error of the mean

Answer 7

sometimes you'll see it as SEm (with the m below the E)

or you might see it as SE with an xbar below the E

Answer 8

sigma/sqrt(n) is basically the standard deviation of the xbar distribution

Answer 9

yah i got that but idk SE = sigma/sqrt(n) are supposed to stand for. non of that looks like the formulas on the examples.

Answer 10

can i use the Wolfram alpha thing to solve this?

Answer 11

yes that's one way to do it

Answer 12

Great! now to learn how to use it. lol

Answer 13

did you figure out what sigma/sqrt(n) is?

Answer 14

nope

Answer 15

sigma = 14
n = 18

Answer 16

so

sigma/sqrt(n) = 14/sqrt(18) = ??

Answer 17

3.299831646

Answer 18

@jim_thompson5910

Answer 19

good, now you use that to find the area under the curve

Answer 20

using wolfram alpha, you would get

http://www.wolframalpha.com/input/?i=normalcdf%28100%2C177%2C182%2C3.29983%29

Answer 21

@jim_thompson5910 Great now the last part say's

Answer 22

change n = 18 to n = 42

Answer 23

and repeat what was done before

Answer 24

\[\frac{ 14 }{ \sqrt{42} }=2.160246899\]

Answer 25

now use that when you type in your normalcdf function

Answer 26

http://www.wolframalpha.com/input/?i=normalcdf%28100%2C177%2C182%2C2.160246899%29

Answer 27

so the answer would be 0.01032?

Answer 28

yep

Answer 29

round to 4 decimal places though

Answer 30

THANKS!! again. lol

Answer 31

yw