Question

see attachment.

Answer 1

@jim_thompson5910

Answer 2

ok for part a), we're going to use a calculator

wolfram alpha is the best but you can use a TI calculator or any other kind of calculator that supports cdf's

Answer 3

remember how to use wolfram alpha to get cdf's?

Answer 4

i never know how to use wolfram alpha

Answer 5

ok the function we're going to use is called "normalcdf"

and it's basic format is

normalcdf(L,U,mu,sigma)

where

L = lower endpoint of interval
U = upper endpoint of interval
mu = mean
sigma = standard deviation


in this case

L = 96
U = 200 (pick any large number that's more than 5 standard deviations away from the mean)
mu = 87
sigma = 18

so we enter this function, with these arguments,

normalcdf(96,200,87,18)

Answer 6

go ahead and do so

tell me what you get on the 4th line in the "probabilities" section

Answer 7

ok on sec

Answer 8

0.3085?

Answer 9

so that's the probability you want because that's essentially the same as saying P(X > 96)

Answer 10

ie 200 is so far out there that P(96 < X < 200) is close enough to P(X > 96)

Answer 11

ok so a) is 0.3085?

Answer 12

yes

Answer 13

for part b, only one thing will change: the standard deviation

Answer 14

it will become sigma/sqrt(n) = 18/sqrt(6) = 7.348469

Answer 15

what do you get when you make this change

Answer 16

0.06681?

Answer 17

hmm I'm not getting the same thing

Answer 18

you only changed the 18 into 7.348469 right?

Answer 19

no i did normalcdf(96,200,87,6)

Answer 20

oh that's not correct, the new standard deviation is not 6

Answer 21

it's 18/sqrt(6) = 7.348469

Answer 22

so i do normalcdf(96,200,87,7.348469)?

Answer 23

yep

Answer 24

0.1103?

Answer 25

better

Answer 26

that's the answer to part b) for similar reasons in part a)

Answer 27

ok so d) is D?

Answer 28

what did you get for C?

Answer 29

oh woops i missed that one

Answer 30

ok so do i do 18/sqrt(18)?

Answer 31

yes

Answer 32

0.01695?

Answer 33

ok i so i still think d) is D.

Answer 34

round that to four places to get your answer

Answer 35

so 0.01695 turns into 0.0170

Answer 36

yes

Answer 37

cause they want the answer rounded to 4 decimal places

Answer 38

ok so d) is D right?

Answer 39

what makes you think it's D

Answer 40

because the probability is decreasing as the samples are increasing.

Answer 41

that part is correct...but is the variability increasing?

Answer 42

as you get bigger and bigger samples, would the variability that's measuring the mean eruption time increase or decrease?

Answer 43

i'm not sure.

Answer 44

xbar estimates mu right?

Answer 45

yes?

Answer 46

xbar is the sample mean
mu is the population mean

so xbar does in fact estimate mu

Answer 47

when you have small samples, the distribution of xbars (or sample means) will be all over the place...but they will be loosely centered around mu

as n increases, they will become more and more centered around mu and they won't be as spread out...this is because your accuracy at estimating mu will increase

Answer 48

this is why bigger samples are always a good idea

Answer 49

so as accuracy increases, variability decreases

Answer 50

so its A?

Answer 51

correct

Answer 52

basically you go from something like this



to something like this