Question

I'm having so trouble with this problem. I keep getting z-score = -3.9, but the z-score doesn't go that high.
Here's the problem:
cars are normally distributed
Mew=88tons
STD=0.9tons
B. What is the probability that 50 cars chosen at random will have a mean load weight X bar of less than 87.5 tons of coal? (Round to 4 decimal places.)

Answer 1

\[\sigma xbar/\sqrt{n} = 0.9/ sq root 50 = 0.127

Answer 2

p(xbar < 87.5) = 87.5 - 88/0.127 = -3.9

Answer 3

what are you using to convert z score to probability?

Answer 4

umm... z = x - mew/ sigma

Answer 5

thats the z score ... how do we use that to find probability?

Answer 6

are you using a stats calculator like a ti83? or tables?

Answer 7

tables

Answer 8

the z score table

Answer 9

is it a table from a book, or something else? there should be a *note on it for high z values.

Answer 10

my teacher gave me the table.... yeah, it says for z values greater than 3.49 use 1.000 to app. the area

Answer 11

is that what you're talking about? what does that mean?

Answer 12

that tells you that for zscores larger than 3.49 the probability is 1 for all practical purposes

what would you say that tells us about zscores less then -3.49?

Answer 13

on the negatiove side it says for values of z less than -3.49 use 0.000 to app. the area

Answer 14

yeah ... so we should use 0.000 right?

Answer 15

i guess so...my teacher never talked about this. so, i have no idea

Answer 16

is -3.9 less then -3.49?

Answer 17

yes

Answer 18

so, based on that statement, it would 0.000

Answer 19

correct

Answer 20

rnd to 4 decimals if your instructions are correct

Answer 21

so, 0.0000 lol

Answer 22

better :)

Answer 23

:) thank you

Answer 24

yw

Answer 25

calculation of normal distribution variable Z = -3.928..... students t = -3.8888 with 49 degrees of freedom.

If you look in the tables, the probability (one sided) for Z < -3.928 is 0... or for t < -3.888 it is 0. So answer is 0... It is correct.

Answer 26

There is no need to be alarmed... As standard deviation of given distribution is very small. The standard error is very small: 0.128..for a mean of 88.. So a mean of 87.5 means dropping by 0.5 which is more than 3 times of standard error... So zero is good.

Conceptually, probability of one truck load being < 87.5 is quite small, say "p". Now probability of each of 50 truck loads going below 87.5 simultaneously is p^50, which is almost 0.

Answer 27

If you want the numerical value, wolfram can find it
http://www.wolframalpha.com/input/?i=-3.928+standard+deviations

Answer 28

it's pretty small
z<-3.928 (left-tailed p-value) is 4.283×10^-5