Question

STATS help!!!

A standard chemistry examination administered nationally by the American Chemical Society has a mean of 500 and a standard deviation of 90. What is the probability that the average of a random sample of examination scores of 25 students will be between 450 and 500?

Answer 1

any thoughts?

Answer 2

not sure :\

Answer 3

well, persoanlly i would work a z score ... maybe a t score

Answer 4

theres usually a flow chart that tells when to use which depending on conditions

i believe this is a zscore tho

Answer 5

do you ahve a ti83 to play with on this?

Answer 6

@amistre64 nope

Answer 7

then tables are your thing

Answer 8

how many standard deviations is 450 from men of 500?

Answer 9

0.56?

Answer 10

50/90 = 5/9 = .56 is fair

now a table z values should be in your material that we can use this with

Answer 11

if I'm looking across the 0.6 z score, which column do i look down?

Answer 12

row(0.5)+ col(.06)

Answer 13

0.7123

Answer 14

well, subtract .5 from it as an adjustment

Answer 15

0.2123?

Answer 16

thats what im getting yes.

i cant say thats the correct answer tho simply becuase i dont have the flow charts that tell us when to sqrt(n) and when not to ...

Answer 17

http://www.statisticshowto.com/when-to-use-a-t-score-vs-z-score/

this gives a little information

Answer 18

We're dealing with the xbar distribution (aka the distribution of sample means) because it says "What is the probability that the `average` of a random sample of examination scores of 25 students will be between 450 and 500?"

keyword: average

this distribution has mean of mu and standard deviation of sigma/sqrt(n)

Answer 19

so we might have to divide .56 by 5 ....

Answer 20

Column 0.6 and row 0.1?

Answer 21

\[\LARGE z = \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}\]

\[\LARGE z = \frac{x-\mu}{\sigma}*\sqrt{n}\]

If you already calculated \(\LARGE \frac{x-\mu}{\sigma}\), then you just need to multiply that result by sqrt(n) = sqrt(25)=5