Question

It's really easy just i don't understand it.
Question: If the average score is 122 and the standard deviation is 35, what percentage of students scored between 52 and 122?

Answer 1

How do i figure out the percentage?

Answer 2

convert 52 and 122 to z-scores using the formula

z-score = (given value-mean)/standard deviation

once you have the z-scores for 52 and 122, you have two options to find the answer

one way would be to use a graphing calculator to find normalcdf(z-score for 52,z-score for 122)

other method would be to use a z-table to find the decimal values for each z-score, then subtract (decimal value for 122 minus decimal value for 52)

Answer 3

hmm

Answer 4

but vocaloid i don't have the given value-mean/sx

Answer 5

the problem gives you the mean and standard deviation, right?

mean = 122
standard deviation = 35

z-score for 52:

(52-122)/35

z-score for 122

(122-122)/35

Answer 6

Wowow I am super dumb haha did not read the problem :/

Answer 7

well, alternatively you could use the empirical rule if you recognize that 52 is two standard deviations below the mean

Answer 8

Sorry about that

Answer 9

no need to apologize

Answer 10

So i'm looking at -2 and 0 for z scores hmm

Answer 11

And I got .0228 from the z table :D

Answer 12

oh wait how do i find the z-score for each value? i had to use both to find one (-2,0)

Answer 13

well, since -2 and 0 are integers that makes our work a bit easier

well, the empirical rule tells us that the percentage is 34%+13.5%

http://www.stat119review.com/_/rsrc/1351288349191/more-material/normal-distribution/empirical-rule/solving-empirical-rule-questions/emprulesmart.bmp

Answer 14

but if you wanted to use a z-table, you would find the table value for 0, then the table value for -2, then subtract the two values

Answer 15

oh ok

Answer 16

thanks