Question

A standardized visual working memory test has a population mean of 60 and a standard deviation of 6. Because the scores are normally distributed, the whole distribution of scores can be converted into a Z distribution. Each raw score in the original distribution has a corresponding Z score in the Z distribution. The Z distribution has a symmetrical bell shape with known properties, so it's possible to mathematically figure out the percentage of scores within any specified area in the distribution.

Question 1:

John has a score of 49. What is John's Z score?

Question 2:

What is the percentage of students that score lower than John?

Question 3:

Based on the Z table, if 1000 students take the test, how many of them would likely score above John's score? (round the answer to a whole number)

Answer 1

(A) population mean of 60 and a standard deviation of 6

x score for John is 49

z = (x-mean)/sd

z = (49-60)/6

z = -1.83

(B) P(z<-1.83)

use z table, find -1.8 in row and 0.03 in column

we get

P(z<-1.83) = 0.0336

Convert to %, 0.0336*100 = 3.36% of students that score lower than John

(C) probability p = 0.0336 and n = 1000

So, expected value = n*p

= 1000*0.0336

= 33.6

= 34