Question

IQ scores (as measured by the Stanford-Binet intelligence test) are normally distributed with a mean of 100 and a standard deviation of 16. Find the approximate number of people in the U.S. (assuming a total population of 280,000,000) with an IQ higher than 110.
Round your answer to the nearest 100,000.

Answer 1

@douglaswinslowcooper @Hero @SithsAndGiggles @whpalmer4 PLEASE HELP

Answer 2

z = (value - mean)/ (standard deviation)
use cumulative normal distribution to find fraction with IQ larger than corresponding to this z-score.

If z = 1, 16% have higher values, so (0.16)(280 x 10^6) or
44, 800,000.

For your question value= 110, mean = 100, SD = 16 {I thought it was 15, check that]

z = (110-100/16 = etc. then look it up in tables or on net

Answer 3

it would then be 1-.63 right?

Answer 4

or no not .63 but .7357

Answer 5

If your z-score indicates the fraction to be 0.63, then 1-.63 would be above, yes. Multiply that by the total population..

But your z-score is 0.63, which the tables for normal distribution indicate 0.736 is the fraction with z<= 0.63.

Use 0.736 times total population, as I see you now realize.

Answer 6

i got 206,080,000 which is wrong by the key

Answer 7

perhaps "nearest 100,000" is key, thus 206,100,000

Answer 8

in the key it says the answer is 74,500,000.. is there perhaps something else I'm leaving out?

Answer 9

I would have done
z = (110-100)/15 = 0.667
p = 0.745
1-p = .255
(0.255) (280x10^6) = 71,400,000
I think this is right.

Answer 10

so do you think he maybe made a typo? because in the key the answer is 74,500,000

Answer 11

all i want to know is how to do it right because i have an exam tomorrow