For a language test with normally distriuted scores, the mean was 70 and the standard deviation was 10. Approximately what percentage of test takers scored a 60 or above?

Question

asked May 7, 2020 164k views

1 Answer

Tdykstra · Answer 1 · 2020-05-11T07:02:07+0000

Answer: 84.13%

Explanation:

Given : The scores in a a language test with normally distributed.

Mean :
$\mu=70$

Standard deviation:
$\sigma= 10$

Let x be the random variable that represents the scores of students.

Formula for z-score:
$z=(x-\mu)/(\sigma)$

For x= 60 , we have

z=(60-70)/(10)=-1

The probability f test takers scored a 60 or above :-

$P(x\geq60)=P(\geq-1)=1-P(z<-1)\\\\=1-0.1586553=0.8413447\approx84.13\%$

Hence, the percentage of test takers scored a 60 or above = 84.13%