Question

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?

Answer 1

Answer: 84.13%

Step-by-step 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: $z=\dfrac{x-\mu}{\sigma}$

For x= 60 , we have

$z=\dfrac{60-70}{10}=-1$

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%