Question

The average score of 100 students taking a statistics final was 70 with a standard deviation of 7. Assuming a normal distribution, what test score separates the top 5% of the students from the lower 95% of students?

Answer 1

Answer:

Hence, the answer is $x>71.1515$

Step-by-step explanation:

We have,

$n=100,\mu=70,\sigma=7$

The top percentage of the students is $5\%$

The  lower percentage of the students is $95\%$

The lowest $95\%$ is the left area from the normal distribution table, the area $0.95$ lies in the z-table

$z=1.645$       from the z- table

$P(z1.645)=0.05\\z>1.645$

$\frac{x-\mu}{\frac{\sigma}{\sqrt{\mu}} } >1.645\\x>1.645\times \frac{\sigma}{\sqrt{n}} +\mu$

   $>1.645\times \frac{7}{\sqrt{100}} +70$

   $\Rightarrow x>71.1515$