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

Hence, the answer is
`x>71.1515`

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(z<1.645)=0.95\\P(z>1.645)=0.05\\z>1.645`

`(x-\mu)/((\sigma)/(√(\mu)) ) >1.645\\x>1.645* (\sigma)/(√(n)) +\mu`

`>1.645* (7)/(√(100)) +70`

`\Rightarrow x>71.1515`