A normal distribution of test scores has a mean of 38 and a

A normal distribution of test scores has a mean of 38 and a standard deviation of 6. Everyone scoring at or above the 80th percentile gets placed in an advanced class. What is the cutoff score to get into the class?

Solution

Let X be the random variable that test scores.

Given that X has Normal distribution with mean = 38 and sd = 6.

Everyone scoring at or above the 80th percentile gets placed in an advanced class.

That is who get 80 or above marks they get placed in an advanced class.

And we have to calculate cut off score to get into class.

that is here we have to calculate x.

probability = 0.80.

P(X <= x) = 0.80

convert x into z-score.

z = (x-mean) / sd

P(Z <=(x-mean)/sd ) = 0.80

This can be done by using EXCEL.

syntax :

NORMSINV(probability)

where probability = 0.80

It will gives us answer 0.8416

Now compare this value with (x-mean) / sd.

That is,

(x-mean) / sd = 0.8416

(x-38) / 6 = 0.8416

x-38 = 0.8416 * 6

x - 38 = 5.0496

x = 38 + 5.0496

x = 43.0496

the cutoff score to get into the class is 43.0496