Question

Suppose scores on a college entrance exam are normally distributed with a mean of 550 and a standard deviation of 100. Find the score that marks the cut-off for the top 16% of the scores. Round to two decimal places.

Answer 1

Solution:

Given, X follows Normal distribution with,

   = 550

= 100

For top 16% data , let x be the required cut-off.

P(X > x) = 16%

P(X > x) = 0.16

P(X < x) = 1 - 0.16

P(X < x) = 0.84

For the standard normal variable z , P(Z < z) = 0.84

Use z table , see where is 0.84 probability and then see the corresponding z value.

P(Z < 0.994) = 0.84

So z = 0.994

Now using z score formula ,

x = + (z * ) = 550 + (0.994 * 100) = 649.40

Required answer : 649.40