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 that,
μ =550, σ=100
By using standard normal distribution ,we have to find the corresponding z-values as
P( Z> z)= 16%
1-p( Z<=z)= 0.16
P( Z<= 0.995 )= 0.84
From the standard normal table.
z= 0.995
Now using the z-score formula, we can find the value of x as follows,
z=(x- μ)/σ
z×σ=( x- μ)
x= μ+( z×σ)
x=550+(0.995×100)
=550+(99.5)
x=649.50

Note:
From Normal distribution table,
P( Z<= 0.99)= 0.8389
P( Z<= 1.0)= 0.8411
So we taken z value as average of 0.99 and 1.0 as 0.995