Question

The SAT Reasoning Test (formerly called the Scholastic Aptitude Test) is perhaps the most widely used standardized test for college admissions in the United States. Scores are based on a normal distribution with a mean of 1500 and a standard deviation of 300. Clinton College would like to offer an honors scholarship to students who score in the top 10% of this test. What is the minimum score that qualifies for the scholarship

Answer 1

Given Information:  

Mean SAT score = μ = 1500

Standard deviation of SAT score = σ = 3 00

Required Information:  

Minimum score in the top 10% of this test that qualifies for the scholarship = ?

Answer:

$\bar{x} = 1884$

Step-by-step explanation:

What is Normal Distribution?

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.  

We want to find out the minimum score that qualifies for the scholarship by scoring in the top 10% of this test.

$P(X > \bar{x} )= P(Z > \bar{x}) = 0.10\\P(X < \bar{x} )= P(Z < \bar{x}) = 1 - 0.10\\P(X < \bar{x} )= P(Z < \bar{x}) = 0.90$

The z-score corresponding to the probability of 0.90 is 1.28 (from the z-table)

$\bar{x} = \mu + z(\sigma) \\\bar{x} = 1500 + 1.28(300)\\\bar{x} = 1500 + 384\\\bar{x} = 1884$

Therefore, you need to score 1884 in order to  qualify for the scholarship.

How to use z-table?

Step 1:

In the z-table, find the probability value of 0.90 and note down the value of the that row which is 1.2

Step 2:

Then look up at the top of z-table and note down the value of the that column which is 0.08

Step 3:

Finally, note down the intersection of step 1 and step 2 which is 1.28