A distribution of scores on an aptitude test has a mean of µ =500 and a standard deviation of σ =100. For the population of test-takers who have taken the aptitude test

Question

A distribution of scores on an aptitude test has a mean of µ =500 and a standard deviation of σ =100. For the population of test-takers who have taken the aptitude test; (10 marks)

a. What proportion have scores greater than 550?

b. What proportion have scores greater than 700?

c. What is the minimum score needed to be in the highest 10% of the population?

d. If a company only accepts people who scored in the top 60%, what is the minimum score needed to be accepted?

Answer 1

a. To find the proportion of scores greater than 550, we first calculate the z-score for 550:

z = (550 - 500) / 100 = 0.5

From a standard normal distribution table, or using a Z-score calculator, the proportion corresponding to z = 0.5 (area to the left of z) is approximately 0.6915. Since we want the proportion of scores greater than 550 (i.e., the area to the right of z = 0.5), we subtract this value from 1:

Proportion = 1 - 0.6915 = 0.3085

So, approximately 30.85% of test-takers scored greater than 550.

b. Similarly, for scores greater than 700:

z = (700 - 500) / 100 = 2

The proportion corresponding to z = 2 is approximately 0.9772. So, the proportion of scores greater than 700 is:

Proportion = 1 - 0.9772 = 0.0228

So, approximately 2.28% of test-takers scored greater than 700.

c. To find the minimum score to be in the highest 10% of the population, we need to find the z-score that corresponds to an area of 0.9 to the left (since the top 10% corresponds to the rightmost 10% of the distribution, or 90% to the left). From a z-table, we find that z is approximately 1.28 for an area of 0.9.

We can then convert this z-score back to a score in the distribution using the formula:

Score = µ + zσ = 500 + 1.28*100 = 628

So, a score of approximately 628 is needed to be in the highest 10% of the population.

d. If a company only accepts people who scored in the top 60% (or equivalently, bottom 40%), we need to find the z-score that corresponds to an area of 0.4 to the left. From a z-table, we find that z is approximately -0.25 for an area of 0.4.

We can then convert this z-score back to a score in the distribution:

Score = µ + zσ = 500 - 0.25*100 = 475

So, a score of approximately 475 is needed to be in the top 60% of the population.

Answer 2

Here is a video guide for this. I hope this helps. If you have any questions, please let me know!

-Dr. C