Question

According to a company’s website, the top 20% of the candidates who take the entrance test will be called for an interview. You have just been called for an interview. The reported mean and standard deviation of the test scores are 70 and 5, respectively. If test scores are normally distributed, what is the minimum score required for an interview? (You may find it useful to reference the z table. Round "z" value to 3 decimal places and final answer to 2 decimal places.)

Answer 1

Answer

Top 20% means z score corresponding to a percentile value of 80% because top 20% is above bottom 80%

Using z distribution table or by using Excel function NORMSINV(probability)

setting probability = 80%= 80/100

we get

z score for top 20% = NORMSINV(0.80) = 0.842

it is given that mean = 70 and standard deviation = 5

Using the formula

Minimum score = mean + (z*standard deviation)

setting the values

= 70 + (0.842+5)

= 70 + 4.21

= 74.21

So, required minimum score for an interview is 74.21