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 67 and 8, 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.)

Minimum Score=

Answer 1

Solution:-

Given that,

mean = = 67

standard deviation = = 8

Using standard normal table,

P(Z > z) = 20%

= 1 - P(Z < z) = 0.20

= P(Z < z) = 1 - 0.20

= P(Z < z ) = 0.80

= P(Z < 0.842 ) = 0.80

z = 0.842

Using z-score formula,

x = z * +

x = 0.842 * 8 + 67

x = 73.736

minimum score = 73.74