Question

According to a company’s website, the top 10% 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 64 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.)

Answer 1

Solution:-

Given that,

mean = = 64

standard deviation = = 8

Using standard normal table,

P(Z > z) = 10%

= 1 - P(Z < z) = 0.10  

= P(Z < z) = 1 - 0.10

= P(Z < z ) = 0.90

= P(Z < 1.282 ) = 0.90  

z = 1.282

Using z-score formula,

x = z * +

x = 1.282 * 8 + 64

x = 74.26