Question

Suppose scores for a test are distributed normally (300,30). a) What percent of test takers can expect to score 250 or above? b) What score is necessary to reach the 60th percentile?

Answer 1

Solution :

Given that ,

a) P(x$\geq$ 250 ) = 1 - P(x  $\leq$ 250)

= 1 - P[(x - $\mu$) / $\sigma$$\leq$ (250 - 300) / 30]

= 1 -  P(z $\leq$ -1.67)

= 1 - 0.0475

= 0.9525

percent = 95.25%

b) Using standard normal table,

P(Z < z) = 60%

= P(Z < 0.253 ) = 0.60

z = 0.253

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 0.253 * 30 + 300

x = 307.59