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?
Solution :
Given that ,
a) P(x 250 ) = 1 - P(x 250)
= 1 - P[(x - ) / (250 - 300) / 30]
= 1 - P(z -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 * +
x = 0.253 * 30 + 300
x = 307.59
Suppose scores for a test are distributed normally (300,30). a) What percent of test takers can...
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