Answer:
A score of 314 is needed to be in the top 25% of students who take this exam.
Step-by-step explanation:
We are given that one of the tests provided by the NAEP assesses the reading skills of twelfth-grade students. In recent years, the national mean score was 289 and the standard deviation was 37.
Let X = <u><em>scores of the tests provided by the NAEP</em></u>
The z-score probability distribution for the normal distribution is given by;
Z =
~ N(0,1)
where,
= national mean score = 289
= standard deviation = 37
Now, we have to find how high a score is needed to be in the top 25% of students who take this exam, that means;
P(X > x) = 0.25 {where x is the required score}
P(
>
) = 0.25
P(Z >
) = 0.25
In the z table, the critial value of z that represents the top 25% of the area is given as 0.6745, that is;


x = 289 + 24.96 = 313.96 ≈ 314
Hence, a score of 314 is needed to be in the top 25% of students who take this exam.