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yarga [219]
3 years ago
9

Many states assess the skills of their students in various grades. One program that is available for this purpose is the Nationa

l Assessment of Educational Progress (NAEP). One of the tests provided by the NAEP assesses the reading skills of twelfth-grade students. In a recent year, the national mean score was 289 and the standard deviation was 37. Assume that these scores are approximately Normally distributed, N(289, 37). How high a score is needed to be in the top 25% of students who take this exam (use technology)?
Mathematics
1 answer:
Nataly [62]3 years ago
7 0

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  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = national mean score = 289

           \sigma = 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( \frac{X-\mu}{\sigma} > \frac{x-289}{37} ) = 0.25

            P(Z > \frac{x-289}{37} ) = 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;

                        \frac{x-289}{37}=0.6745

                        {x-289}=0.6745\times 37

                        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.

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