Question

An art history professor assigns letter grades on a test according to the following scheme. A: Top 6% of scores B: Scores below the top 6% and above the bottom 63% C: Scores below the top 37% and above the bottom 16% D: Scores below the top 84% and above the bottom 8% F: Bottom 8% of scores Scores on the test are normally distributed with a mean of 77.4 and a standard deviation of 9.6. Find the minimum score required for an A grade. Round your answer to the nearest whole number, if necessary.

Answer 1

Answer:

The minimum score required for an A grade is 92.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 77.4, \sigma = 9.6$

Find the minimum score required for an A grade.

Top 6%, so at least the 100-6 = 94th percentile, which is the value of X when Z has a pvalue of 0.94. So X when Z = 1.555. So

$Z = \frac{X - \mu}{\sigma}$

$1.555 = \frac{X - 77.4}{9.6}$

$X - 77.4 = 1.555*9.6$

$X = 92$

The minimum score required for an A grade is 92.

Answer 2

Answer:

The minimum score required for an A grade is 92.5.

Step-by-step explanation:

We are given that an art history professor assigns letter grades on a test according to the following scheme. A: Top 6% of scores B: Scores below the top 6% and above the bottom 63% C: Scores below the top 37% and above the bottom 16% D: Scores below the top 84% and above the bottom 8% E: Bottom 8% of scores

Also, Scores on the test are normally distributed with a mean of 77.4 and a standard deviation of 9.6.

Let X = Scores on a test

So, X ~ N($\mu=77.4,\sigma^{2} = 9.6^{2}$)

The z score probability distribution is given by;

       Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean

           $\sigma$ = standard deviation

Now, we have to find the minimum score required for an A grade, i.e.; Top 6% of scores.

So, Probability that the test separate the top 6% of scores is given by;

        P(X > x) = 0.06

        P( $\frac{X-\mu}{\sigma}$ > $\frac{x-77.4}{9.6}$ ) = 0.06

        P(Z > $\frac{x-77.4}{9.6}$ ) = 0.06

So, the critical value of x in z table which separate the top 6% is given as 1.5722, which means;

                    $\frac{x-77.4}{9.6}$  = 1.3543

                 $x-77.4 = 9.6 \times 1.5722$

                           $x$  = 77.4 + 15.09312 = 92.5

Therefore, minimum score required for an A grade that represent top 6% of scores is 92.5.