Question

A humanities professor assigns letter grades on a test according to the following scheme. A: Top 11% of scores B: Scores below the top 11% and above the bottom 64% C: Scores below the top 36% and above the bottom 22% D: Scores below the top 78% and above the bottom 9% F: Bottom 9% of scores Scores on the test are normally distributed with a mean of 70.6 and a standard deviation of 9.3. 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 82.

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:

$\mu = 70.6, \sigma = 9.3$

Find the minimum score required for an A grade.

Top 11%, which is at least 100-11 = 89th percentile. So the value of X when Z has a pvalue of 0.89. So X when Z = 1.225.

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

$1.225 = \frac{X - 70.6}{9.3}$

$X - 70.6 = 1.225*9.3$

$X = 82$

The minimum score required for an A grade is 82.

Answer 2

Answer:

The minimum score required for an A grade is 82.

Step-by-step explanation:

We are given that a humanities professor assigns letter grades on a test according to the following scheme. A: Top 11% of scores, B: Scores below the top 11% and above the bottom 64%, C: Scores below the top 36% and above the bottom 22%, D: Scores below the top 78% and above the bottom 9% and E: Bottom 9% of scores.

Also, Scores on the test are normally distributed with a mean of 70.6 and a standard deviation of 9.3.

Let X = Scores on a test

So, X ~ N($\mu=70.6,\sigma^{2} = 9.3^{2}$)

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 11% of scores.

So, Probability that the test separate the top 11% of scores is given by;

         P(X > x) = 0.11

         P( $\frac{X-\mu}{\sigma}$ > $\frac{x-70.6}{9.3}$ ) = 0.11

         P(Z > $\frac{x-70.6}{9.3}$ ) = 0.11

So, the critical value of x in z table which separate the top 11% is given as 1.2326, which means;

                   $\frac{x-70.6}{9.3}$ = 1.3543

                $x-70.6 = 9.3 \times 1.2326$

                       $x$  = 70.6 + 11.4632 = 82.1 ≈ 82

Therefore, minimum score required for an A grade that represent top 1% of scores is 82.