Question

Final numeric grades in a certain course are normally distributed with a mean of 72.3 and a standard deviation of 6.4. The professor curves the grades so that the top 8% of students will receive an A. What is the minumum numeric grade you have to earn to obtain an A?

Answer 1

Answer:

The minumum numeric grade you have to earn to obtain an A is 81.29.

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:

$\mu = 72.3, \sigma = 6.4$

The professor curves the grades so that the top 8% of students will receive an A. What is the minumum numeric grade you have to earn to obtain an A?

The minimum numeric value is the value of X when Z has a pvalue of 1-0.08 = 0.92. So it is X when Z = 1.405.

So

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

$1.405 = \frac{X - 72.3}{6.4}$

$X - 72.3 = 1.405*6.4$

$X = 81.29$

The minumum numeric grade you have to earn to obtain an A is 81.29.