Question

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Part 1
The final grades in a math class have a mean of 71 and a standard deviation of 5.31. If the professor plans to give an A to the top 6​% of the​ class, what is the cutoff for an​ A?
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Part 1
The cutoff for an A is
  
enter your response here.
​(Simplify your answer. Round to the nearest whole number as​ needed.)

Answer 1

To find the cutoff for an A, we first need to find the z-score that corresponds to the top 6% of students. Since we are looking for the top 6%, we want to find the z-score that corresponds to the 94th percentile (100% - 6% = 94%).

We can use a standard normal distribution table (z-table) or a calculator to find the z-score for the 94th percentile. The z-score associated with the 94th percentile is approximately \( z \approx 1.5548 \).

Next, we can use the z-score formula to find the cutoff score for an A.

The z-score formula is given by:

\[
z = \frac{X - \mu}{\sigma}
\]

Where:
- \( z \) is the z-score,
- \( X \) is the value we are solving for,
- \( \mu \) is the mean,
- \( \sigma \) is the standard deviation.

Rearranging the formula to solve for \( X \):

\[
X = \mu + z \cdot \sigma
\]

Now plug in the values:

- \( \mu = 71 \)
- \( \sigma = 5.31 \)
- \( z \approx 1.5548 \)

Substituting these values in:

\[
X = 71 + (1.5548 \cdot 5.31)
\]
\[
X = 71 + 8.256
\]
\[
X \approx 79.256
\]

Rounding to the nearest whole number:

\[
X \approx 79
\]

Therefore, the cutoff for an A is **79**.