Question

A psychology professor assigns letter grades on a test according to the following scheme. a: top 9% 9 % of scores b: scores below the top 9% 9 % and above the bottom 63% 63 % c: scores below the top 37% 37 % and above the bottom 17% 17 % d: scores below the top 83% 83 % and above the bottom 8% 8 % f: bottom 8% 8 % of scores scores on the test are normally distributed with a mean of 75.8 75.8 and a standard deviation of 8.1 8.1 . find the minimum score required for an a grade. round your answer to the nearest whole number, if necessary.

Answer 1

To find the minimum score for an A, we used the standard normal distribution to find the z-score corresponding to the top 9%, which is typically around 1.34. Multiplying this by the standard deviation and adding to the mean, we get approximately 87.06, which we round to 87 as the minimum A score.

Step-by-step explanation:

To find the minimum score required for an A grade, we need to determine the score that corresponds to the top 9% of a normally distributed set of test scores. Given that the mean is 75.8 and the standard deviation is 8.1, we use the standard normal distribution (z-score) to find this value.

The z-score corresponding to the top 9% can be found using a z-table or technology such as the TI-83/84 calculator. Typically, a z-score for the 91st percentile (since 100% - 9% = 91%) is approximately 1.34. Now we apply the formula for a z-score in the context of a normal distribution:

Z = (X - μ) / σ

Rearranging the formula to solve for X (the minimum A score), we get:

X = Z*σ + μ

Therefore:

X = 1.34 * 8.1 + 75.8

X ≈ 87.06

So, we round the score to the nearest whole number, which gives us 87 as the minimum score for an A grade.