Question

A college professor has accumulated a record of grades achieved by his students over a multi-year period. The mean over the period is 83.5, with a sample standard deviation of 5.5. What is the probability that:

a) A student achieves a grade greater than 97.
b) A student achieves a grade greater than 80.7.
c) A student achieves a grade less than 82.7?
d) A student achieves a grade between 81 and 86.

Answer 1

a) Probability of a grade greater than 97: Approximately 0.69%

b) Probability of a grade greater than 80.7: Approximately 69.36%

c) Probability of a grade less than 82.7: Approximately 44.43%

d) Probability of a grade between 81 and 86: Approximately 33.36%

Step-by-step explanation:

To answer these probability questions, we need to use the concept of the standard normal distribution. Given that the mean grade over the period is 83.5 and the sample standard deviation is 5.5, we can use the z-score formula to standardize the grades.

Mean (μ) = 83.5

Standard deviation (σ) = 5.5

a) For a grade greater than 97:

To find the probability, we'll use the z-score formula. Z-score = (Value - Mean) / Standard Deviation.

For 97: Z = (97 - 83.5) / 5.5 ≈ 2.455

Looking up this z-score in a standard normal distribution table, the probability is approximately 0.0069 or 0.69%.

b) For a grade greater than 80.7:

Z = (80.7 - 83.5) / 5.5 ≈ -0.509

Using the z-score, the probability of a grade greater than 80.7 is roughly 69.36% or 0.6936.

c) For a grade less than 82.7:

Z = (82.7 - 83.5) / 5.5 ≈ -0.145

The probability of a grade less than 82.7 is approximately 44.43% or 0.4443.

d) For a grade between 81 and 86:

Z for 81 ≈ -0.455

Z for 86 ≈ 0.455

The probability of a grade between 81 and 86 is around 33.36% or 0.3336.