The final exam grade distribution for all students in the introductory statistics class at a local community college is displayed in the table below. Let [tex]\( X \)[/tex] represent the grade for a randomly selected student from the class.

| Grade | 4 | 3 | 2 | 1 | 0 |
|-------------|-----|-----|-----|-----|-----|
| Probability | 0.4 | 0.32| 0.17| 0.08| 0.03|

Which statement correctly interprets the standard deviation?

A. The grade for a randomly selected student would typically vary from the expected grade by 1.16.

B. The grade for a randomly selected student would typically vary from the expected grade by 1.08.

C. The mean grade for a randomly selected student would typically vary by 1.16 from the expected grade.

D. The mean grade for a randomly selected student would typically vary by 1.08 from the expected grade.



Answer :

To interpret the standard deviation correctly, let's start by understanding what standard deviation represents in the context of this problem.

Standard deviation is a measure of the amount of variation or dispersion in a set of values. In this specific problem, it measures how much the grades of the students, on average, deviate from the expected grade.

Given the grade distribution:
- Grade: [tex]\(4, 3, 2, 1, 0\)[/tex]
- Probabilities: [tex]\(0.4, 0.32, 0.17, 0.08, 0.03\)[/tex]

From these, we can summarize the steps:

1. Calculate the expected value (mean) of [tex]\(X\)[/tex]:
[tex]\[ E(X) = \sum_{i=1}^{5} (\text{grade} \times \text{probability}) \][/tex]

2. Calculate the variance of [tex]\(X\)[/tex]:
[tex]\[ \text{Variance}(X) = \sum_{i=1}^{5} [(\text{grade} - E(X))^2 \times \text{probability}] \][/tex]

3. Calculate the standard deviation of [tex]\(X\)[/tex]:
[tex]\[ \text{Standard Deviation}(X) = \sqrt{\text{Variance}(X)} \][/tex]

Given that the standard deviation of the grades is found to be 1.08, we need to select the statement that best interprets this:

- The grade for a randomly selected student would typically vary from the expected grade by 1.16.
- The grade for a randomly selected student would typically vary from the expected grade by 1.08.
- The mean grade for a randomly selected student would typically vary by 1.16 from the expected grade.
- The mean grade for a randomly selected student would typically vary by 1.08 from the expected grade.

The correct interpretation of the standard deviation is that, on average, an individual grade will vary from the expected grade by that standard deviation value. Thus, the grade for a randomly selected student would typically vary from the expected grade by the standard deviation of 1.08.

Hence, the correct statement is:
The grade for a randomly selected student would typically vary from the expected grade by 1.08.