A Calculus Instructor Uses Computer-Aided Instruction And AI
A Calculus Instructor Uses Computer Aided Instruction And Allows Stude
A calculus instructor uses computer aided instruction and allows students to take the midterm exam as many times as needed until a passing grade is obtained. The record indicates the number of students in a class of 40 who took the test a specified number of times, along with the probabilities associated with each number of attempts. The data is as follows:
| Number of Tests | Number of Students | Probability |
|-----------------|-------------------|--------------|
| 1 | 20 | 0.075 |
| 2 | ??? | ??? |
| 3 | ??? | ??? |
| ... | ... | ... |
(Note: The original data provided is incomplete; assume the probabilities for each number of tests sum to 1 and sum to 40 students in total.)
Calculate the expected value, variance, and standard deviation of the number of tests taken by students.
Paper For Above instruction
The problem involves analyzing the distribution of the number of attempts students make to pass a midterm exam under an adaptive testing policy. Since students can retake the exam as many times as necessary until they achieve a passing grade, the distribution of the number of tests taken per student is crucial for understanding the effectiveness and burden of this policy. This analysis requires computing the expected value, variance, and standard deviation of the number of tests taken, based on the given probabilities.
Understanding the Distribution
The fundamental approach is to treat the number of tests taken as a discrete random variable, denoted as \(X\). The probability distribution shows the likelihood of a student taking \(k\) tests, expressed as \(P(X=k)\). For comprehensive analysis, the sum of all probabilities must equal 1, and the total number of students sums to 40.
Assuming the data provided:
- 20 students took 1 test
- Remaining students took 2 or more tests (with given probabilities)
Given only partial data, let's hypothesize assignments based on typical probability distributions. For example, if the probability of taking only 1 test is 0.075 for each student, then the total students taking 1 test is \(20\).
Calculating the expected value requires summing the products of each number of tests and corresponding probabilities:
\[
E(X) = \sum_{k} k \times P(X=k)
\]
Similarly, variance is computed using:
\[
Var(X) = \sum_{k} (k - E(X))^2 \times P(X=k)
\]
or equivalently:
\[
Var(X) = E(X^2) - [E(X)]^2
\]
where:
\[
E(X^2) = \sum_{k} k^2 \times P(X=k)
\]
Step 1: Estimating Probabilities
Suppose for illustration, the distribution is:
| Number of Tests | Number of Students | Probability |
|----------------|-------------------|----------------------|
| 1 | 3 (or 20 students) | 0.075 (per student) |
| 2 | 10 | ??? |
| 3 | 7 | ??? |
| 4 or more | Remaining students | ??? |
Given the total students (40), and assuming the data is:
- 20 students for 1 test
- 10 students for 2 tests
- 7 students for 3 tests
- 3 students for 4 tests
Corresponding probabilities:
\[
P(X=1) = \frac{20}{40} = 0.5
\]
\[
P(X=2) = \frac{10}{40} = 0.25
\]
\[
P(X=3) = \frac{7}{40} = 0.175
\]
\[
P(X=4) = \frac{3}{40} = 0.075
\]
Check sum: \(0.5 + 0.25 + 0.175 + 0.075 = 1.0\).
Step 2: Calculating Expected Value
\[
E(X) = 1 \times 0.5 + 2 \times 0.25 + 3 \times 0.175 + 4 \times 0.075
\]
\[
E(X) = 0.5 + 0.5 + 0.525 + 0.3 = 1.825
\]
Step 3: Calculating \(E(X^2)\)
\[
E(X^2) = 1^2 \times 0.5 + 2^2 \times 0.25 + 3^2 \times 0.175 + 4^2 \times 0.075
\]
\[
E(X^2) = 1 \times 0.5 + 4 \times 0.25 + 9 \times 0.175 + 16 \times 0.075
\]
\[
E(X^2) = 0.5 + 1 + 1.575 + 1.2 = 4.275
\]
Step 4: Computing Variance
\[
Var(X) = E(X^2) - [E(X)]^2 = 4.275 - (1.825)^2
\]
\[
Var(X) = 4.275 - 3.33 = 0.945
\]
Step 5: Computing Standard Deviation
\[
\text{Std Dev} = \sqrt{Var(X)} = \sqrt{0.945} \approx 0.972
\]
Conclusion
The expected number of tests taken by students is approximately 1.825, with a variance of approximately 0.945, and a standard deviation of about 0.972. These results suggest that most students tend to pass within roughly two attempts, with moderate variability around this average.
The implications are significant for evaluating the policy’s efficiency: while allowing unlimited retakes encourages mastery, it also results in a fluctuating workload for the instructor and resource considerations for the institution.
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