The Length Of Time X Needed To Complete A State Teachers Exa

The Length Of Time X Needed To Complete A State Teachers Examinati

The length of time X needed to complete a state teachers' examination is normally distributed with a mean of 80 minutes and a standard deviation of 15 minutes. For each question below, draw a diagram indicating known values and what needs to be found.

(a) What proportion of candidates complete the exam in an hour or less?

(b) How much time should be allotted so that 90% of examinees complete the test?

(c) If the time limit was set at 1 hour and 45 minutes (105 minutes), what proportion of candidates would not complete the exam?

Paper For Above instruction

The problem revolves around understanding the properties of the normal distribution and applying them to real-world testing scenarios. Given that the time X to complete a teacher’s examination follows a normal distribution with a mean (μ) of 80 minutes and a standard deviation (σ) of 15 minutes, we can employ standard Z-scores to determine the probabilities and the necessary time thresholds.

Part (a): Proportion of Candidates Completing in 60 Minutes or Less

To find the proportion of candidates finishing within 60 minutes, we first calculate the Z-score corresponding to 60 minutes:

Z = (X - μ) / σ = (60 - 80) / 15 = -20 / 15 ≈ -1.33

Using the standard normal distribution table, the cumulative probability for Z = -1.33 is approximately 0.0918. This indicates that about 9.18% of candidates complete the exam in an hour or less.

Part (b): Time to Cover 90% of Candidates

To find the time T where 90% of students finish, we identify the Z-score associated with the 90th percentile, which is approximately 1.28. Then, we convert this Z-score back to the original units:

T = μ + Z σ = 80 + 1.28 15 ≈ 80 + 19.2 = 99.2 minutes

Thus, approximately 99.2 minutes should be allocated to ensure that 90% of examinees finish the test.

Part (c): Proportion of Candidates Not Completing in 1 Hour 45 Minutes (105 Minutes)

Calculate the Z-score for 105 minutes:

Z = (105 - 80) / 15 = 25 / 15 ≈ 1.67

The cumulative probability up to Z = 1.67 is roughly 0.9525. Therefore, the proportion of candidates who do not complete the exam within 105 minutes is:

1 - 0.9525 ≈ 0.0475 or 4.75%

Implications

These calculations assist in setting appropriate time limits and understanding candidate performance. For example, knowing that only about 9.18% of candidates finish within an hour helps in evaluating whether a shorter exam duration is feasible. Conversely, setting a time limit around 99 minutes ensures that the vast majority (90%) will conclude the exam comfortably, optimizing fairness and efficiency.

Distribution of Student Scores and Cutting Points

The next problem involves selecting cutpoints based on the normal distribution of exam scores and the percentages associated with each grade category.

Part (a): Percentage of Students in Each Grade Category

Assuming the scores are normally distributed, the cutpoints are derived from Z-scores relative to their standard deviations from the mean. The grade categories are:

  • A: more than 1.5 SDs above the mean
  • B: between 0.5 and 1.5 SDs above the mean
  • C: within 0.5 SDs of the mean (both sides)
  • D: more than 0.5 SDs below the mean

Using standard Z-value probabilities:

  • Z > 1.5: area to the right is approximately 0.0668, so percentage for A is about 6.68%.
  • Z between 0.5 and 1.5: cumulative probabilities for Z = 0.5 (0.6915), Z = 1.5 (0.9332); B percentage = 93.32% - 69.15% = 24.17%
  • Z within ±0.5: for Z = 0.5 (0.6915), so within area is 69.15%; C percentage = 69.15% - 6.68% (above 1.5 SD) + rest within ±0.5 (twice, as both sides), but since mixed, the most straightforward is to note the central region within ±0.5 SD is approximately 38.3% (double the 19.2% for Z = 0.5 to 0). Assuming symmetry, the percentages are approx:
  • D: less than -0.5 SD: area to the left of Z = -0.5 is about 0.3085 or 30.85%.

Note: Actual precise calculations may vary slightly but generally, the approximate percentages are:

  • A: 6.7%,
  • B: 24.2%,
  • C: 38.3%,
  • D: 30.8%

Part (b): Corresponding Test Scores in a Given Distribution

Given a mean of 55 and standard deviation of 10, the cutpoints are calculated as:

  • A: mean + 1.5 SD = 55 + 1.5 10 = 55 + 15 = 70
  • B: between 0.5 and 1.5 SDs above the mean: 55 + 0.5 * 10 = 60 and 70 respectively
  • C: within ±0.5 SDs: 50 to 60
  • D: less than 50

Similarly, scores for the lower bounds are:

  • Below 50 for D (Z = -0.5)
  • Between 50 and 60 for C
  • Between 60 and 70 for B
  • Above 70 for A

Part (c): Setting Cut-Scores for Given Percentages

To allocate percentages (15% A, 30% B, 35% C, remainder D), the Z-scores corresponding to cumulative probabilities are:

  • A: 85th percentile corresponding to Z ≈ 1.04
  • B: 55th percentile (for cumulative 55%), Z ≈ 0.13
  • C: cumulative 20%, Z ≈ -0.84

These Z-scores define the cutpoints in standard deviation units, which can be translated back to raw scores using:

Score = mean + Z * standard deviation

Thus, the cutpoints are approximately:

  • Upper bound for C: mean + 0.84 * SD = mean + 8.4
  • Upper bound for B: mean + 0.13 * SD = mean + 1.3
  • Upper bound for A: mean + 1.04 * SD = mean + 10.4

These approximate cutpoints partition the scores into the desired percentage categories.

Self-Esteem Scores and Hypothesis Testing

The final scenario involves a hypothesis test comparing the mean self-esteem scores of domestic violence victims with the general population. The general population mean is 48, with a standard deviation of 12. A sample of 56 victims has a mean of 44.1.

Calculating the probability of observing a sample mean of 44.1 or less, assuming the population mean is 48, involves the standard error (SE):

SE = σ / √n = 12 / √56 ≈ 12 / 7.48 ≈ 1.60

Z = (sample mean - population mean) / SE = (44.1 - 48) / 1.60 ≈ -3.9 / 1.60 ≈ -2.44

The probability corresponding to Z = -2.44 is approximately 0.0073, or 0.73%. This indicates a very low likelihood that such a low mean would occur if the actual mean was 48.

In hypothesis testing, such a small p-value suggests that we can reject the null hypothesis at conventional significance levels, supporting the alternative that domestic violence victims may indeed have lower self-esteem than the general female population.

This conclusion aligns with the psychologist's hypothesis, providing statistical evidence of a significant difference.

Conclusion

Overall, these statistical analyses demonstrate the usefulness of the normal distribution in evaluating test scores and candidate performance, as well as in testing hypotheses about population parameters. Accurately calculating probabilities and setting appropriate cutpoints are vital components of educational assessment and psychological research, aiding in informed decision-making and policy development.

References

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