John Jayro Ofmotah Pharmaceuticals Found That The Scores

John Jayro Ofmotah Pharmaceuticals Found That The Scores On A Trai

John & Jayro, of Motah Pharmaceuticals, found that the scores on a training exam for their new trainees were normally distributed with a mean score of 67 and a standard deviation of 8. Use this information in answering “a” through “c”, draw bell curves.

a. About what percentage of the exam scores were above 85?

b. A failing grade on the exam was anything 1.5 or more standard deviations below the mean. What was the cutoff for a failing score?

c. What are the standard scores that correspond to trainee scores between 66 and 79 on the exam?

Paper For Above instruction

The distribution of exam scores among trainees at Motah Pharmaceuticals provides valuable insights into student performance and the effectiveness of training methods. Given that these scores are normally distributed with a mean of 67 and a standard deviation of 8, we can assess the percentage of students scoring above certain thresholds, determine failing scores, and convert raw scores into standard scores (z-scores).

Part a: Percentage of scores above 85

To find the percentage of exam scores above 85, we first calculate the z-score for 85 using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where \(X = 85\), \(\mu = 67\), and \(\sigma = 8\).

Calculating this:

\[ z = \frac{85 - 67}{8} = \frac{18}{8} = 2.25 \]

Referring to standard normal distribution tables or using statistical software, the cumulative probability up to z = 2.25 is approximately 0.9878. This means about 98.78% of students scored below 85.

Therefore, the percentage of students scoring above 85 is:

\[ 1 - 0.9878 = 0.0122 \text{ or } 1.22\% \]

This indicates that approximately 1.22% of the trainees scored higher than 85 on the exam.

Part b: Failing score cutoff

A failing grade is defined as a score 1.5 or more standard deviations below the mean. The z-score corresponding to this threshold is:

\[ z = -1.5 \]

Using the z-table or software, the probability of a z-score less than -1.5 is approximately 0.0668.

Calculating the raw score:

\[ X = \mu + z \times \sigma = 67 + (-1.5) \times 8 = 67 - 12 = 55 \]

Thus, the cutoff score for failing on the exam is 55. Any score below or equal to this value indicates a failing grade.

Part c: Standard scores for scores between 66 and 79

To find the z-scores corresponding to raw scores 66 and 79:

- For 66:

\[ z = \frac{66 - 67}{8} = \frac{-1}{8} = -0.125 \]

- For 79:

\[ z = \frac{79 - 67}{8} = \frac{12}{8} = 1.5 \]

Hence, scores between 66 and 79 correspond to z-scores between approximately -0.125 and 1.5.

Conclusion

These calculations elucidate the performance distribution within this training cohort. The small percentage of students scoring above 85 suggests that exceptional performance was rare. The failing score cutoff at 55 points provides a clear benchmark for identifying underperformers. Lastly, the z-scores associated with scores between 66 and 79 demonstrate the relative standing of students within the distribution.

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