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The Results Of A Statewide Exam For Assessing the Mathematic

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The Results Of A Statewide Exam For Assessing the Mathematics Skills O

The results of a statewide exam for assessing the mathematics skills of realtors were normally distributed with a mean score of 73 and a standard deviation of 11. The realtors who scored in the top 15% are to receive a special certificate, while those in the bottom 10% will be required to attend a remedial workshop. a. What minimum score does a realtor need in order to avoid the workshop? b. What score does a realtor need in order to receive the certificate? Suppose x is a normally distributed random variable with µ = 12 and σ = 2. Find each of the following probabilities. A) P(x greater than or equal to 14.5) B) P(x less than or equal to 9) C) P(13.44 less than or equal to x less than or equal to 17.42) D) P(6.56 less than or equal to x less than or equal to 14.62)

Paper For Above instruction

Introduction

The analysis of exam scores using statistical methods provides valuable insights into the performance and distribution of scores among populations. This paper addresses two related statistical problems involving a statewide mathematics exam for realtors and the properties of a normally distributed variable, focusing on finding specific score thresholds and calculating probabilities using the normal distribution.

Part 1: Exam Score Analysis for Realtors

The statewide exam scores are normally distributed with a mean (µ) of 73 and a standard deviation (σ) of 11. Understanding the distribution allows us to determine cut-off scores corresponding to specific percentiles, namely the top 15% and the bottom 10%.

A. Minimum score to avoid the remedial workshop

To find the minimum score needed to avoid the remedial workshop, we must calculate the score that corresponds to the 10th percentile of the normal distribution. This percentile indicates the score below which 10% of the scores fall.

First, we convert the percentile to a z-score. The 10th percentile (P10) corresponds to a cumulative probability of 0.10. Using the standard normal distribution table or a calculator, the z-score for P10 is approximately -1.28.

Next, the score (X) corresponding to this z-score is calculated as:

\[X = \mu + z\sigma = 73 + (-1.28)(11) = 73 - 14.08 = 58.92\]

Therefore, any realtor scoring above approximately 58.92 will avoid the remedial workshop.

B. Score needed to receive the certificate

Similarly, to find the cutoff score for the top 15%, we identify the 85th percentile (since the top 15% exceed this score). The cumulative probability for this percentile is 0.85.

The z-score corresponding to 0.85 is approximately 1.04.

The score (X) at this percentile is:

\[X = 73 + (1.04)(11) = 73 + 11.44 = 84.44\]

Thus, realtors scoring above approximately 84.44 will receive the certificate.

Part 2: Probabilities for a Normal Variable X with µ = 12 and σ = 2

Given \(X \sim N(\mu=12, \sigma=2)\), compute the following:

A. \(P(X \geq 14.5)\)

Convert 14.5 to a z-score:

\[z = \frac{14.5 - 12}{2} = 1.25\]

Using the standard normal table, \(P(Z \geq 1.25) = 1 - P(Z \leq 1.25) \approx 1 - 0.8944 = 0.1056\).

B. \(P(X \leq 9)\)

Convert 9 to a z-score:

\[z = \frac{9 - 12}{2} = -1.5\]

\(P(Z \leq -1.5) \approx 0.0668\).

C. \(P(13.44 \leq X \leq 17.42)\)

Convert both bounds:

\[

z_1 = \frac{13.44 - 12}{2} = 0.72 \\

z_2 = \frac{17.42 - 12}{2} = 2.71

\]

Find probabilities: \[

P(Z \leq 2.71) \approx 0.9966 \\

P(Z \leq 0.72) \approx 0.7642

\]

Thus, the probability:

\[

P(13.44 \leq X \leq 17.42) = 0.9966 - 0.7642 = 0.2324

\]

D. \(P(6.56 \leq X \leq 14.62)\)

Convert bounds:

\[

z_1 = \frac{6.56 - 12}{2} = -2.72 \\

z_2 = \frac{14.62 - 12}{2} = 1.31

\]

Corresponding probabilities:

\[

P(Z \leq 1.31) \approx 0.9049 \\

P(Z \leq -2.72) \approx 0.0033

\]

Probability between bounds:

0.9049 - 0.0033 = 0.9016

Conclusion

Applying the properties of the normal distribution, we identified the minimum scores for certificates and remediation workshops among realtors. Additionally, the probability calculations for the specified ranges demonstrate how standard normal distribution tables and z-score conversions facilitate understanding of data within a normal distribution context. These statistical tools are vital for educational assessments and various applications across disciplines.

References

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