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The instructions require helping a student walk through solving statistical problems step by step. The problems involve understanding normal distribution, calculating probabilities, finding z-scores, and interpreting areas under the curve. The solutions should include detailed explanations of each step, correct formatting of variables and equations, and proper use of statistical terminology. The goal is to clarify the reasoning process and ensure the student understands how to approach and solve each problem correctly.

Paper For Above instruction

The following paper provides detailed solutions and explanations to the statistical problems presented, demonstrating thorough understanding of the concepts and proper application of methods such as standard normal distribution calculations, z-score computations, and probability interpretations.

Introduction

Statistics, particularly the study of normal distribution, is fundamental in understanding how data behaves in various scientific and practical contexts. This paper walks through six specific problems, illustrating the detailed steps necessary to compute probabilities, z-scores, and other relevant statistical measures. Each problem is approached with clarity, emphasizing the reasoning behind each step to facilitate comprehension and application in similar problems.

Problem 1: Probability of a Bone Density Score between -1.53 and 1.98

Given that bone density scores follow a standard normal distribution, the task is to find the probability that a randomly selected subject’s score is between -1.53 and 1.98.

First, we recall that the standard normal distribution has a mean of 0 and a standard deviation of 1. To find these probabilities, we use the standard normal table or software like Excel. The Student's calculations indicate:

  • P(Z
  • P(Z

To find the probability that Z is between -1.53 and 1.98, subtract the cumulative probability at -1.53 from that at 1.98:

Step 1: Write the probability expression:

P(-1.53

Step 2: Plug in the values from the table:

= 0.9761 - 0.0630 = 0.9131

Thus, there is approximately a 91.31% chance that a subject's bone density score falls between -1.53 and 1.98.

Problem 2: Percentage of Women Meeting Height Requirements

The U.S. Airforce requires pilots to have a height between 64 in. and 77 in. The heights are normally distributed with a mean of 65 in. and a standard deviation of 3.5 in. We aim to find the percentage of women whose heights meet this criterion.

Step 1: Convert the height bounds to z-scores:

  • For 64 in.:

    z = (64 - 65) / 3.5 = -1 / 3.5 ≈ -0.2857

  • For 77 in.:

    z = (77 - 65) / 3.5 = 12 / 3.5 ≈ 3.4286

Step 2: Find the cumulative probabilities for these z-scores:

Using standard normal tables or software:

  • P(Z
  • P(Z

Step 3: Calculate the probability that a woman’s height falls between 64 and 77 inches:

P(64

Expressed as a percentage, approximately 61.09% of women meet the height requirement.

Problem 3: Z-Score for a Pulse Rate of 66

Women’s pulse rates are normally distributed with a mean of 69.4 beats per minute and a standard deviation of 11.3 beats per minute. To find the z-score corresponding to a pulse rate of 66:

Step 1: Apply the z-score formula:

z = (X - μ) / σ = (66 - 69.4) / 11.3 ≈ -3.4 / 11.3 ≈ -0.3018

This indicates that a pulse rate of 66 bpm is approximately 0.302 standard deviations below the mean.

Problem 4: Area Under the Curve for a Z-Score of -0.875

To find the cumulative area to the left of z = -0.875:

  • Consult a standard normal table or software. P(Z

The area on the right of z = -0.875 is:

1 - 0.1908 = 0.8092

This means approximately 19.08% of the distribution lies to the left, and 80.92% lies to the right of this z-score.

Problem 5: Z-Score Corresponding to an Area of 0.6573 from the Right

The area to the right under the standard normal curve is 0.6573. To find the corresponding z-score, first determine the area to the left:

Area to the left = 1 - 0.6573 = 0.3427

Using Excel or a z-score table, apply the inverse standard normal function:

=NORM.INV(0.3427, 0, 1) ≈ -0.41

Thus, the z-score with 65.73% area to the right is approximately -0.41, indicating a position below the mean.

Problem 6: Men Fitting into a Manhole Cover of 22.5 Inches

Men’s shoulder widths are normally distributed with a mean of 18.2 in. and a standard deviation of 2.09 in. The question is: what percentage of men have shoulder widths less than 22.5 in., thus fitting into the manhole?

Step 1: Calculate the z-score for 22.5 inches:

z = (22.5 - 18.2) / 2.09 ≈ 4.3 / 2.09 ≈ 2.057

Step 2: Find the cumulative probability for this z-score:

P(Z

Therefore, approximately 98.04% of men will fit into a manhole with a diameter of 22.5 inches.

This high percentage reflects that the vast majority of men's shoulder widths are below this threshold, making the manhole suitable for most individuals.

Conclusion

This detailed analysis demonstrates how to approach normal distribution problems systematically. Calculations of z-scores, use of standard normal tables or software, and understanding of areas under the curve are essential skills in statistics. Correctly interpreting these results allows for informed decision-making and understanding data distribution in real-world contexts.

References

  • Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences (8th ed.). Brooks/Cole.
  • Moore, D. S., McCabe, G. P., & Craig, B. A. (2017). Introduction to the Practice of Statistics (9th ed.). W. H. Freeman.
  • Newbold, P., Carlson, W. L., & Thorne, B. (2013). Statistics for Business and Economics (8th ed.). Pearson.
  • Wasserman, L. (2004). All of Statistics: A Concise Course in Statistical Inference. Springer.
  • Agresti, A., & Franklin, C. (2017). Statistics: The Art and Science of Learning from Data. Pearson.
  • Johnson, R. A., & Wichern, D. W. (2014). Applied Multivariate Statistical Analysis. Pearson.
  • Ott, R. L., & Longnecker, M. (2010). An Introduction to Statistical Methods and Data Analysis. Brooks/Cole.
  • Rice, J. A. (2007). Mathematical Statistics and Data Analysis. Cengage Learning.
  • Willmot, C. J. (2012). Statistical Methods for Quality Improvement. John Wiley & Sons.
  • Zar, J. H. (2010). Biostatistical Analysis. Pearson.

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