Assuming Normally Distributed Bone Density Scores And Others

Assuming normally distributed bone density scores and other statistical problems

Assuming normally distributed bone density scores and other statistical problems

The instructions involve analyzing six different statistical problems related to normal distribution, probability calculations, and standard normal Z-scores using Excel functions. Students are asked to identify errors in partially completed solutions, complete missing steps, and explain their reasoning throughout. The tasks include calculating probabilities for given Z-scores, converting probabilities to Z-scores, and interpreting areas under the normal curve, all utilizing Excel functions such as NORM.S.DIST and NORM.S.INV. Several problems also involve understanding the context of the data, such as bone density scores, heights of pilots, magnitudes of earthquakes, and their implications within the normal distribution framework.

Paper For Above instruction

The following analysis addresses six statistical problems involving the application of normal distribution concepts, probability calculations, and the use of Excel functions. Each problem is examined in detail, with correct steps and explanations provided to ensure understanding and proper application of statistical methods.

Problem 1: Probability of a bone density score between -1.03 and 2.01

Given that bone density scores follow a standard normal distribution (mean = 0, standard deviation = 1), we are asked to find the probability that a randomly selected subject's score falls between -1.03 and 2.01. To find this probability, we need to calculate the cumulative area (probability) to the left of 2.01 and subtract the cumulative area to the left of -1.03.

Using Excel functions, the calculations are as follows:

=P( Z < 2.01 ) = NORM.S.DIST(2.01, TRUE)


=P( Z < -1.03 ) = NORM.S.DIST(-1.03, TRUE)

Executing these in Excel provides:

  • NORM.S.DIST(2.01, TRUE) ≈ 0.9778
  • NORM.S.DIST(-1.03, TRUE) ≈ 0.1515

Therefore, the probability that the score lies between -1.03 and 2.01 is:

0.9778 - 0.1515 = 0.8263

Thus, there is approximately an 82.63% chance that a randomly selected subject has a bone density score between -1.03 and 2.01.

Problem 2: Percentage of women meeting height requirements

The U.S. Airforce requires pilot heights between 62 inches and 76 inches. Women's heights are normally distributed with a mean of 63.8 inches and a standard deviation of 2.9 inches. To find the percentage of women meeting this requirement, calculate the probability that a woman's height falls between 62 and 76 inches.

Converting the heights to Z-scores:

Z1 = (62 - 63.8) / 2.9 ≈ -0.6207


Z2 = (76 - 63.8) / 2.9 ≈ 4.8966

The probability that a woman's height is less than 76 inches is:

=NORM.S.DIST(4.8966, TRUE) ≈ 1 (since Z is very high, practically 1)

The probability that a woman's height is less than 62 inches is:

=NORM.S.DIST(-0.6207, TRUE) ≈ 0.2675

The percentage of women with heights between 62 and 76 inches is:

(1 - 0.2675) ≈ 0.7325 or 73.25%

Hence, approximately 73.25% of women meet the height requirement for pilots.

Problem 3 and 4: Cumulative area for a Z-score of -0.18

To find the cumulative area from the left under the curve for a Z-score of -0.18, use:

=NORM.S.DIST(-0.18, TRUE) ≈ 0.4292

This represents the probability that a Z-score is less than -0.18. The area on the right of the Z-score is simply:

=1 - 0.4292 ≈ 0.5708

Thus, approximately 42.92% of the distribution lies to the left of Z = -0.18, and about 57.08% lies to the right.

Problem 5: Z-score corresponding to an area of 0.6630 on the right

The student incorrectly used =NORM.S.INV(0.6630), leading to a Z-score of approximately 0.42. The mistake here is that NORM.S.INV returns the Z-score corresponding to a given left-tail probability. Since the question states that the area to the right under the curve is 0.6630, the left-tail area (cumulative area to the left) is:

1 - 0.6630 = 0.3370

Therefore, to find the correct Z-score, we should input this left-tail area into =NORM.S.INV(0.3370):

=NORM.S.INV(0.3370) ≈ -0.42

This matches the negative Z-score of approximately -0.42, indicating the value on the standard normal distribution with 66.30% area on the right is Z ≈ -0.42.

Problem 6: Percentage of earthquakes with magnitude less than 2.00

The Richter scale magnitudes are normally distributed with mean 1.192 and standard deviation 0.615. To find the percentage of microearthquakes (magnitudes less than 2.00), we first convert 2.00 to a Z-score:

Z = (2.00 - 1.192) / 0.615 ≈ 1.250

Using Excel:

=NORM.S.DIST(1.250, TRUE) ≈ 0.8944

This indicates that about 89.44% of earthquakes have magnitudes less than 2.00. Therefore, approximately 89.44% of earthquakes qualify as microearthquakes, which are generally unfelt by humans.

Conclusion

All six problems demonstrate core principles of normal distribution analysis—calculating probabilities from Z-scores, converting probabilities to Z-scores, and understanding the areas under the normal curve. Proper use of Excel functions like NORM.S.DIST and NORM.S.INV is essential in accurately solving these types of problems. Recognizing common pitfalls, such as misinterpreting the direction of areas or misusing the inverse function, helps ensure correct solutions and deepens understanding of statistical concepts in practical contexts.

References

  • Devore, J. L. (2015). Probability and Statistics for Engineering and the Sciences. Cengage Learning.
  • Moore, D. S., Notz, W., & Flinger, M. (2013). The Basic Practice of Statistics. W.H. Freeman and Company.
  • Ott, R. L. (2017). An Introduction to Statistical Methods and Data Analysis. Cengage Learning.
  • Agresti, A., & Franklin, C. (2017). Statistics: The Art and Science of Data. Pearson.
  • U.S. Census Bureau. (2022). Height and Weight Data Sets. https://www.census.gov
  • National Earthquake Information Center. (2023). Earthquake Magnitude Data. https://earthquake.usgs.gov
  • Alonso, J. M., et al. (2012). "Statistical Methods in Psychology and Education." Applied Psychological Measurement, 36(4), 273–281.
  • Johnson, R. A., & Wichern, D. W. (2014). Applied Multivariate Statistical Analysis. Pearson.
  • Chin, F. (1986). Donald Duk. University of California Press.
  • Chang, D. (1993). The Other Side of the Postcard. Wesleyan University Press.

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