Deliverable 02 Worksheet Instructions The Following W 842639

Deliverable 02worksheetinstructionsthe Following Worksheet Is Shown

The worksheet presents a series of statistical problems related to normal distribution, involving probability calculations, z-scores, and interpretation of areas under the standard normal curve. The problems include calculating probabilities between z-scores, finding percentage of population meeting height requirements, calculating z-scores for given data points, and interpreting areas under the normal curve. Several answers are provided with corrections and explanations needed to clarify and complete the step-by-step solutions accurately.

Paper For Above instruction

Understanding and Solving Normal Distribution Problems: A Step-by-Step Approach

Normal distribution problems are fundamental in statistics because many real-world variables follow a bell-shaped, symmetric distribution. Properly solving these problems involves understanding the relationship between raw scores (like height, pulse rate, or bone density), z-scores, and the corresponding areas under the standard normal curve. This paper presents a detailed explanation of each problem provided, correcting errors where necessary, and elucidating each step to ensure clarity and comprehension.

Problem 1: Probability That a Bone Density Test Score Falls Between Two Z-Scores

The problem involves a bone density test where scores are normally distributed with a mean of zero and a standard deviation of one (standard normal distribution). The task is to find the probability that a randomly selected subject's score falls between -1.93 and 2.37.

The solution involves two primary steps: finding the cumulative probability for each z-score and then subtracting these probabilities to find the probability of being between them.

First, using a standard normal distribution table or a calculator like Excel's NORM.DIST function, the probabilities are obtained:

  • P(Z
  • P(Z

Since the probability that Z falls between -1.93 and 2.37 is the difference of these two cumulative probabilities, we perform:

0.9911 - 0.0268 = 0.9643

This is the corrected probability, indicating approximately 96.43% of subjects have a bone density score within this range.

Problem 2: Percentage of Women Meeting Height Requirements

The U.S. Airforce requires women to have a height between 64 inches and 77 inches. Women's heights are normally distributed with a mean of 63.8 inches and a standard deviation of 2.6 inches. The goal is to determine what percentage of women meet this requirement.

The steps involve calculating the z-scores for the bounds of the interval:

  • Z for 64 inches: (64 - 63.8) / 2.6 ≈ 0.077
  • Z for 77 inches: (77 - 63.8) / 2.6 ≈ 5.077

Using Excel or standard normal tables:

  • P(Z
  • P(Z

The probability that a woman's height falls between these bounds is approximately:

1.0000 - 0.5308 ≈ 0.4692

However, note that the upper z-score is extremely high, so the probability of being less than 77 inches is close to 1, and the total probability between the two bounds is close to 53%. Nonetheless, the original solution provided a higher percentage (96.80%), which suggests an error in calculating the z-scores or interpreting the distribution. Correct calculation confirms approximately 53% of women meet the height requirement.

Problem 3: Calculating the Z-Score for a Given Pulse Rate

Given the mean pulse rate of 69.4 beats per minute and a standard deviation of 11.3, find the z-score for a pulse of 66 bpm.

Using the z-score formula:

Z = (X - μ) / σ = (66 - 69.4) / 11.3 ≈ -3.4 / 11.3 ≈ -0.30

The student's answer is correct; the z-score is approximately -0.30. This indicates that a pulse rate of 66 bpm is 0.30 standard deviations below the mean.

Problem 4: Cumulative Area for a Z-Score of -1.645 and Its Complement

The question asks for the cumulative probability from the left under the curve for Z = -1.645 and the area to the right.

Using Excel's NORM.DIST function:

=NORMDIST(-1.645, 0, 1, TRUE) ≈ 0.04998

Thus, the area under the curve to the left of Z = -1.645 is approximately 0.04998.

The area to the right is the complement:

1 - 0.04998 ≈ 0.95002

This means approximately 4.998% of the distribution lies below Z = -1.645, and about 95.002% lies above it.

Problem 5: Z-Score Corresponding to a Right Tail Area of 0.8980

If the area to the right under the normal curve is 0.8980, then the area to the left is:

1 - 0.8980 = 0.1020

To find the z-score corresponding to this cumulative area, we use Excel's NORM.INV function:

=NORM.INV(0.1020, 0, 1) ≈ -1.27

Therefore, the z-score corresponding to an area of 0.8980 from the right is approximately -1.27, which aligns with the student's answer.

Problem 6: Percentage of Men Fitting into a Manhole of Diameter 22 Inches

Men's shoulder widths are normally distributed with a mean of 18.2 inches and a standard deviation of 2.09 inches. The question asks for the percentage of men whose shoulder width is less than 22 inches, so they will fit into the manhole.

Calculating the z-score:

Z = (22 - 18.2) / 2.09 ≈ 3.8 / 2.09 ≈ 1.82

Using Excel's NORM.DIST:

=NORMDIST(22, 18.2, 2.09, TRUE) ≈ 0.9655

Multiplying by 100% gives:

96.55%

This indicates that approximately 96.55% of men have shoulder widths less than 22 inches and will fit through the manhole.

Conclusion

These problems exemplify key concepts in understanding the normal distribution, including calculating probabilities, z-scores, and interpreting areas under the curve. Correct application of formulas, accurate use of statistical tools like Excel, and careful interpretation of results are essential for solving such statistical problems reliably.

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