Normal Probability Worksheet March 160 Statistics Version 1 ✓ Solved

Normal Probability Worksheet Mth 160 Statisticsversion 1 0913read

Read the following scenario and complete each of the four problem sets below: A. Use the z-table to determine the following probabilities. Sketch a normal curve for each problem with the appropriate probability area shaded. 1. P(z > 2.34) 2. P(z

Read the following scenario: Park Rangers in a Yellowstone National Park have determined that fawns less than 6 months old have a body weight that is approximately normally distributed with a mean µ = 26.1 kg and standard deviation σ = 4.2 kg. Let x be the weight of a fawn in kilograms. Complete each of the following steps for the word problems below:

  • Rewrite each of the following word problems into a probability expression, such as P(x>30).
  • Convert each of the probability expressions involving x into probability expressions involving z, using the information from the scenario.
  • Sketch a normal curve for each z probability expression with the appropriate probability area shaded.
  • Solve the problem.

1. What is the probability of selecting a fawn less than 6 months old in Yellowstone that weighs less than 25 kilograms? 2. What is the probability of selecting a fawn less than 6 months old in Yellowstone that weighs more than 19 kilograms? 3. What is the probability of selecting a fawn less than 6 months old in Yellowstone that weighs between 30 and 38 kilograms? 4. If a fawn less than 6 months old weighs 16 pounds, would you say that it is an unusually small animal? Explain and verify your answer mathematically. 5. What is the weight of a fawn less than 6 months old that corresponds with a 20% probability of being randomly selected? Explain and verify your answer mathematically.

Sample Paper For Above instruction

Introduction

Understanding the probabilities associated with normally distributed data is essential in statistical analysis, especially in biological and ecological studies such as those involving animal weights. The following paper applies the principles of normal distribution, Z-scores, and probability calculations to a real-world scenario involving the weights of fawns in Yellowstone National Park.

Part A: Normal Distribution and Z-Values

Given the problem, we first identify key parameters: mean (μ) = 26.1 kg, standard deviation (σ) = 4.2 kg. Utilizing the Z-table, we interpret probabilities and create corresponding sketches of the normal curve for each given probability.

1. P(z > 2.34)

This probability indicates the area under the normal curve to the right of z = 2.34. Consulting the Z-table, the cumulative probability P(Z 2.34) = 1 - 0.9904 = 0.0096.

Sketch: Shade the area to the right of z=2.34.

2. P(z

The cumulative probability for Z

Sketch: Shade the area to the left of z=-1.56.

3. P(z = 1.23)

Since a point probability in a continuous distribution is effectively zero, P(z=1.23) = 0.

Sketch: No area; point probability.

4. P(-1.82

Find P(Z

Sketch: Shade the area between Z=-1.82 and Z=0.79.

5. Z-score for 67% probability

To find Z associated with 67% probability, locate the Z-value where P(Z

Part B: Real-world Application - Fawn Weights

Using the normal distribution model, the problems below involve converting weights to Z-scores and interpreting probabilities.

1. Probability of a fawn weighing less than 25 kg

Probability expression: P(x

Compute Z: z = (x - μ)/σ = (25 - 26.1)/4.2 ≈ -0.26

Look up P(Z

Sketch: Shade the area to the left of z=-0.26 on the normal curve.

2. Probability of a fawn weighing more than 19 kg

Probability expression: P(x > 19)

Z-score: z = (19 - 26.1)/4.2 ≈ -1.69

P(Z > -1.69): 1 - P(Z

Sketch: Shade area to the right of z=-1.69.

3. Probability of a fawn weighing between 30 and 38 kg

Expressed as: P(30

Z(30): (30 - 26.1)/4.2 ≈ 0.93

Z(38): (38 - 26.1)/4.2 ≈ 2.88

P(0.93

Sketch: Shade between Z=0.93 and Z=2.88.

4. Is a 16-pound fawn unusually small?

Convert 16 pounds to kilograms: 16 lbs * 0.453592 = 7.26 kg.

Z = (7.26 - 26.1)/4.2 ≈ -4.55.

P(Z

Conclusion: Yes, a 7.26 kg fawn is extremely small, below typical variation — considered unusually small.

5. Weight corresponding to 20% probability

The Z-score for 20% probability (P(Z

Use z = (x - μ)/σ: x = z*σ + μ = (-0.84)(4.2) + 26.1 ≈ 22.23 kg.

Therefore, a fawn weighing about 22.23 kg represents the 20th percentile in weight.

Conclusion

This analysis demonstrates how normal distribution and Z-score calculations facilitate understanding of variation in biological measurements, such as animal weights. By transforming raw data into standardized Z-scores, researchers and practitioners can estimate probabilities, identify outliers, and make informed decisions about animal health and ecology.

References

  • Altman, D. G. (1991). Practical Statistics for Medical Research. Chapman & Hall.
  • DeGroot, M. H., & Schervish, M. J. (2012). Probability and Statistics. Addison-Wesley.
  • Hart, J. (2001). Introduction to Biostatistics. CRC Press.
  • Sokal, R. R., & Rohlf, F. J. (2012). Biometry. W. H. Freeman.
  • Moore, D. S., McCabe, G. P., & Craig, B. A. (2012). Introduction to the Practice of Statistics. W.H. Freeman.
  • Newcombe, R. G. (1998). Two-sided confidence intervals for the single proportion: Comparison of seven methods. Canadian Journal of Statistics.
  • Norman, G., & Streiner, D. (2008). Biostatistics: The Bare Essentials. BC Decker Inc.
  • Weiss, N. A. (2012). Introductory Statistics. Pearson.
  • Zar, J. H. (2010). Biostatistical Analysis. Pearson.
  • Zwillinger, D. (2003). CRC Standard Probability and Statistics Tables and Formulae. CRC Press.