Normal Distribution and Inverse Normal Calculations

Normal Distribution Practice Worksheet

Overview of Inverse Normal Distribution Calculations

  • The normal distribution is defined by two primary parameters:
    • Mean: μ\mu (or sample mean xˉ\bar{x}), which represents the center of the symmetric bell curve.
    • Standard deviation: σ\sigma, which represents the spread or variability of the distribution.
  • The inverse normal function, typically executed via the graphing calculator command invNorm\text{invNorm}, is used to calculate the specific boundary value (xx) corresponding to a given cumulative probability or area under the normal curve:
    • General syntax: invNorm(Area,μ,σ,[Tail])\text{invNorm}(\text{Area}, \mu, \sigma, [\text{Tail}])
    • In standard configurations without tail specification, the function assumes cumulative area measured from the extreme left up to the cutoff score (Left\text{Left} tail).
    • Advanced calculator operating systems allow tail orientation arguments: Left\text{Left}, Right\text{Right}, or Center\text{Center}.
  • Standard transformation relationship between the raw score (xx), mean (μ\mu), standard deviation (σ\sigma), and standard normal score (zz):
    • z=x−μσz = \frac{x - \mu}{\sigma}
    • x=μ+z×σx = \mu + z \times \sigma
    • Where z=invNorm(Area,0,1)z = \text{invNorm}(\text{Area}, 0, 1).

Problem 5: Wechsler Adult Intelligence Scale Subtest Scores

  • Distribution Parameters:
    • Model: The raw scores on the subtest form a normal distribution: X∼N(35,62)X \sim N(35, 6^2).
    • Mean: xˉ=35\bar{x} = 35
    • Standard deviation: σ=6\sigma = 6
  • Part (a): Determination of the 28th Percentile
    • Objective: Determine the raw score cutoff that represents the 28th percentile (cumulative area=0.28\text{cumulative area} = 0.28).
    • Calculator Command: invNorm(0.28,35,6,Left)\text{invNorm}(0.28, 35, 6, \text{Left})
    • Mathematical Calculation:
    • Standard normal cutoff: z0.28=invNorm(0.28,0,1)≈−0.5828z_{0.28} = \text{invNorm}(0.28, 0, 1) \approx -0.5828
    • Raw score evaluation: x=35+(−0.5828)×6=35−3.497≈31.50x = 35 + (-0.5828) \times 6 = 35 - 3.497 \approx 31.50 (recorded calculation on worksheet: 31.3131.31)
    • Curve Sketch Characteristics:
    • Symmetrical bell-shaped curve centered at xˉ=35\bar{x} = 35.
    • A vertical cutoff line placed in the lower half of the distribution at approximately 31.3131.31.
    • The region beneath the curve to the left of 31.3131.31 is shaded, representing the lowest 28%28\% of the distribution.
    • Written Conclusion: 31.3131.31 is the 28th percentile for intelligence score.
  • Part (b): Determination of the 90th Percentile
    • Objective: Determine the raw score cutoff that represents the 90th percentile (cumulative area=0.90\text{cumulative area} = 0.90).
    • Calculator Command: invNorm(0.90,35,6,Left)\text{invNorm}(0.90, 35, 6, \text{Left})
    • Mathematical Calculation:
    • Standard normal cutoff: z0.90=invNorm(0.90,0,1)≈1.28155z_{0.90} = \text{invNorm}(0.90, 0, 1) \approx 1.28155
    • Raw score evaluation: x=35+(1.28155)×6=35+7.6893=42.6893≈42.69x = 35 + (1.28155) \times 6 = 35 + 7.6893 = 42.6893 \approx 42.69
    • Curve Sketch Characteristics:
    • Symmetrical bell-shaped curve centered at xˉ=35\bar{x} = 35.
    • A vertical cutoff line positioned in the upper tail at 42.6942.69.
    • The entire region beneath the curve to the left of 42.6942.69 is shaded, encompassing 90%90\% of the total area.
    • Written Conclusion: 42.6942.69 is the 90th percentile intelligence score.

Problem 6: SAT Math Section Score Distribution

  • Distribution Parameters:
    • Model: The SAT Math section scores are approximately normally distributed: X∼N(500,1002)X \sim N(500, 100^2).
    • Mean: xˉ=500\bar{x} = 500
    • Standard deviation: σ=100\sigma = 100
  • Part (a): Minimum Score for the Top 15%
    • Objective: Determine the minimum score required to place in the top 15%15\% of test takers.
    • Area Interpretation and Calculator Commands:
    • Being in the top 15%15\% implies that 85%85\% of scores fall below this cutoff (cumulative left area=1.00−0.15=0.85\text{cumulative left area} = 1.00 - 0.15 = 0.85).
    • Left-tail command: invNorm(0.85,500,100,Left)\text{invNorm}(0.85, 500, 100, \text{Left})
    • Right-tail command: invNorm(0.15,500,100,Right)\text{invNorm}(0.15, 500, 100, \text{Right})
    • Uncorrected evaluation for the lower 15th percentile: invNorm(0.15,500,100,Left)=396.36\text{invNorm}(0.15, 500, 100, \text{Left}) = 396.36 (which designates the bottom 15%15\% rather than the top 15%15\%).
    • Mathematical Calculation for Upper Cutoff:
    • Standard normal cutoff: z0.85≈1.0364z_{0.85} \approx 1.0364
    • Raw score: x=500+(1.0364)×100=603.64x = 500 + (1.0364) \times 100 = 603.64
    • Curve Sketch Characteristics:
    • Symmetrical bell curve centered at 500500.
    • Shaded upper tail above the cutoff threshold representing the top 15%15\% of scores (or lower tail below 396.36396.36 for the bottom 15%15\%).
    • Concluding Sentence: 396.36396.36 represents the bottom 15%15\% of math scores, whereas a score of 603.64603.64 is required to reach the top 15%15\%.
  • Part (b): Score Range for the Middle 80%
    • Objective: Identify the interval of scores containing the central 80%80\% of the distribution.
    • Tail Distribution:
    • Total excluded area: 1.00−0.80=0.201.00 - 0.80 = 0.20.
    • Area per tail: 0.202=0.10\frac{0.20}{2} = 0.10 in each tail (lower 10%10\% and upper 10%10\%).
    • Cumulative left area for lower cutoff: 0.100.10.
    • Cumulative left area for upper cutoff: 0.10+0.80=0.900.10 + 0.80 = 0.90.
    • Calculator Commands:
    • Center tail mode: invNorm(0.80,500,100,Center)\text{invNorm}(0.80, 500, 100, \text{Center})
    • Separate bounds mode: invNorm(0.10,500,100,Left)\text{invNorm}(0.10, 500, 100, \text{Left}) and invNorm(0.90,500,100,Left)\text{invNorm}(0.90, 500, 100, \text{Left})
    • Mathematical Calculation:
    • Critical values: z=±1.28155z = \pm 1.28155
    • Lower limit: x1=500−(1.28155)×100=371.8445≈371.84x_1 = 500 - (1.28155) \times 100 = 371.8445 \approx 371.84 (recorded on worksheet as 371.69371.69 / 371.84371.84)
    • Upper limit: x2=500+(1.28155)×100=628.1555≈628.16x_2 = 500 + (1.28155) \times 100 = 628.1555 \approx 628.16
    • Curve Sketch Characteristics:
    • Normal curve centered at 500500.
    • Two vertical boundary lines at 371.84371.84 and 628.16628.16.
    • Symmetrically shaded central region comprising 80%80\% of the total distribution area.
    • Written Conclusion: Middle 80%80\% of SAT scores are between 371.84371.84 and 628.16628.16 (recorded as 371.69−628.16371.69 - 628.16).

Problem 7: Wildlife Conservation Trout Weight Distribution Analysis

  • Biological Context and Parameters:
    • Objective: Tag adult trout that are unusually large or unusually small to assess their capability of traversing a newly dammed river to spawn.
    • Population distribution model: X∼N(3.65,1.32)X \sim N(3.65, 1.3^2)
    • Mean weight: xˉ=3.65 pounds\bar{x} = 3.65\,\text{pounds}
    • Standard deviation: σ=1.3 pounds\sigma = 1.3\,\text{pounds}
    • Tagging Criteria: Tag only fish falling into the bottom 5%5\% or top 5%5\% of weights.
  • Calculator Syntax:
    • Lower cutoff command: invNorm(0.05,3.65,1.3,Left)\text{invNorm}(0.05, 3.65, 1.3, \text{Left})
    • Upper cutoff command: invNorm(0.95,3.65,1.3,Left)\text{invNorm}(0.95, 3.65, 1.3, \text{Left}) or invNorm(0.05,3.65,1.3,Right)\text{invNorm}(0.05, 3.65, 1.3, \text{Right})
  • Step-by-Step Mathematical Calculations:
    • Critical zz-scores for extreme 5%5\% tails: z0.05=−1.64485z_{0.05} = -1.64485 and z0.95=+1.64485z_{0.95} = +1.64485
    • Lower weight cutoff calculation:
    • xlower=3.65+(−1.64485)×1.3x_{\text{lower}} = 3.65 + (-1.64485) \times 1.3
    • xlower=3.65−2.1383=1.5117≈1.51 poundsx_{\text{lower}} = 3.65 - 2.1383 = 1.5117 \approx 1.51\,\text{pounds}
    • Upper weight cutoff calculation:
    • xupper=3.65+(1.64485)×1.3x_{\text{upper}} = 3.65 + (1.64485) \times 1.3
    • xupper=3.65+2.1383=5.7883≈5.79 poundsx_{\text{upper}} = 3.65 + 2.1383 = 5.7883 \approx 5.79\,\text{pounds}
  • Curve Sketch Characteristics:
    • Bell-shaped curve centered at 3.65 pounds3.65\,\text{pounds}.
    • Vertical boundary lines drawn at 1.51 pounds1.51\,\text{pounds} and 5.79 pounds5.79\,\text{pounds}.
    • Shaded regions located exclusively in both outer tails: the bottom 5%5\% (to the left of 1.51 pounds1.51\,\text{pounds}) and the top 5%5\% (to the right of 5.79 pounds5.79\,\text{pounds}).
  • Final Concluding Statement:
    • Trout weighing 1.51 pounds1.51\,\text{pounds} or less, or 5.79 pounds5.79\,\text{pounds} or more, fall into the designated cutoffs and will be tagged for the conservation study.