Backward Normal Calculations and Standard Normal Distribution Applications

Standard Normal Distribution: Backward Calculations Overview

• Overview of calculation directions in standard normal distribution problems:
  • Forward calculations (review): Starting with a known observation value $x$, computing its standardized $z$-score, and finding the proportion/probability of observations falling below, above, or between specified values using Table A.
  • Backward calculations (new concept): Starting with a target proportion, percentile, or percentage, and working in reverse to determine the corresponding $z$-score and observation value $x$.
• Practical real-world applications of backward calculations:
  • Determining the exact score threshold required to rank in the top $10\%$ of students on an examination.
  • Finding score requirements for the top $5\%$ or top $8\%$ of applicants for university admissions.
  • Identifying values associated with specific distribution percentiles or quartiles.
• Source of values in backward problem solving:
  • Final observation values $x$ are calculated algebraically rather than read directly from standard normal tables.
  • Table A is utilized at the beginning of the process to identify the standardized $z$-score matching the specified area/proportion under the curve.

Step-by-Step Procedure for Backward Normal Calculations

• Complete step-by-step process for working backward from a proportion to an observation value $x$:
  1. State the problem and clearly specify the given proportion or percentage.
  2. Sketch the standard normal curve:
  • Plot the mean at $z = 0$.
  • Shade the specified region corresponding to the target proportion.
  • Use visual inspection relative to $z = 0$ ($50\%$ cumulative mark) to determine whether the target $z$-score must be positive or negative.
  1. Re-express or rework the target proportion into cumulative area to the left (below $z$):
  • Standard normal tables (Table A) exclusively report cumulative area falling below a given $z$-score ($P(Z < z)$).
  • If given an area above a threshold (e.g., top $15\%$), calculate the corresponding area below as:        $\text{Area Below} = 1 - \text{Area Above}$
  1. Locate the corresponding $z$-score in Table A:
  • Search the interior body of Table A to find the probability value numerically closest to the target proportion.
  • Evaluate absolute differences between candidate table entries and the target proportion:        $\text{Absolute Difference} = |\text{Table Entry} - \text{Target Proportion}|$
  • Select the entry with the smallest absolute difference (regardless of whether it is slightly above or below the target value).
  • Trace back to the corresponding row (ones and tenths digits) and column (hundredths digit) to determine the exact $z$-score.
  1. Back-solve for the original observation value $x$:
  • Rearrange the standard $z$-score formula:        $z = \frac{x - \mu}{\sigma}$
  • Algebraically solve for $x$ to obtain the un-standardizing equation:        $x = \mu + z \times \sigma$
  • Substitute the known population mean $\mu$, population standard deviation $\sigma$, and the identified $z$-score into the equation to determine $x$.

Table A Lookup Strategy and Symmetry Properties

• Table lookup mechanics for standard normal probabilities:
  • On the negative side of Table A ($z < 0$): Scan down the left column and search across rows past the target value to evaluate entry differences.
  • On the positive side of Table A ($z > 0$): Scan across rows and evaluate entries near the target proportion.
• Rounding rules and precision standards:
  • Standardized $z$-scores MUST ALWAYS be rounded to exactly two decimal places ($2\,\text{decimal places}$) because Table A is constructed to two-decimal-place precision.
  • Proportions and probabilities are recorded to four decimal places ($4\,\text{decimal places}$) as provided in Table A.
  • Final observation values $x$ retain the precision resulting from exact calculation or as specified by context.
• Symmetry of the normal distribution:
  • The standard normal curve is perfectly symmetric about $z = 0$
  • The cumulative proportion falling below a negative $z$-score ($-z$) is exactly equal to the proportion falling above the corresponding positive $z$-score ($+z$):     $P(Z < -z) = P(Z > +z)$
  • The first quartile ($Q_1$, corresponding to $25\%$ or $0.2500$ of observations below) always corresponds to:     $z = -0.67$
  • The third quartile ($Q_3$, corresponding to $75\%$ or $0.7500$ of observations below) always corresponds to:     $z = +0.67$

Practice Problem 1: Standard Normal Variable z Lookups (Question 13.13)

• Problem Statement: Use Table A to find the value $z$ of a standard normal variable satisfying specified conditions, selecting the $z$ value from Table A that comes closest to satisfying each condition.

• Part A: Find the point $z$ such that $75\%$ ($0.7500$) of observations fall below it ($P(Z < z) = 0.7500$).

  • Visualization and reasoning:
  • Mean at $z = 0$ splits cumulative area into $50\%$ below and $50\%$ above.
  • Since $75\% > 50\%$, the required $z$-score must be positive ($z > 0$).
  • Table search process:
  • Search positive side of Table A for entries near $0.7500$
  • Scan row $0.6$ across columns:
    • Entry at row $0.6$, column $0.07$: $0.7486$
    • Entry at row $0.6$, column $0.08$: $0.7517$
  • Evaluate absolute numerical differences:
    • Difference for $0.7486$:         $|0.7500 - 0.7486| = 0.0014$
    • Difference for $0.7517$:         $|0.7500 - 0.7517| = 0.0017$
  • Selection: $0.7486$ is numerically closer to $0.7500$
  • Result: Tracing back to row $0.6$ and column $0.07$ gives:     $z = +0.67$
  • Universal principle: Any scenario requiring exactly $75\%$ cumulative area below a threshold in a normal distribution always corresponds to $z = +0.67$

• Part B: Find the point $z$ such that $15\%$ ($0.1500$) of observations fall above it ($P(Z > z) = 0.1500$).

  • Table constraint and re-working:
  • Table A only provides cumulative areas below values.
  • Convert "area above" to "area below":       $\text{Area Below} = 1.0000 - 0.1500 = 0.8500$
  • Table search process:
  • Search positive side of Table A for entries near $0.8500$
  • Scan row $1.0$ across columns:
    • Entry at row $1.0$, column $0.03$: $0.8485$
    • Entry at row $1.0$, column $0.04$: $0.8508$
  • Evaluate absolute numerical differences:
    • Difference for $0.8485$:         $|0.8500 - 0.8485| = 0.0015$
    • Difference for $0.8508$:         $|0.8500 - 0.8508| = 0.0008$
  • Selection: $0.8508$ is numerically closer to $0.8500$
  • Result: Tracing back to row $1.0$ and column $0.04$ gives:     $z = +1.04$

Practice Problem 2: Thorax Length in Fruit Flies

• Population Parameters:

  • Variable $x$: Thorax length in a population of fruit flies.
  • Distribution shape: Approximately normal distribution.
  • Population mean: $\mu = 0.8$
  • Population standard deviation: $\sigma = 0.078$

• Part A: What proportion of flies have thorax lengths less than $0.75$ ($x < 0.75$)?

  • Calculation of $z$-score:     $z = \frac{x - \mu}{\sigma} = \frac{0.75 - 0.8}{0.078} = \frac{-0.05}{0.078} = -0.641025... \rightarrow -0.64$
  • Table A lookup:
  • Locate row $-0.6$ and column $0.04$
  • Table value: $0.2611$
  • Result: Since the problem asks for area less than $0.75$:     $\text{Proportion} = 0.2611$

• Part B: What proportion of flies have thorax lengths greater than $0.92$ ($x > 0.92$)?

  • Calculation of $z$-score:     $z = \frac{x - \mu}{\sigma} = \frac{0.92 - 0.8}{0.078} = \frac{0.12}{0.078} = 1.53846... \rightarrow 1.54$
  • Table A lookup:
  • Locate row $1.5$ and column $0.04$
  • Table value (area below): $0.9382$
  • Adjustment for "greater than":     $\text{Area Above} = 1 - 0.9382 = 0.0618$
  • Result:     $\text{Proportion} = 0.0618$

• Part C: What proportion of flies have thorax lengths between $0.7$ and $1.0$ ($0.7 < x < 1.0$)?

  • Calculation of $z$-scores for both bounds:
  • For lower bound $x_1 = 0.7$:       $z_1 = \frac{0.7 - 0.8}{0.078} = \frac{-0.1}{0.078} = -1.28205... \rightarrow -1.28$
  • For upper bound $x_2 = 1.0$:       $z_2 = \frac{1.0 - 0.8}{0.078} = \frac{0.2}{0.078} = 2.56410... \rightarrow 2.56$
  • Table A lookups:
  • Upper $z$-score ($z_2 = 2.56$): Locate row $2.5$ and column $0.06$ $\rightarrow$ Table value = $0.9948$
  • Lower $z$-score ($z_1 = -1.28$): Locate row $-1.2$ and column $0.08$ $\rightarrow$ Table value = $0.1003$
  • Calculate area between:     $\text{Proportion} = 0.9948 - 0.1003 = 0.8945$

• Part D: What length is the first quartile ($Q_1$)?

  • Definition of First Quartile ($Q_1$):
  • The observation value below which $25\%$ ($0.2500$) of observations fall ($P(Z < z) = 0.2500$).
  • Visualization and sign determination:
  • Since $25\% < 50\%$, the required $z$-score must be negative ($z < 0$).
  • Table A lookup on negative side for entry closest to $0.2500$:
  • Candidate entries in row $-0.6$:
    • Row $-0.6$, column $0.07$: $0.2514$ (difference: $|0.2500 - 0.2514| = 0.0014$)
    • Row $-0.6$, column $0.08$: $0.2483$ (difference: $|0.2500 - 0.2483| = 0.0017$)
  • Selection: $0.2514$ is closer to $0.2500$
  • Identified $z$-score: $z = -0.67$
  • Application of back-solving formula:     $x = \mu + z \times \sigma$$x = 0.8 + (-0.67) \times 0.078 = 0.8 - 0.05226 = 0.747738 \rightarrow 0.7477$
  • Academic grading criteria note for exams:
  • Question value: $3\,\text{points}$ total.
  • $2\,\text{points}$ awarded for correctly identifying the correct $z$-score (including proper positive or negative sign).
  • $1\,\text{point}$ awarded for correctly setting up and substituting values into the equation $x = \mu + z \times \sigma$
  • Test strategy recommendation: If unable to find the correct $z$-score, state a substitute $z$-score value explicitly and apply the formula to secure partial credit.

• Part E: How large a length is needed to be in the lowest $15\%$?

  • Target condition: Lowest $15\%$ corresponds to an area below of $0.1500$ ($P(Z < z) = 0.1500$).
  • Table A lookup on negative side for entry closest to $0.1500$:
  • Candidate entries in row $-1.0$:
    • Row $-1.0$, column $0.03$: $0.1515$ (difference: $|0.1500 - 0.1515| = 0.0015$)
    • Row $-1.0$, column $0.04$: $0.1492$ (difference: $|0.1500 - 0.1492| = 0.0008$)
  • Selection: $0.1492$ is closer to $0.1500$
  • Identified $z$-score: $z = -1.04$
  • Application of back-solving formula:     $x = \mu + z \times \sigma$$x = 0.8 + (-1.04) \times 0.078 = 0.8 - 0.08112 = 0.71888 \rightarrow 0.7189$

Practice Problem 3: MCAT Examination Scores

• Population Parameters:

  • Distribution: Approximately normal distribution.
  • Population mean: $\mu = 500$
  • Population standard deviation: $\sigma = 10.6$

• Part A: What proportion of students taking the MCAT had a score over $510$ ($x > 510$)?

  • Calculation of $z$-score:     $z = \frac{x - \mu}{\sigma} = \frac{510 - 500}{10.6} = \frac{10}{10.6} = 0.943396... \rightarrow 0.94$
  • Table A lookup:
  • Row $0.9$, column $0.04$ $\rightarrow$ Table value = $0.8264$
  • Area greater than calculation:     $\text{Proportion} = 1 - 0.8264 = 0.1736$

• Part B: What proportion of students had scores between $505$ and $515$ ($505 < x < 515$)?

  • Calculation of $z$-scores for both bounds:
  • For $x_1 = 505$:       $z_1 = \frac{505 - 500}{10.6} = \frac{5}{10.6} = 0.471698... \rightarrow 0.47$
  • For $x_2 = 515$:       $z_2 = \frac{515 - 500}{10.6} = \frac{15}{10.6} = 1.415094... \rightarrow 1.42$
  • Table A lookups:
  • Upper $z$-score ($z_2 = 1.42$): Row $1.4$, column $0.02$ $\rightarrow$ Table value = $0.9222$
  • Lower $z$-score ($z_1 = 0.47$): Row $0.4$, column $0.07$ $\rightarrow$ Table value = $0.6808$
  • Calculate area between:     $\text{Proportion} = 0.9222 - 0.6808 = 0.2414$
  • Error-checking principle: Computed proportions must always fall within the closed interval $[0, 1]$. Obtaining a negative proportion indicates a calculation error (such as subtracting in reverse order).

• Part C: What proportion of students had a score below $475$ ($x < 475$)?

  • Calculation of $z$-score:     $z = \frac{x - \mu}{\sigma} = \frac{475 - 500}{10.6} = \frac{-25}{10.6} = -2.35849... \rightarrow -2.36$
  • Table A lookup:
  • Row $-2.3$, column $0.06$ $\rightarrow$ Table value = $0.0091$
  • Result:     $\text{Proportion} = 0.0091$

• Part D: What score is the third quartile ($Q_3$)?

  • Target condition: Third quartile ($Q_3$) represents the score below which $75\%$ ($0.7500$) of observations lie ($P(Z < z) = 0.7500$).
  • Table A lookup for entry closest to $0.7500$:
  • Entry at row $0.6$, column $0.07$ ($0.7486$) is closer to $0.7500$ than $0.7517$
  • Identified $z$-score: $z = +0.67$
  • Application of back-solving formula:     $x = \mu + z \times \sigma$$x = 500 + (0.67) \times 10.6 = 500 + 7.102 = 507.102$

• Part E: How high a score is needed to be in the highest $8\%$?

  • Target condition and re-working:
  • Highest $8\%$ corresponds to an area above of $0.0800$
  • Convert to area below:       $\text{Area Below} = 1.0000 - 0.0800 = 0.9200$
  • Table A lookup on positive side for entry closest to $0.9200$:
  • Scan row $1.4$ candidate entries:
    • Row $1.4$, column $0.00$: $0.9192$ (difference: $|0.9200 - 0.9192| = 0.0008$)
    • Row $1.4$, column $0.01$: $0.9207$ (difference: $|0.9200 - 0.9207| = 0.0007$)
  • Selection: $0.9207$ is closer to $0.9200$
  • Identified $z$-score: $z = +1.41$
  • Application of back-solving formula:     $x = \mu + z \times \sigma$$x = 500 + (1.41) \times 10.6 = 500 + 14.946 = 514.946$
  • Institutional admissions rounding note: In contexts requiring integer test scores, score cutoffs would round up to $515$ or higher to ensure applicants meet or exceed the top $8\%$ threshold.

Practice Problem 4: Body Mass Index (BMI) of Young Men

• Population Parameters:
  • Variable $x$: Body Mass Index (BMI) of young men.
  • Distribution shape: Approximately normal distribution.
  • Population mean: $\mu = 26.8$
  • Population standard deviation: $\sigma = 5.2$
• Structure of practice problem set:
  • Parts A, B, and C require calculating proportions given specific BMI values $x$ (forward direction).
  • Parts D and E require finding exact BMI values $x$ corresponding to specified percentiles or cutoffs (backward direction).
• Reference: Solutions to Practice Problem 3 are published in online course materials.

Administrative Details and Upcoming Class Topics

• Homework Assignment Schedule:
  • The current homework assignment is due on Tuesday.
  • Assignment structure matches practice problems: Parts A–C require proportions; Parts D–E require value calculations.
• Upcoming Class Schedule and Topics:
  • Next Tuesday's lecture covers Chapters 4 and 5.
  • Topics: Linear Regression and Correlation.
  • Mandatory Equipment: A handheld calculator is strictly required for class on Tuesday; performing linear regression manually without a calculator is excessively time-consuming.

Student Questions and Classroom Exchanges

• Percentile scale and conversion exchange:
  • Clarification provided regarding percentages versus decimal proportions (e.g., $100\%$ scale versus proportions from $0$ to $1$).
• Visual learning strategies:
  • Student confirmation on using colored pens/highlighters to visually differentiate areas under the standard normal curve (above, below, between).
  • Instructor approval: Any personalized visual notation or calculator expression layout that facilitates understanding is encouraged.
• Attendance and identification verification:
  • Student confirmation of individual identification number: $39797$.