Ch6 Comprehensive Study Notes: Normal Distribution & Standard Normal Table

Midterm Exam Logistics and Course Overview

• Exam Schedule and Format:

  • The midterm exam takes place on Thursday of Week 7 during regular class time.
  • The exam is strictly paper-based. No Excel software is permitted or required on the exam.
  • The exam consists of 7 main questions, each containing multiple sub-parts (e.g., parts A through F).
  • The full-class review session is scheduled for Tuesday of Week 7, where the midterm review packet is covered in depth.

• Scope of Content:

  • Covers all course sections completed from the beginning of the term through Chapter 6.
  • Questions are structured similarly to homework and in-class practice problems, emphasizing showing step-by-step work and writing out mathematical formulas.
  • Students may be required to compute summary statistics such as the sample/population mean or standard deviation by hand.

• Allowed Examination Materials:

  • Reference Sheet: Students may prepare one standard-sized piece of paper ($8.5 \times 11$\text{ inches}), front and back. Any content is allowed, including formulas, worked practice problems, copied questions from the review packet, and personal notes.
  • Calculator: A scientific calculator is required for the exam. The instructor has two spare scientific calculators available on a first-come, first-served basis for students who forget theirs.
  • Provided Tables: The Standard Normal Distribution Table (z-table) will be printed and distributed by the instructor during the exam.

• Post-Exam Policies, Grading, and Corrections:

  • Exams will be graded over Fall Break following the Thursday test date.
  • Graded exams will be returned after Fall Break.
  • Students have the opportunity to complete take-home exam corrections to earn points back on their score. Students may consult with the instructor during office hours for guidance on corrections.

• Calculator Policy on Assessments:

  • Calculator shortcuts (such as binompdf or binomcdf) are permitted on quizzes and exams provided that the student explicitly writes down the function command and inputs used on their paper to show their work.

Introduction to the Normal Distribution

• Definition and Importance:

  • The normal distribution is a continuous probability distribution and is considered the most fundamental distribution in statistics.
  • It serves as a foundational model across diverse fields, including psychology, business, economics, natural sciences, nursing, and mathematics.

• Mathematical Notation:

  • A normally distributed random variable $X$ with mean $\mu$ and standard deviation $\sigma$ is expressed as:     $X \sim N(\mu, \sigma)$
  • Inside the parentheses, the first term $\mu$ denotes the population mean, and the second term $\sigma$ denotes the population standard deviation.

• Fundamental Properties:

  • Symmetry: The probability density curve is perfectly symmetric about a vertical line drawn through its center at the mean $\mu$. The left and right halves are mirror images.
  • Total Area: The total area underneath the probability density curve equals exactly $1$ ($\text{Total Area} = 1.00$).
  • Position of the Center: The mean $\mu$ always lies in the exact middle of the distribution curve.

Z-Scores and Data Standardization

• Definition and Interpretation:

  • A z-score (or standardized value) measures the exact number of standard deviations an observation $x$ lies away from its mean $\mu$.
  • Directionality:
  • Positive z-score: Indicates the value $x$ lies to the right of (above) the mean.
  • Negative z-score: Indicates the value $x$ lies to the left of (below) the mean.
  • Magnitude: Represents the distance in units of standard deviation. When interpreting a negative z-score's distance, the negative sign represents the direction ("to the left" or "below"), while the absolute value represents the distance.

• Mathematical Formula:   $z = \frac{x - \mu}{\sigma}$

• Utility:

  • Standardizing data into z-scores enables direct comparison between datasets or observations that are measured on different scales or with different units.

• Rounding Conventions:

  • Z-scores are rounded to $2$ decimal places to align directly with standard normal probability tables.
  • Assessment instructions explicitly state the required rounding precision for final calculated answers (e.g., $4$ decimal places on exam problems).

• Lecture Examples and Step-by-Step Calculations:

  • Conceptual Example 1:

  • Given $X \sim N(5, 2)$ and an observation $x = 11$:       $11 = 5 + 3(2)$

  • The value $11$ is $3$ standard deviations above the mean, giving $z = 3.00$.

  • Conceptual Example 2:

  • Given $X \sim N(5, 6)$, where $\mu = 5$ and $\sigma = 6$:

    • Case 1 ($x = 17$):         $z = \frac{17 - 5}{6} = \frac{12}{6} = 2.00$         Interpretation: $x = 17$ is $2$ standard deviations above the average ($2$ standard deviations to the right of the mean).
    • Case 2 ($x = 1$):         $z = \frac{1 - 5}{6} = \frac{-4}{6} \approx -0.67$         Interpretation: $x = 1$ is $0.67$ standard deviations below the average ($0.67$ standard deviations to the left of the mean).

  • Student Practice 1:

  • Given $X \sim N(12, 3)$, calculate the z-score for $x = 1$:       $z = \frac{1 - 12}{3} = \frac{-11}{3} \approx -3.67$

  • Weight Loss Application Example:

  • Context: Doctors suggest a person can lose an average of $5\,\text{lbs}$ per month through reduced fat intake and consistent exercise. Weight loss is normally distributed with standard deviation $\sigma = 2\,\text{lbs}$. Distribution: $X \sim N(5, 2)$.

  • Part A ($x = 10\,\text{lbs}$ weight loss):       $z = \frac{10 - 5}{2} = \frac{5}{2} = 2.50$       Interpretation: A weight loss of $10\,\text{lbs}$ corresponds to $z = 2.50$, meaning $x = 10$ is $2.5$ standard deviations to the right of (above) the mean of $5\,\text{lbs}$.

  • Part B (A person gains $3\,\text{lbs}$, represented as negative weight loss $x = -3$):       $z = \frac{-3 - 5}{2} = \frac{-8}{2} = -4.00$       Interpretation: $z = -4.00$. The value $x = -3$ is $4$ standard deviations to the left of (below) the mean of $5\,\text{lbs}$.

  • Comparing Distinct Variables Example:

  • Given random variables $X \sim N(13, 2)$ and $Y \sim N(2, 1)$.

  • Find $z$ when $y = 4$ using distribution $Y$ ($\mu = 2, \sigma = 1$):       $z = \frac{4 - 2}{1} = 2.00$

  • Student Practice 2 (Basketball Scoring):

  • Context: Jerome averages $16\,\text{points/game}$ with a standard deviation of $4\,\text{points}$. Distribution: $X \sim N(16, 4)$.

  • Find the z-score for $x = 10\,\text{points}$:       $z = \frac{10 - 16}{4} = \frac{-6}{4} = -1.50$

  • Interpretation: $z = -1.50$. The observation $x = 10$ is $1.5$ standard deviations to the left of (below) the mean game score of $16$.

The Empirical Rule (68-95-99.7 Rule)

• Definition and Scope:

  • Applies specifically to bell-shaped, normally distributed random variables.

  • Quantifies the approximate percentage of data falling within $1$, $2$, or $3$ standard deviations of the mean \mu$.\n\n- The Three Rules:\n - 68\%$Rule: Approximately$68\%$of all data points lie within$1 standard deviation of the mean:\n    [\mu - 1\sigma, \mu + 1\sigma] \quad \text{corresponding to} \quad z \in [-1, 1]\n - 95\%$Rule: Approximately$95\%$of all data points lie within$2 standard deviations of the mean:\n    [\mu - 2\sigma, \mu + 2\sigma] \quad \text{corresponding to} \quad z \in [-2, 2]\n - 99.7\%$Rule: Approximately$99.7\%$of all data points lie within$3 standard deviations of the mean:\n    [\mu - 3\sigma, \mu + 3\sigma] \quad \text{corresponding to} \quad z \in [-3, 3]\n\n- Chilean Male Heights Application Example:\n - Dataset 1 (1984–1985 males aged 15–18): Y \sim N(172.36, 6.34)\,\text{cm}.\n - Dataset 2 (2009–2010 males aged 15–18): X \sim N(170, 6.28)\,\text{cm}.\n\n - Part 1: Z-score calculations and comparison for x = 160.58\,\text{cm}$and$y = 162.85\,\text{cm}:\n    z_x = \frac{160.58 - 170}{6.28} = \frac{-9.42}{6.28} = -1.50\n    z_y = \frac{162.85 - 172.36}{6.34} = \frac{-9.51}{6.34} = -1.50\n    Interpretation: Both individual heights sit exactly 1.5 standard deviations to the left of (below) their respective generational population means.\n\n - Part 2: Applying Empirical Rule to Distribution Y \sim N(172.36, 6.34)\,\text{cm}:\n - Find interval containing 68\%$of height values ($z = \pm 1):\n      \text{Lower Bound} = 172.36 - 1(6.34) = 166.02\,\text{cm}\n      \text{Upper Bound} = 172.36 + 1(6.34) = 178.70\,\text{cm}\n      Range: 166.02\,\text{cm}$to$178.70\,\text{cm}.\n\n - Find interval containing 95\%$of height values ($z = \pm 2):\n      \text{Lower Bound} = 172.36 - 2(6.34) = 159.68\,\text{cm}\n      \text{Upper Bound} = 172.36 + 2(6.34) = 185.04\,\text{cm}\n      Range: 159.68\,\text{cm}$to$185.04\,\text{cm}.\n\n - Find interval containing 99.7\%$of height values ($z = \pm 3):\n      \text{Lower Bound} = 172.36 - 3(6.34) = 153.34\,\text{cm}\n      \text{Upper Bound} = 172.36 + 3(6.34) = 191.38\,\text{cm}\n      Range: 153.34\,\text{cm}$to$191.38\,\text{cm}.\n\n- Student Practice (2012 SAT Verbal Exam Scores):\n - Context: 1,600,000$students took the SAT in 2012. Verbal section scores follow$X \sim N(496, 140).\n - Calculate z-score for x_1 = 325:\n    z_1 = \frac{325 - 496}{140} = \frac{-171}{140} = -1.50\n - Calculate z-score for x_2 = 366.21:\n    z_2 = \frac{366.21 - 496}{140} = \frac{-129.79}{140} \approx -1.14\n - Interpretation: Score x_1 = 325$lies$1.50$standard deviations below the mean ($z = -1.50$). Score$x_2 = 366.21$lies$1.14$standard deviations below the mean ($z = -1.14$). The score of$366.21 represents a better score because it is closer to the average performance.\n\n\n# Calculating Normal Probabilities Using the Standard Normal Table\n\n- Standard Normal Probability Table Structure:\n - The table provides cumulative probabilities representing the area under the standard normal curve strictly to the left of a given z-score (P(Z < z)).\n - Layout: Row headers display the integer and tenths digit of the z-score. Column headers display the hundredths digit. The intersection gives the area/probability.\n - Negative Z-Scores: Located on the front page (-3.99 \le z \le 0.00).\n - Positive Z-Scores: Located on the reverse page (0.00 \le z \le 3.99).\n\n- Off-the-Table Extreme Value Rules:\n - If a positive z-score exceeds the table limits (z > 3.99$), the cumulative left-tail area approaches$1.0000$($100\%).\n - If a negative z-score is below table limits (z < -3.99$), the cumulative left-tail area approaches$0.0000$($0\%).\n\n- Probability Conversion Rules:\n - Cumulative Left-Tail ("Less Than"): P(X < x)\n - Calculate z = \frac{x - \mu}{\sigma}.\n - Read cumulative probability directly from table body.\n - Right-Tail ("Greater Than" / "More Than"): P(X > x)\n - Standard normal tables only report left-tail areas. Rewrite using the complement rule:\n      P(X > x) = 1 - P(X < x)\n - Interval ("Between Two Values"): P(a < X < b)\n - Rewrite as the difference between two cumulative left-tail probabilities:\n      P(a < X < b) = P(X < b) - P(X < a)\n\n- Comprehensive Problems and Solutions:\n\n - Statistics Final Exam Problem: Distribution X \sim N(63, 5).\n - Part A: Find the probability that a student scores more than 65$($P(X > 65)):\n      P(X > 65) = 1 - P(X < 65)\n      z = \frac{65 - 63}{5} = \frac{2}{5} = 0.40\n      Table lookup for z = 0.400$gives cumulative probability$0.65542$.       $P(X > 65) = 1 - 0.65542 = 0.34458$

  • Part B: Find the probability that a student scores less than $85$ ($P(X < 85)$):       $z = \frac{85 - 63}{5} = \frac{22}{5} = 4.40$       Since $z = 4.40$ exceeds $3.99$, it is off the positive side of the table:       $P(X < 85) = 1.0000 \quad (100\%$\text{ probability})\n\n - Part C: Find the 90th Percentile (P_{90}) Exam Score:\n - Target: Find score x$such that$90\%$of scores lie below it ($P(X < x) = 0.9000).\n - Working Backwards: Search body of normal table for entry closest to 0.9000$. The value$0.89973$is closest, corresponding to$z = 1.28$.

    • Set up z-score equation and solve for $x$:         $1.28 = \frac{x - 63}{5}$$x - 63 = 1.28 \times 5 = 6.4$$x = 63 + 6.4 = 69.4$
    • An exam score of $69.4$ marks the 90th percentile.

  • Practice Problem (Golf Team Scores):

  • Distribution: $X \sim N(68, 3)$. Find $P(X < 65)$.

  • Step 1: Calculate z-score:       $z = \frac{65 - 68}{3} = \frac{-3}{3} = -1.00$

  • Step 2: Table lookup for $z = -1.00$ yields cumulative area 0.15866$.\n - Step 3: Since this is a "less than" statement, no complement subtraction is needed:\n      P(X < 65) = 0.15866\n\n - Household PC Entertainment Usage Example:\n - Context: Household PC entertainment usage time follows X \sim N(2, 0.5)$hours per day ($\mu = 2\,\text{hours}, \sigma = 0.5\,\text{hours}).\n\n - Part A: Find probability usage is between 1.8$and$2.75$hours per day ($P(1.8 < X < 2.75)):\n      P(1.8 < X < 2.75) = P(X < 2.75) - P(X < 1.8)\n - Upper value x_1 = 2.75:\n        z_1 = \frac{2.75 - 2}{0.5} = \frac{0.75}{0.5} = 1.50\n        Table lookup for z_1 = 1.50$gives area$0.93319$.

    • Lower value $x_2 = 1.8$:         $z_2 = \frac{1.8 - 2}{0.5} = \frac{-0.2}{0.5} = -0.40$         Table lookup for $z_2 = -0.40$ gives area 0.34458$.\n - Compute probability:\n        P(1.8 < X < 2.75) = 0.93319 - 0.34458 = 0.58861\n\n - Part B: Find maximum daily entertainment hours for the bottom quartile (1st Quartile / 25th Percentile, Q_1):\n - The 1st quartile represents the 25th percentile (P(X < x) = 0.2500).\n - Working Backwards: Search normal table body on negative side for probability closest to 0.2500$. The value$0.25143$is closest, corresponding to$z = -0.67$.
    • Solve for $x$:         $-0.67 = \frac{x - 2}{0.5}$$x - 2 = -0.67 \times 0.5 = -0.335$$x = 2 - 0.335 = 1.665\,\text{hours per day}$

Questions and Discussion

• Question: Is the midterm review packet representative of the actual exam length?

  • Answer: The review packet is significantly longer than the actual exam. The midterm consists of 7 main questions with multiple sub-parts (e.g., parts A through F), making it much shorter than the review packet.

• Question: What specific materials can be included on the exam reference sheet?

  • Answer: Anything can be put on the single reference sheet (front and back of an $8.5 \times 11$\text{ inch} paper), including formulas, worked practice problems, copied questions directly from the review packet, or written notes.

• Question: Are extra notes permitted besides the reference sheet?

  • Answer: No additional notes are allowed beyond the single double-sided reference sheet.

• Question: How many decimal places should answers be rounded to on the midterm exam?

  • Answer: Specific rounding instructions will be provided on the exam for each problem (typically 4 decimal places for final calculated probabilities). Standard z-scores are always rounded to 2 decimal places to match the standard normal distribution table.

• Question: Will students be provided with the z-table during the exam?

  • Answer: Yes, printed standard normal tables will be handed out by the instructor during the exam.

• Question: When will exam grades be finalized and returned?

  • Answer: Grades will be processed over Fall Break. Returned exams will be handed back during class after returning from break.

• Question: What is the policy for earning back points on the exam?

  • Answer: Students will have the opportunity to complete take-home exam corrections after tests are handed back to earn points back on their grade. Instructor assistance is available during office hours to review corrections.

• Question: Are calculator shortcuts (such as binompdf or binomcdf) acceptable on quizzes and exams?

  • Answer: Yes, calculator function shortcuts are permitted as long as the student writes down the function syntax used and presents supporting work on paper.