10/1/26: Continuous Random Variables and the Normal Distribution

Review of Discrete Random Variables and the Binomial Distribution

• Discrete random variables represent outcomes that take on distinct, countable values $k \in \{0, 1, 2, 3, \dots\}$.

• A primary discrete distribution is the binomial distribution, denoted $X \sim B(n, p)$, where $n$ represents the number of trials and $p$ represents the probability of success on a single trial.

• Key parameters and formulas for the binomial distribution include:

  • Mean (expected value): $\mu = n \times p$

  • Standard deviation: $\sigma = \sqrt{n \times p \times (1 - p)}$

• As the sample size $n$ in a binomial distribution becomes sufficiently large ($n \rightarrow \infty$), the discrete distribution smooths out and closely approximates a continuous normal distribution curve.

Characteristics of the Normal Distribution

• Continuous random variables can take on any real value within a continuous range, unlike discrete random variables.

• The normal distribution is a continuous probability distribution defined by two key parameters: mean ($\mu$) and standard deviation ($\sigma$), denoted as $X \sim N(\mu, \sigma)$.

• Graphical Properties of the Normal Curve:

  • Shape: The density curve is symmetric and bell-shaped, commonly referred to as the bell curve.

  • Center: The exact center and peak of the curve occurs at the value $x = \mu$.

  • Measures of Central Tendency: Because the distribution is perfectly symmetric across the vertical line $x = \mu$, the mean, median, and mode are all equal and located at $x = \mu$.

• Effects of Parameters on the Graph:

  • Horizontal Position: Changing the value of $\mu$ shifts the entire distribution left or right along the horizontal axis without altering its shape.

  • Vertical Stretch and Spread: Changing the value of $\sigma$ controls how spread out or clustered the data is:

  • A smaller standard deviation ($\sigma$) produces a taller, narrower curve tightly clustered around $\mu$.

  • A larger standard deviation ($\sigma$) produces a flatter, wider curve that tapers off more slowly towards the tails.

• Total Area and Probability Representation:

  • The total area under the density curve over its entire domain equals exactly $1$ ($100\%$).

  • Probability for a continuous distribution corresponds to the area under the curve over a specified interval, where $P(X < x)$ represents the area under the curve to the left of x$.\n\n# The Standard Normal Distribution and Z-Scores\n\n* The standard normal distribution is a specific normal distribution with a mean of zero and a standard deviation of one, denoted N(0, 1).\n* The Z$-score transforms any normal random variable$X \sim N(\mu, \sigma)$into a standard normal variable$Z \sim N(0, 1) via the standardizing formula:\n  Z = \frac{x - \mu}{\sigma}\n* Relationship in Standard Normal:\n * For the standard normal distribution N(0, 1)$, substituting$\mu = 0$and$\sigma = 1 yields:\n    Z = \frac{x - 0}{1} = x\n * Thus, in N(0, 1)$, the$Z$-score is identical to the raw$x value.\n* Interpretation of Z-Score:\n * A Z$-score represents the exact number of standard deviation step sizes ($\sigma$) that a raw value$x$lies away from its mean$\mu.\n * A positive Z$-score ($Z > 0$) indicates that the raw value$x$is greater than the mean ($x > \mu).\n * A negative Z$-score ($Z < 0$) indicates that the raw value$x$is less than the mean ($x < \mu).\n * A Z$-score of zero ($Z = 0$) indicates that the raw value$x$is equal to the mean ($x = \mu).\n* Converting Z-Scores back to Raw Values:\n * To find the raw data value x$given a$Z-score, rearrange the formula:\n    x = \mu + Z \times \sigma\n\n# Z-Score Calculations and Conversions\n\n* Calculating Z-Scores from Raw Values:\n * Example 1: Given x = 7$from a distribution with$\mu = 5.8$and$\sigma = 2:\n    Z = \frac{7 - 5.8}{2} = \frac{1.2}{2} = 0.6\n    This means x = 7$lies$0.6$standard deviations above the mean$\mu = 5.8$.

  • Example 2: Given $y = 15$ from a distribution with $\mu = 18$ and $\sigma = 1.4$:     $Z = \frac{15 - 18}{1.4} = \frac{-3}{1.4} \approx -2.14$     This means $y = 15$ lies approximately $2.14$ standard deviations below the mean \mu = 18$.\n * Example 3: Given p = 4.1$from a distribution with$\mu = 3.25$and$\sigma = 2.4:\n    Z = \frac{4.1 - 3.25}{2.4} = \frac{0.85}{2.4} \approx 0.354\n    Since p > \mu$, the resulting$Z-score is positive.\n* Calculating Raw Values from Z-Scores:\n * Example: Find x$given$Z = -2.3$for a distribution with$\mu = 80$and$\sigma = 5.7:\n    x = 80 + (-2.3) \times 5.7 = 80 - 13.11 = 66.89\n    Because Z < 0$, the calculated$x$value ($66.89$) is smaller than the mean ($80).\n\n# The Empirical Rule (68-95-99.7 Rule)\n\n* For any bell-shaped, symmetric distribution (such as a normal distribution), the empirical rule describes the approximate proportion of data falling within specified standard deviation intervals around the mean:\n * Approximately 68\%$of the data lies within$1 standard deviation of the mean:\n    P(\mu - 1\sigma \le X \le \mu + 1\sigma) = P(-1 \le Z \le 1) \approx 0.68\n * Approximately 95\%$of the data lies within$2 standard deviations of the mean:\n    P(\mu - 2\sigma \le X \le \mu + 2\sigma) = P(-2 \le Z \le 2) \approx 0.95\n * Approximately 99.7\%$of the data lies within$3 standard deviations of the mean:\n    P(\mu - 3\sigma \le X \le \mu + 3\sigma) = P(-3 \le Z \le 3) \approx 0.997\n\n# Standard Normal Probability Tables and Reading Z-Tables\n\n* Standard Normal Cumulative Probability Tables (Z-Tables) display the left-tail probability P(Z < z)$for standard normal$Z-scores up to two decimal places:\n * The left column lists the integer and tenths digits of Z.\n * The top row lists the hundredths digit of Z.\n * The intersection cell contains the cumulative area under the curve to the left of Z$.

• Key Reference Values in Z-Tables:

  • For $Z = 0.00$, the table area is $0.50000$ ($50\%$), reflecting symmetry about the mean \mu = 0$.\n * For Z = -3.24$, combining row$-3.2$and column$0.04$yields a cumulative area of$0.00060$.

  • For $Z = 0.67$, combining row $0.6$ and column $0.07$ yields a cumulative area of 0.74857$.\n * For Z = 2.19$, the cumulative left-tail area is high, reaching$0.98574$.

Rules for Calculating Probabilities in Continuous Distributions

• Left-Tail Probability:

  • The probability that $Z$ is less than a value $a$ is read directly from the Z-table:     $P(Z < a)$

  • Example: $P(Z < 0.67) = 0.74857$

• Right-Tail Probability:

  • The probability that $Z$ is greater than a value $a$ uses the complement rule, since total area equals $1$:     $P(Z > a) = 1 - P(Z < a)$

  • Example: For $Z = 1.77$, with $P(Z < 1.77) = 0.96164$:     $P(Z > 1.77) = 1 - 0.96164 = 0.03836$

• Between-Two-Values (Interval) Probability:

  • The probability that $Z$ lies between two values $a$ and $b$ (where $a < b$) is found by subtracting the smaller left-tail area from the larger left-tail area:     $P(a < Z < b) = P(Z < b) - P(Z < a)$

  • Example: For $P(-1.61 < Z < 1.00)$:     $P(Z < 1.00) = 0.84134$     $P(Z < -1.61) = 0.05370$     $P(-1.61 < Z < 1.00) = 0.84134 - 0.05370 = 0.78764$

• Exact Point Probability:

  • For any continuous distribution, the probability of obtaining an exact single value is identically zero:     $P(Z = a) = 0$

  • Reason: A single point has zero width, resulting in zero area under the continuous density curve:     $P(Z = a) = P(Z \le a) - P(Z < a) = 0$

Inverse Normal Calculations (Finding Z-Scores from Probabilities)

• Inverse normal calculations involve finding the $Z$-score associated with a given cumulative left-tail probability $P(Z < z)$:

  • Step 1: Compare the target probability to $0.5000$:

  • If target area $< 0.5000$, $z$ is negative.

  • If target area $> 0.5000$, $z$ is positive.

  • Step 2: Locate the value closest to the target probability inside the body of the Z-table.

  • Step 3: Read off the corresponding row and column header to determine z$.\n* Examples:\n * Given cumulative probability 0.41294:\n * Search standard table for 0.41294 on the negative side.\n * Intersects at row -0.2$and column$0.02$, giving$Z = -0.22$.

  • Given cumulative probability $0.74857$:

  • Search standard table for $0.74857$ on the positive side.

  • Intersects at row $0.6$ and column $0.07$, giving Z = 0.67$.\n * Given cumulative probability 0.99963:\n * Surrounding table entries are P(Z < 2.99) = 0.99861$and$P(Z < 3.00) = 0.99865$.

  • Estimating between table limits yields Z \approx 3.375$.\n\n# Real-World Applications and Problem-Solving Procedures\n\n* Standard Three-Step Procedure for Normal Distribution Problems:\n * Step 1: Write out the required probability expression in terms of the original variable X$.

  • Step 2: Convert the boundary $X$ values to standard $Z$-scores using $Z = \frac{x - \mu}{\sigma}$.

  • Step 3: Look up the cumulative probabilities in the Z-table (or evaluate using calculator functions such as normalCDF) and complete the required arithmetic.

• Application 1: Giraffe Heights

  • Scenario: Heights of giraffes are normally distributed with mean $\mu = 18\,\text{ft}$ and standard deviation $\sigma = 0.5\,\text{ft}$, denoted $X \sim N(18, 0.5)$.

  • Goal: Calculate the probability that a randomly selected giraffe is between $19\,\text{ft}$ and $20\,\text{ft}$ tall ($P(19 < X < 20)$).

  • Step 1 (Convert boundaries to Z-scores):     $Z_1 = \frac{19 - 18}{0.5} = \frac{1}{0.5} = 2.00$     $Z_2 = \frac{20 - 18}{0.5} = \frac{2}{0.5} = 4.00$

  • Step 2 (Rewrite probability in Z-scores):     $P(19 < X < 20) = P(2.00 < Z < 4.00) = P(Z < 4.00) - P(Z < 2.00)$

  • Step 3 (Evaluate probabilities):     $P(Z < 4.00) \approx 0.99997$     $P(Z < 2.00) = 0.97725$     $P(2.00 < Z < 4.00) = 0.99997 - 0.97725 = 0.02272$

  • Conclusion: There is approximately a $2.27\%$ chance ($0.02272$) of selecting a giraffe between $19\,\text{ft}$ and $20\,\text{ft}$ tall.

• Application 2: Exam Completion Times

  • Scenario: Exam completion times are normally distributed with mean $\mu = 100\,\text{min}$ and standard deviation $\sigma = 15\,\text{min}$, with a fixed exam limit of $120\,\text{min}$.

  • Goal: Calculate the probability that a student fails to finish within the $120\,\text{min}$ time limit ($P(T > 120)$).

  • Step 1 (Convert boundary to Z-score):     $Z = \frac{120 - 100}{15} = \frac{20}{15} \approx 1.33$

  • Step 2 (Apply right-tail rule):     $P(T > 120) = P(Z > 1.33) = 1 - P(Z < 1.33)$

  • Step 3 (Evaluate probability):     $P(Z < 1.33) = 0.90824$     $P(Z > 1.33) = 1 - 0.90824 = 0.09176$

  • Conclusion: Approximately $9.18\%$ (roughly $10\%$) of students will not finish the exam within the allotted $120\,\text{min}$.

• Application 3: Percentiles and Minimum Thresholds (Joey's Rat Weight)

  • Scenario: Male rat weights are normally distributed with mean $\mu = 440\,\text{g}$ and standard deviation $\sigma = 25\,\text{g}$. A rat is stated to be in the top $1\%$ ($99\text{th}$ percentile).

  • Goal: Determine the minimum weight $x$ required to place a rat in the $99\text{th}$ percentile ($P(X < x) = 0.9900$).

  • Step 1 (Find required Z-score for $99\text{th}$ percentile):

  • Search Z-table interior for cumulative probability 0.9900$.\n * Closest matching entry yields Z = 2.33$.

  • Step 2 (Convert Z-score back to weight $x$):     $x = \mu + Z \times \sigma = 440 + 2.33 \times 25 = 440 + 58.25 = 498.25\,\text{g}$

  • Conclusion: A rat must weigh at least $498.25\,\text{g}$ to be in the top $1\%$ of male rat weights.

• Application 4: Warranty Lifetime Calculations

  • Scenario: An appliance lifespan is normally distributed with mean $\mu = 2.8\,\text{years}$ and standard deviation $\sigma = 1.3\,\text{years}$. The company wants $85\%$ of appliances to outlast the warranty period (meaning only $15\%$ require warranty coverage).

  • Goal: Find warranty threshold duration $w$ such that $P(X > w) = 0.85$, which corresponds to cumulative left-tail probability P(X < w) = 0.1500$.\n * Step 1 (Find Z-score for 15\text{th} percentile):\n * Search Z-table interior for cumulative area 0.1500$.

  • Closest matching entry yields $Z \approx -1.036$ (or $-1.04$).

  • Step 2 (Convert Z-score back to appliance age $w$):     $w = \mu + Z \times \sigma = 2.8 + (-1.036) \times 1.3 = 2.8 - 1.3468 \approx 1.4532\,\text{years}$

  • Conclusion: Setting the warranty duration to approximately $1.45\,\text{years}$ ensures that $85\%$ of appliances outlast the warranty, limiting claims to $15\%$ of customers.