6.2: Using the Normal Distribution

Foundations of Normal Distributions and Z-Score Standardization

• Standardization of Normal Random Variables:

  • If a random variable $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$, denoted as $X \sim N(\mu, \sigma)$, any individual data value $x$ from $X$ can be standardized and compared to the standard normal distribution $Z \sim N(0, 1)$ by calculating its corresponding z-score:     $z = \frac{x - \mu}{\sigma}$

• Probability as Area Under the Density Curve:

  • For continuous random variables such as the normal distribution, the probability of obtaining a value strictly less than a specific threshold $x$, denoted $P(X < x)$, is evaluated by calculating the cumulative area under the probability density function curve to the left of x$.\n\n- **Computational Methods for Continuous Probabilities**:\n - Evaluating cumulative areas under continuous curves requires calculus-based integration.\n - In practice, probabilities are obtained through statistical computer applications or standard cumulative distribution tables.\n - The standard normal table (accessible at `https://math.arizona.edu/~rsims/ma464/standardnormaltable.pdf`) provides the exact cumulative area under the standard normal curve to the left of a specified z-score z$, enabling direct computation of standard normal probabilities$P(Z < z).\n\n# Evaluating Standard Normal Probabilities\n\n- **Symmetric Baseline Probability**:\n - Problem: Determine P(Z < 0).\n - Theoretical Reasoning: Since the standard normal distribution curve is completely symmetric around its mean \mu = 0$, exactly half ($50\%$) of the total area under the curve lies to the left of$z = 0.\n - Solution:\n    P(Z < 0) = 0.5000\n\n- **Direct Cumulative Standard Normal Lookup Examples**:\n - Standard cumulative probability to the left of z = 0.67:\n    P(Z < 0.67)\n - Standard cumulative probability to the left of z = -1.22:\n    P(Z < -1.22)\n - Standard cumulative probability to the left of z = 2.19:\n    P(Z < 2.19)\n\n- **Modified Region Probabilities (Intervals, Equality, and Right-Tails)**:\n - Interval Probability P(-1.61 < Z < 1):\n - To find the probability that Z falls between two values, subtract the cumulative area to the left of the lower bound from the cumulative area to the left of the upper bound:\n      P(-1.61 < Z < 1) = P(Z < 1) - P(Z < -1.61)\n - Exact Point Probability P(Z = 2):\n - For any continuous probability distribution, the probability assigned to any single individual point is precisely zero because the area under a curve above a single point has zero width:\n      P(Z = 2) = 0\n - Upper-Tail / Right-Tail Probability P(Z \ge 1.77):\n - To determine the probability that Z is greater than or equal to a specified value, apply the complement rule using the cumulative left-tail area:\n      P(Z \ge 1.77) = 1 - P(Z < 1.77)\n\n- **Inverse Cumulative Lookups (Determining Z-Scores from Probabilities)**:\n - Identifying the z-score corresponding to a target cumulative left-tail probability P(Z < z) = \text{Target Area}:\n - Target Area 0.41294$: Locate$0.41294$inside the table body to identify the exact threshold z-score$z$where$P(Z < z) = 0.41294\n - Target Area 0.74857$: Locate$0.74857$inside the table body to identify the exact threshold z-score$z$where$P(Z < z) = 0.74857\n - Target Area 0.99963$: Locate$0.99963$inside the table body to identify the exact threshold z-score$z$where$P(Z < z) = 0.99963\n\n# Applied Normal Distribution Problem Solving\n\n- **Giraffe Height Range Analysis**:\n - Context and Parameters:\n - Heights of giraffes follow a normal distribution with mean \mu = 18\,\text{feet}$and standard deviation$\sigma = 0.5\,\text{feet}.\n - Objective:\n - Calculate the probability that a randomly selected giraffe has a height between 19\,\text{feet}$and$20\,\text{feet}$, represented as$P(19 < X < 20).\n - Step-by-Step Procedure:\n - Convert lower bound x_1 = 19\,\text{feet}$to its corresponding z-score$z_1:\n      z_1 = \frac{19 - 18}{0.5} = \frac{1}{0.5} = 2.00\n - Convert upper bound x_2 = 20\,\text{feet}$to its corresponding z-score$z_2:\n      z_2 = \frac{20 - 18}{0.5} = \frac{2}{0.5} = 4.00\n - Express the probability interval in standard normal terms:\n      P(19 < X < 20) = P(2.00 < Z < 4.00) = P(Z < 4.00) - P(Z < 2.00)\n\n- **Exam Completion Time and Failure-to-Finish Rate**:\n - Context and Parameters:\n - An exam has a strict two-hour time limit (120\,\text{minutes}).\n - Completion time X$is normally distributed with mean$\mu = 100\,\text{minutes}$and standard deviation$\sigma = 15\,\text{minutes}.\n - Objective:\n - Find the probability that a student fails to finish the exam within the allowed duration (takes longer than 120\,\text{minutes}$), represented as$P(X > 120).\n - Step-by-Step Procedure:\n - Convert time limit x = 120\,\text{minutes}$to its corresponding z-score$z:\n      z = \frac{120 - 100}{15} = \frac{20}{15} = \frac{4}{3} \approx 1.33\n - Calculate the right-tail probability using the complement rule:\n      P(X > 120) = P(Z > 1.33) = 1 - P(Z < 1.33)\n\n- **Percentile Threshold for Pet Rat Weight**:\n - Context and Parameters:\n - Claim: A pet rat named Joey is in the "top 1% of all rats" in terms of weight.\n - Distribution: Weight W$of a male rat follows$W \sim N(440, 25)$, where mean$\mu = 440\,\text{grams}$and standard deviation$\sigma = 25\,\text{grams}.\n - Objective:\n - Find the minimum weight threshold w$required for Joey's rat to be in the top$1\%\n - Step-by-Step Procedure:\n - Being in the top 1\%$corresponds to the$99\text{th}$percentile, meaning$99\% of male rats weigh less:\n      P(W < w) = 0.9900\n - Find the z-score z$associated with a cumulative area of$0.9900 from the standard normal table.\n - Convert the target z-score back to raw weight w using the inverse standardization formula:\n      w = \mu + z \times \sigma = 440 + z \times 25\n\n- **Small Appliance Lifespan and Warranty Duration Determination**:\n - Context and Parameters:\n - A small appliance manufacturer wants to set a warranty duration such that 85\% of appliances last longer than the warranty span (preventing financial loss from high claim rates).\n - Appliance lifespan X$is normally distributed with mean$\mu = 2.8\,\text{years}$and standard deviation$\sigma = 1.3\,\text{years}.\n - Objective:\n - Determine the acceptable lifespan time duration t$that$85\% of appliances will exceed.\n - Step-by-Step Procedure:\n - If 85\%$of appliances last longer than time$t$, then the lowest$15\%$of lifespans fall below time$t:\n      P(X < t) = 0.1500\n - Locate the z-score z$corresponding to a lower-tail cumulative probability of$0.1500 in the standard normal table.\n - Convert the target z-score back to time in years using the inverse formula:\n      t = \mu + z \times \sigma = 2.8 + z \times 1.3$$