Statistics Practice Problems: Normal Distribution and Empirical Rule Study Guide

The Empirical Rule (also known as the $68-95-99.7$ rule) provides a statistical guideline for data that follows a normal distribution.
Standard Distribution Segments: * The area within one standard deviation ( $\mu \pm 1\sigma$ ) of the mean encompasses approximately $68\%$ of the data. This is divided into two segments of $34\%$ on either side of the mean ( $\mu$ ). * The area within two standard deviations ( $\mu \pm 2\sigma$ ) of the mean encompasses approximately $95\%$ of the data. The additional area between $1\sigma$ and $2\sigma$ (on both the positive and negative ends) accounts for $13.5\%$ each ( $95\% - 68\% = 27\%$ , and $\frac{27\%}{2} = 13.5\%$ ). * The area within three standard deviations ( $\mu \pm 3\sigma$ ) of the mean encompasses approximately $99.7\%$ of the data. The additional area between $2\sigma$ and $3\sigma$ accounts for approximately $2.35\%$ on each side ( $99.7\% - 95\% = 4.7\%$ , and $\frac{4.7\%}{2} = 2.35\%$ ). * The area beyond three standard deviations (the tails) accounts for the remaining $0.3\%$ , which is divided into $0.15\%$ in each tail ( $\mu - 3\sigma$ and $\mu + 3\sigma$ ).

Given Parameters: * Distribution Type: Normal Distribution. * Mean ( $\mu$ ): $40\,\text{minutes}$ . * Standard Deviation ( $\sigma$ ): $5\,\text{minutes}$ .
Objective: Calculate the percentage of employees with a commute time between $35$ and $50\,\text{minutes}$ .
Step-by-Step Breakdown: * The lower bound of $35\,\text{minutes}$ is equal to $\mu - 1\sigma$ ( $40 - 5 = 35$ ). * The upper bound of $50\,\text{minutes}$ is equal to $\mu + 2\sigma$ ( $40 + (2 \times 5) = 50$ ). * Summing the distribution percentages from the Empirical Rule: * From $\mu - 1\sigma$ to $\mu$ : $34\%$ * From $\mu$ to $\mu + 1\sigma$ : $34\%$ * From $\mu + 1\sigma$ to $\mu + 2\sigma$ : $13.5\%$ * Final Calculation: $34\% + 34\% + 13.5\% = 81.5\%$ . * Conclusion: Approximately $81.5\%$ of employees have a commute time between $35$ and $50\,\text{minutes}$ .

Given Parameters: * Total Sample Size ( $n$ ): $600\,\text{teenagers}$ . * Mean ( $\mu$ ): $14\,\text{hours/week}$ . * Standard Deviation ( $\sigma$ ): $2.5\,\text{hours/week}$ .
Objective: Determine how many teenagers in the group of $600$ spend between $9$ and $19\,\text{hours}$ watching TV per week.
Distribution Analysis: * Lower bound calculation: $9\,\text{hours} = 14 - (2 \times 2.5) = \mu - 2\sigma$ . * Upper bound calculation: $19\,\text{hours} = 14 + (2 \times 2.5) = \mu + 2\sigma$ . * According to the Empirical Rule, the area within $2$ standard deviations of the mean encompasses $95\%$ of the population.
Numerical Estimation: * Total expected number = $95\% \times 600$ . * Calculation: $0.95 \times 600 = 570$ .
Conclusion: We would expect $570\,\text{teenagers}$ to spend between $9$ and $19\,\text{hours}$ watching TV each week.

Given Parameters: * Distribution Type: Normal Distribution. * Mean Score ( $\mu$ ): $95\,\text{points}$ . * Standard Deviation ( $\sigma$ ): $2\,\text{points}$ . * Observed Performance ( $x$ ): $89\,\text{points}$ .
Performance Assessment (Standard Deviations): * Calculate the Z-score (the number of standard deviations the value is from the mean). * Formula: $Z = \frac{x - \mu}{\sigma}$ * Calculation: $Z = \frac{89 - 95}{2} = \frac{-6}{2} = -3$ .
Interpretation: * A performance of $89$ points is exactly $3\,\text{standard deviations}$ below the mean score ( $\mu - 3\sigma$ ). * Conclusion: The team performed significantly below average. In a normal distribution, only $0.15\%$ of the population scores lower than $3$ standard deviations below the mean, making this an extremely poor performance compared to the average match.

Scenario Part A: Group Expectations: * Sample Size ( $n$ ): $400\,\text{participants}$ . * Mean ( $\mu$ ): $50\,\text{push-ups}$ . * Standard Deviation ( $\sigma$ ): $8\,\text{push-ups}$ . * Target Range: $42$ to $58\,\text{push-ups}$ . * Range Analysis: $42 = 50 - 8 = \mu - 1\sigma$ and $58 = 50 + 8 = \mu + 1\sigma$ . * Empirical Rule Percentage: $68\%$ of the population falls within one standard deviation. * Calculation: $0.68 \times 400 = 272$ . * Result: Approximately $272\,\text{participants}$ are expected to complete between $42$ and $58\,\text{push-ups}$ .
Scenario Part B: Medal Eligibility (Top 2.5%): * Objective: Determine if someone doing $67\,\text{push-ups}$ deserves a medal. * Criteria: Exceed expectations and perform in the top $2.5\%$ . * Statistical Threshold: In a normal distribution, the area beyond $2\,\text{standard deviations}$ on the high side is $2.35\% + 0.15\% = 2.5\%$ . * Upper limit for top $2.5\%$ : $\mu + 2\sigma = 50 + (2 \times 8) = 50 + 16 = 66\,\text{push-ups}$ . * Evaluation: The participant performed $67\,\text{push-ups}$ , which is greater than the threshold of $66$ . * Conclusion: Yes, the participant deserves a medal because their performance of $67$ push-ups places them within the top $2.5\%$ of the group.

Given Parameters: * Mean Daily Intake ( $\mu$ ): $2,200\,\text{calories}$ . * Standard Deviation ( $\sigma$ ): $300\,\text{calories}$ .
Task 1: Percentage between 1,900 and 3,100 calories: * Lower limit: $1,900 = 2,200 - 300 = \mu - 1\sigma$ . * Upper limit: $3,100 = 2,200 + (3 \times 300) = \mu + 3\sigma$ . * Empirical Rule Summation: $\text{Percent} = (\text{area from } \mu-1\sigma \text{ to } \mu) + (\text{area from } \mu \text{ to } \mu+3\sigma)$ . * Calculation: $34\% + (34\% + 13.5\% + 2.35\%) = 34\% + 49.85\% = 83.85\%$ . * Result: Roughly $83.85\%$ of adults consume between $1,900$ and $3,100\,\text{calories}$ per day.
Task 2: Expected headcount out of 500 adults: * Calculation: $0.8385 \times 500 = 419.25$ . * Result: We would expect approximately $419\,\text{adults}$ out of $500$ to be within this range.
Task 3: Assessing an intake of 1,300 calories: * Standard Deviation Calculation: $Z = \frac{1,300 - 2,200}{300} = \frac{-900}{300} = -3$ . * This intake is exactly $3\,\text{standard deviations}$ below the mean ( $\mu - 3\sigma$ ). * Percentage ranking: Only $0.15\%$ of the population consumes less than this amount. * Conclusion: Someone eating $1,300\,\text{calories}$ per day would be considered on the extreme "low side." It is not considered "okay" or average as it represents a significant statistical outlier compared to the typical intake of that region.