Normal Distribution: Properties, Applications & Real-World Use

TAKEAWAYS:

• The normal distribution is a symmetric, bell-shaped curve defined by its mean (μ) and standard deviation (σ), where 68% of data falls within ±1σ, 95% within ±2σ, and 99.7% within ±3σ.
• Z-scores standardize data to compare values across different distributions using the formula $z=\frac{X-\mu }{\sigma }$.
• The empirical rule (68-95-99.7 rule) helps estimate probabilities for continuous data without complex calculations.
• Applications include financial risk assessment (e.g., loan defaults), quality control (e.g., manufacturing defects), and natural phenomena (e.g., human heights).
• Limitations: Assumes data is continuous, symmetric, and follows a bell curve—real-world data often deviates (e.g., skewed distributions).
• Business use: Companies like Ncell (customer churn analysis), Daraz (delivery time predictions), and Nepal Rastra Bank (inflation modeling) rely on normal distribution for decision-making.

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1. Definition and Properties of Normal Distribution

The normal distribution (or Gaussian distribution) is a continuous probability distribution characterized by:

• A bell-shaped curve symmetric about the mean (μ).
• Asymptotic tails: The curve approaches but never touches the x-axis.
• Two parameters:
  • Mean (μ): Center of the distribution.
  • Standard deviation (σ): Measures spread (width of the curve).

Key Properties:

• Total area under the curve = 1 (represents total probability).
• Mean = Median = Mode (perfect symmetry).
• 68% of data lies within μ ± σ, 95% within μ ± 2σ, and 99.7% within μ ± 3σ (empirical rule).

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2. The Empirical Rule (68-95-99.7 Rule)

The empirical rule provides a quick way to estimate probabilities for normally distributed data.

Example 1: Daraz Delivery Times Suppose Daraz’s delivery times in Kathmandu follow a normal distribution with:

• Mean (μ) = 48 hours
• Standard deviation (σ) = 6 hours

Question: What percentage of orders are delivered within 42 to 54 hours? Solution:

• 42 hours = μ – σ (48 – 6)
• 54 hours = μ + σ (48 + 6)
• By the empirical rule, 68% of orders fall within this range.

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3. Standard Normal Distribution (Z-Scores)

The standard normal distribution (Z-distribution) is a normal distribution with:

• μ = 0
• σ = 1

Z-score formula: $$z=\frac{X-\mu }{\sigma }$$ Purpose: Converts any normal distribution to the standard normal for probability calculations using Z-tables.

Example 2: Ncell Customer Churn Ncell tracks customer retention times (in months) with:

• μ = 24 months
• σ = 5 months

Question: What is the probability a customer stays longer than 30 months? Solution:

1. Calculate Z-score: $$z=\frac{30-24}{5}=1.2$$
2. Look up P(Z > 1.2) in the Z-table:
   • P(Z ≤ 1.2) = 0.8849
   • P(Z > 1.2) = 1 – 0.8849 = 0.1151 (11.51%)

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4. Applications in Business and Finance

A. Financial Risk Assessment (Loan Defaults)

Banks (e.g., Nepal Bank Limited) use normal distribution to model loan defaults.

• Assumption: Default rates follow a normal distribution.
• Example: If the average default rate is 5% (μ) with σ = 1.5%, what is the probability of a default rate > 8%? $$z=\frac{8-5}{1.5}=2$$ P(Z > 2) = 1 - 0.9772 = 0.0228 \text{ (2.28%)}

B. Quality Control (Manufacturing Defects)

Factories (e.g., Himalayan Glass Industries) use normal distribution to monitor product defects.

• Example: If a factory produces light bulbs with a mean lifespan of 1000 hours (μ) and σ = 50 hours, what % of bulbs last > 1100 hours? $$z=\frac{1100-1000}{50}=2$$ $$P(Z>2)=2.28\%$$

C. Stock Market Returns (NEPSE Index)

Investors use normal distribution to predict stock returns.

• Example: If NEPSE’s monthly returns have μ = 2% and σ = 4%, what is the probability of a loss (> -5%)? $$z=\frac{-5-2}{4}=-1.75$$ P(Z < -1.75) = 0.0401 \text{ (4.01%)}

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5. Limitations of Normal Distribution

While powerful, the normal distribution has limitations:

Example 3: Khalti Transaction Fees Khalti’s transaction fees (in NPR) are not normally distributed (skewed right).

• Solution: Use log-transformation before applying normal distribution.

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6. Comparing Normal Distribution with Other Distributions

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7. Solving Problems Step-by-Step

Worked Example 4: NTC Internet Speed

NTC claims its internet speeds in Kathmandu follow a normal distribution with:

• μ = 50 Mbps
• σ = 10 Mbps

Question: What % of users get speeds between 40 and 60 Mbps? Solution:

1. Convert to Z-scores: $$z_1=\frac{40-50}{10}=-1$$ $$z_2=\frac{60-50}{10}=1$$
2. Find probabilities:
   • P(Z < 1) = 0.8413
   • P(Z < -1) = 0.1587
3. Calculate the range: P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826 \text{ (68.26%)}

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8. Real-World Case Study: Pathao Driver Earnings

Pathao’s driver earnings (in NPR/day) are normally distributed:

• μ = 3000 NPR
• σ = 500 NPR

Scenario: Pathao wants to ensure top 5% of drivers earn > X NPR/day. Find X. Solution:

1. Find the Z-score for top 5%: $$P(Z>z)=0.05⟹z=1.645$$
2. Convert back to earnings: $$X=\mu +z\cdot \sigma =3000+1.645\times 500=3822.5\text{ NPR}$$ Conclusion: Pathao should target 3823 NPR/day for the top 5% earners.

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In the Real World

1. eSewa (Digital Payments)

   • Idea Used: Normal distribution models transaction success rates.
   • How: eSewa assumes payment failures follow a normal curve to predict system downtimes. For example, if μ = 2% failure rate and σ = 0.5%, they can estimate the probability of > 3% failures during festivals.
   • Calculation: $$z=\frac{3-2}{0.5}=2⟹P(Z>2)=2.28\%$$ This helps eSewa prepare for high-demand periods.

2. Daraz (Logistics & Delivery)

   • Idea Used: Normal distribution predicts delivery delays.
   • How: Daraz tracks delivery times and uses the empirical rule to set SLA (Service Level Agreements). If μ = 48 hours and σ = 6 hours, they guarantee 95% of orders arrive within 60 hours (μ + 2σ).
   • Visual:

3. Nepal Rastra Bank (Inflation Forecasting)

   • Idea Used: Normal distribution models inflation rates.
   • How: NRB uses historical inflation data (assumed normal) to predict future trends. If μ = 6% inflation and σ = 1.5%, they can estimate the chance of inflation > 9%: $$z=\frac{9-6}{1.5}=2⟹P(Z>2)=2.28\%$$
   • Impact: Helps in setting monetary policies (e.g., interest rates).

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Exam Tip

What Examiners Look For

1. Correct Formula Application

   • Always show the Z-score formula and empirical rule steps.
   • Example: For a question on "probability within a range," write: $$P(a<X<b)=P\left(\frac{a-\mu }{\sigma }<Z<\frac{b-\mu }{\sigma }\right)$$

2. Interpretation Over Calculation

   • Examiners reward real-world context. For example:
     • "If σ increases, the curve flattens, meaning more variability in data (e.g., Daraz delivery times become less predictable)."

3. Graphical Representation

   • Draw the normal curve for every question involving probabilities.
   • Shade the relevant area (use the empirical rule for quick sketches).

4. Common Pitfalls to Avoid

   • Forgetting to convert to Z-scores before using tables.
   • Misapplying the empirical rule (e.g., confusing 68% with 95%).
   • Ignoring units (always label μ and σ with units, e.g., "hours," "NPR").

5. Shortcut for Quick Marks

   • Memorize the empirical rule percentages (68%, 95%, 99.7%) and Z-table values for common Z-scores (e.g., Z = 1.96 for 95% confidence).

Sample Exam Question & Answer

Question: The heights of male students in TU follow a normal distribution with μ = 170 cm and σ = 10 cm. What is the probability that a randomly selected student is: a) Taller than 185 cm? b) Between 160 cm and 190 cm?

Model Answer: a) P(X > 185)

1. Calculate Z-score: $$z=\frac{185-170}{10}=1.5$$
2. Look up P(Z > 1.5) = 1 – 0.9332 = 0.0668 (6.68%).

b) P(160 < X < 190)

1. Convert to Z-scores: $$z_1=\frac{160-170}{10}=-1$$ $$z_2=\frac{190-170}{10}=2$$
2. Find probabilities: P(-1 < Z < 2) = P(Z < 2) - P(Z < -1) = 0.9772 - 0.1587 = 0.8185 \text{ (81.85%)}

Exam Tip: Always label your Z-scores and shade the area in your answer—this shows structured thinking.