Normal Distribution – Definition, Properties, Z‑Scores, Empirical Rule & Applications

1. Definition and Key Properties

A continuous random variable $X$ follows a normal distribution if its probability density function (pdf) is

$$f(x)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{(x-\mu {)}^2}{2{\sigma }^2}},-\infty <x<\infty$$

where

• $\mu$ = mean (also the median and mode because of symmetry)
• $\sigma >0$ = standard deviation (controls spread)

Properties

2. Standard Normal Distribution

When $\mu =0$ and $\sigma =1$ the distribution is called standard normal and denoted $Z\sim N(0,1)$.

The Z‑score of a raw observation $x$ is

$$Z=\frac{x-\mu }{\sigma }$$

It tells how many standard deviations $x$ lies from the mean.

2.1 Using the Z‑table

A Z‑table (or standard normal cumulative table) gives

$$P(Z\le z)=\Phi (z)$$

For negative $z$ we use symmetry: $\Phi (-z)=1-\Phi (z)$.

3. Worked Example – Probability of a Transaction Amount

Problem: eSewa records the amount (in thousands of rupees) of daily online transactions. Historical data show a mean of $\mu =12$ kRs and a standard deviation of $\sigma =3$ kRs. What is the probability that a randomly chosen day’s total transaction amount exceeds 15 kRs?

Solution steps

1. Compute the Z‑score:

$$Z=\frac{15-12}{3}=1.0$$

2. From the Z‑table, $\Phi (1.0)=0.8413$.

3. Required probability

$$P(X>15)=1-\Phi (1.0)=1-0.8413=0.1587$$

Interpretation: About 16 % of days the transaction total is above 15 kRs.

4. Normal Approximation to the Binomial

When a binomial variable $X\sim B(n,p)$ has large $n$ and $p$ not too close to 0 or 1,

$$X\approx N(\mu =np,{\sigma }^2=np(1-p))$$

A continuity correction of ±0.5 is applied because the binomial is discrete.

Example – Daraz Order Queue

Daraz receives on average 120 orders per hour ($p=0.6$ of 200 potential customers). Approximate the probability that in a given hour they receive at most 110 orders.

1. Binomial parameters: $n=200,p=0.6$ → $\mu =np=120,\sigma =\sqrt{np(1-p)}=\sqrt{200\cdot 0.6\cdot 0.4}=6.93$.

2. Apply continuity correction: $X\le 110$ → $X\le 110.5$.

3. Z‑score:

$$Z=\frac{110.5-120}{6.93}=-1.37$$

4. $\Phi (-1.37)=1-\Phi (1.37)=1-0.9147=0.0853$.

Result: Only about 8.5 % of hours will have ≤110 orders.

5. Empirical Rule (68‑95‑99.7)

For any normal distribution:

Real‑World Use – Ncell Signal Strength

Ncell measures signal strength (dBm) across Kathmandu. Mean = –85 dBm, σ = 5 dBm.

• Within 1σ (–90 to –80 dBm) ≈ 68 % of locations have acceptable signal.
• Beyond 2σ (≤ –95 dBm or ≥ –75 dBm) ≈ 5 % of locations experience poor or unusually strong signal.

6. Comparison with Other Distributions

Advantages of Normal Approximation

• Simple calculations using a single Z‑table.
• Works well for large samples due to Central Limit Theorem.

Disadvantages

• Poor fit for heavily skewed or bounded data.
• Requires continuity correction for discrete variables.

7. Applications in Business Contexts

8. Worked Example – Determining Safety Stock

A warehouse manager knows that daily demand for a popular gadget follows $N(\mu =200,\sigma =30)$ units. To achieve a 95 % service level (i.e., stockout probability ≤ 5 %), what safety stock should be kept?

1. 95 % service level → $z_{0.95}=1.645$.
2. Safety stock = $z\sigma =1.645\times 30\approx 49$ units.

Thus, keep ≈ 50 units as safety stock.

9. Limitations and When Not to Use

• Heavy tails: Financial returns often exhibit kurtosis > 3; normal underestimates extreme losses.
• Bounded variables: Percentages (0–100 %) are better modelled by beta distribution.
• Small samples: Central Limit Theorem requires moderate to large n; with n < 30 normal approximation may be inaccurate.

10. Summary

• Normal distribution is the cornerstone of continuous probability modelling.
• Z‑scores convert any normal variable to the standard form, enabling lookup of probabilities.
• Empirical rule provides quick estimates of spread.
• Normal approximation simplifies binomial calculations when conditions are met, but continuity correction is essential.
• Real‑world business problems—from e‑payment volumes to delivery time forecasts—rely on normal‑based reasoning.

In the real world

1. eSewa: The daily total of transaction amounts (in thousands of rupees) is modeled as $N(12,3^2)$. The probability of exceeding 15 kRs (≈ 16 %) helps the company plan liquidity.
2. Daraz: Order arrivals per hour (≈ 120 on average, σ≈7) are approximated by a normal curve; the 95 % service‑level safety stock for a fast‑moving item is computed as $1.645\sigma \approx 12$ units.
3. Ncell: Signal strength across Kathmandu follows a normal distribution (µ = –85 dBm, σ = 5 dBm). The empirical rule tells the marketing team that about 95 % of the city experiences signal between –95 and –75 dBm, guiding tower placement.

Exam tip

• Memorise the Z‑table layout (area to the left of z).
• Always apply continuity correction when a normal approximation is asked for a discrete variable.
• Write the Z‑score formula first, then substitute numbers; keep track of sign (positive → right of mean, negative → left).
• For the empirical rule, remember the three percentages (68‑95‑99.7) and the corresponding σ‑multiples; many questions ask “approximately what proportion lies between µ ± 1.5σ?” – interpolate between the 68 % and 95 % rows.
• Show work: even if the answer is a single decimal, write each step (mean, σ, Z, table lookup) to earn full marks.