Learning Objectives

By the end of this chapter, you will be able to:

  • Describe the properties of the normal distribution
  • Standardize values using z-scores
  • Use the standard normal table
  • Calculate probabilities for normal distributions
  • Apply the normal distribution to real-world problems

What is the Normal Distribution?

The normal distribution (also called Gaussian distribution) is the most important continuous probability distribution in statistics. Many administrative measurements—processing times around a typical value, exam scores, measurement errors—are unimodal and roughly symmetric. Even when a single observation is skewed (for example, household income), sample means of large random samples tend toward normality by the Central Limit Theorem (Unit 4). That is why z-scores, confidence intervals, and many hypothesis tests build on the normal curve.

\[X \sim N(\mu, \sigma^2)\]

Where:

  • $\mu$ = mean
  • $\sigma^2$ = variance
  • $\sigma$ = standard deviation

The Normal Curve

flowchart LR
    A["Bell-shaped curve"]
    B["Symmetric about mean"]
    C["Mean = Median = Mode"]
    D["Extends from -∞ to +∞"]

Properties of Normal Distribution

1. Bell-Shaped and Symmetric

  • The curve is perfectly symmetric around the mean
  • The highest point is at the mean

2. Mean, Median, and Mode are Equal

\[\mu = \text{Median} = \text{Mode}\]

3. Asymptotic

  • The tails never touch the x-axis
  • They approach but never reach zero

4. Total Area = 1

\[\int_{-\infty}^{+\infty} f(x) \, dx = 1\]

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

flowchart TD
    A["Empirical Rule"] --> B["68% within ± 1σ"]
    A --> C["95% within ± 2σ"]
    A --> D["99.7% within ± 3σ"]
Range Percentage of Data
$\mu \pm 1\sigma$ 68.26%
$\mu \pm 2\sigma$ 95.44%
$\mu \pm 3\sigma$ 99.74%

Example 1: Applying Empirical Rule

Problem: Heights of adults are normally distributed with mean 170 cm and standard deviation 8 cm. What percentage of adults have heights between 154 cm and 186 cm?

Solution:

  • $\mu = 170$ cm, $\sigma = 8$ cm
  • 154 = 170 - 16 = 170 - 2(8) = $\mu - 2\sigma$
  • 186 = 170 + 16 = 170 + 2(8) = $\mu + 2\sigma$

Since the range is $\mu \pm 2\sigma$:

Answer: Approximately 95.44% of adults have heights between 154 cm and 186 cm.

Standard Normal Distribution

The standard normal distribution has:

  • Mean $\mu = 0$
  • Standard deviation $\sigma = 1$
\[Z \sim N(0, 1)\]

Why Use Standard Normal?

  • All normal distributions can be converted to standard normal
  • Single table covers all normal distributions
  • Easier to calculate probabilities

Z-Score (Standardization)

Different PA indicators use different units: days, NPR, minutes, scores out of 100. A z-score puts any normal measurement on a common scale—“how many SDs from its own mean?”—so one z-table serves salaries, wait times, and test scores alike.

A z-score tells how many standard deviations a value is from the mean:

\[Z = \frac{X - \mu}{\sigma}\]

Where:

  • $X$ = raw score
  • $\mu$ = population mean
  • $\sigma$ = population standard deviation
  • $Z$ = standardized score

Interpreting Z-Scores

Z-Score Meaning
Z = 0 At the mean
Z = +1 1 SD above mean
Z = -1 1 SD below mean
Z = +2 2 SD above mean
Z = -2.5 2.5 SD below mean 
flowchart LR
    A["-3"] --> B["-2"] --> C["-1"] --> D["0<br/>(mean)"] --> E["+1"] --> F["+2"] --> G["+3"]

Step-by-Step Example 2: Calculating Z-Score

Problem: Test scores are normally distributed with mean 75 and standard deviation 10. A student scores 85. What is their z-score?

Solution:

\[Z = \frac{X - \mu}{\sigma} = \frac{85 - 75}{10} = \frac{10}{10} = 1\]

Answer: Z = 1 (The student scored 1 standard deviation above the mean)

Using the Standard Normal Table (Z-Table)

The Z-table gives $P(Z < z)$, the area to the LEFT of z.

Partial Z-Table

z 0.00 0.01 0.02 0.03 0.04 0.05
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531
1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394
2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798

How to Read the Table

For z = 1.53:

  1. Go to row 1.5
  2. Go to column 0.03
  3. Read value: 0.9370

So $P(Z < 1.53) = 0.9370$

Types of Probability Questions

flowchart TD
    A[Probability Question Type]
    A --> B["P(Z < z)<br/>Left tail"]
    A --> C["P(Z > z)<br/>Right tail"]
    A --> D["P(a < Z < b)<br/>Between two values"]

    B --> B1["Read directly from table"]
    C --> C1["= 1 - P(Z < z)"]
    D --> D1["= P(Z < b) - P(Z < a)"]

Step-by-Step Example 3: Finding Probabilities

Problem: For standard normal distribution, find: (a) P(Z < 1.25) (b) P(Z > 0.84) (c) P(-1.5 < Z < 1.5)

Solution:

(a) P(Z < 1.25)

From z-table: P(Z < 1.25) = 0.8944

(b) P(Z > 0.84)

\(P(Z > 0.84) = 1 - P(Z < 0.84)\) \(= 1 - 0.7995 = 0.2005\)

Answer: 0.2005

(c) P(-1.5 < Z < 1.5)

\(P(-1.5 < Z < 1.5) = P(Z < 1.5) - P(Z < -1.5)\) \(= 0.9332 - 0.0668 = 0.8664\)

Answer: 0.8664

Step-by-Step Example 4: Complete Problem

Problem: Monthly electricity consumption in households is normally distributed with mean 250 kWh and standard deviation 40 kWh. Find the probability that a randomly selected household consumes: (a) Less than 300 kWh (b) More than 200 kWh (c) Between 220 and 280 kWh

Solution:

Given: $\mu = 250$ kWh, $\sigma = 40$ kWh

(a) P(X < 300)

Step 1: Convert to z-score \(z = \frac{300 - 250}{40} = \frac{50}{40} = 1.25\)

Step 2: Find probability \(P(X < 300) = P(Z < 1.25) = 0.8944\)

Answer: 89.44%

(b) P(X > 200)

Step 1: Convert to z-score \(z = \frac{200 - 250}{40} = \frac{-50}{40} = -1.25\)

Step 2: Find probability \(P(X > 200) = P(Z > -1.25) = 1 - P(Z < -1.25)\) \(= 1 - 0.1056 = 0.8944\)

Answer: 89.44%

(c) P(220 < X < 280)

Step 1: Convert both values to z-scores \(z_1 = \frac{220 - 250}{40} = -0.75\) \(z_2 = \frac{280 - 250}{40} = 0.75\)

Step 2: Find probability \(P(220 < X < 280) = P(-0.75 < Z < 0.75)\) \(= P(Z < 0.75) - P(Z < -0.75)\) \(= 0.7734 - 0.2266 = 0.5468\)

Answer: 54.68%

Step-by-Step Example 5: Exam-Style Problem

Problem: Annual salaries of government employees are normally distributed with mean NPR 600,000 and standard deviation NPR 80,000.

(a) What percentage earn less than NPR 500,000? (b) What percentage earn more than NPR 750,000? (c) What salary level marks the top 10% of earners? (d) What is the IQR (interquartile range) of salaries?

Solution:

Given: $\mu = 600,000$, $\sigma = 80,000$

(a) P(X < 500,000)

\[z = \frac{500,000 - 600,000}{80,000} = \frac{-100,000}{80,000} = -1.25\] \[P(Z < -1.25) = 0.1056\]

Answer: 10.56% earn less than NPR 500,000

(b) P(X > 750,000)

\[z = \frac{750,000 - 600,000}{80,000} = 1.875\] \[P(Z > 1.875) = 1 - P(Z < 1.875) = 1 - 0.9696 = 0.0304\]

Answer: 3.04% earn more than NPR 750,000

(c) Top 10% threshold

Find z such that P(Z > z) = 0.10, meaning P(Z < z) = 0.90

From table: z ≈ 1.28

\(X = \mu + z\sigma = 600,000 + (1.28)(80,000)\) \(= 600,000 + 102,400 = 702,400\)

Answer: NPR 702,400 marks the top 10%

(d) Interquartile Range (IQR)

Q1: P(Z < z₁) = 0.25 → z₁ ≈ -0.675 Q3: P(Z < z₃) = 0.75 → z₃ ≈ +0.675

\(Q_1 = 600,000 + (-0.675)(80,000) = 546,000\) \(Q_3 = 600,000 + (0.675)(80,000) = 654,000\)

\[IQR = Q_3 - Q_1 = 654,000 - 546,000 = 108,000\]

Answer: IQR = NPR 108,000

Finding X-Values from Probabilities (Reverse Problems)

When given a probability, find the corresponding X value:

\[X = \mu + Z \times \sigma\]

Example 6: Reverse Problem

Problem: Scores are normally distributed with mean 70 and standard deviation 12. What score is needed to be in the top 5%?

Solution:

Step 1: Find z for top 5% P(Z < z) = 0.95 → z = 1.645

Step 2: Calculate X \(X = \mu + z\sigma = 70 + (1.645)(12) = 70 + 19.74 = 89.74\)

Answer: A score of at least 89.74 (or 90) is needed

Important Z-Values to Remember

Percentile Z-value
1% -2.33
5% -1.645
10% -1.28
25% (Q1) -0.675
50% 0
75% (Q3) +0.675
90% +1.28
95% +1.645
99% +2.33

Common Critical Values

Confidence Level Z-value
90% 1.645
95% 1.96
99% 2.576

Normal Approximation to Binomial

Counting successes in n independent yes/no trials is binomial (note-12). When n is large and p is not too close to 0 or 1, the binomial histogram looks like a normal bell—so you can use z-table methods instead of summing many binomial terms. If $np$ or $nq$ is small, keep the exact binomial (or use software); the normal shortcut then misleads.

When n is large and p is not too extreme, binomial can be approximated by normal:

\[X \sim B(n, p) \approx N(np, npq)\]

Rule of Thumb

Use normal approximation when:

  • $np \geq 5$ AND $nq \geq 5$

Continuity Correction

Since binomial is discrete and normal is continuous, add/subtract 0.5:

Binomial Normal (with correction)
$P(X = k)$ $P(k-0.5 < X < k+0.5)$
$P(X \leq k)$ $P(X < k+0.5)$
$P(X \geq k)$ $P(X > k-0.5)$

Practice Problems

Problem 1

Heights of students are normally distributed with mean 165 cm and SD 7 cm. Find: (a) Probability a student is taller than 175 cm (b) Probability a student is between 158 and 172 cm (c) Height below which 90% of students fall

Answer / solution sketch (a) $z=(175-165)/7\approx 1.43$; $P(Z>1.43)\approx 0.0764$ (b) $z=(158-165)/7=-1$, $z=(172-165)/7=1$; $P(-1<Z<1)\approx 0.6827$ (c) $z_{0.90}\approx 1.28$; $X=165+1.28(7)\approx 173.96$ cm

Problem 2

Processing times are normally distributed with mean 45 minutes and SD 8 minutes. What proportion of tasks take: (a) Less than 30 minutes? (b) More than 60 minutes? (c) Between 40 and 55 minutes?

Answer / solution sketch (a) $z=(30-45)/8=-1.875$; $P\approx 0.0304$ (b) $z=(60-45)/8=1.875$; $P\approx 0.0304$ (c) $z=(40-45)/8=-0.625$, $z=(55-45)/8=1.25$; $P\approx 0.628$

Problem 3

IQ scores follow N(100, 225). Find: (a) P(IQ > 130) (b) The IQ threshold for the top 2% (c) P(90 < IQ < 120)

Answer / solution sketch $\sigma=\sqrt{225}=15$. (a) $z=(130-100)/15=2$; $P\approx 0.0228$ (b) $z_{0.98}\approx 2.05$; $X=100+2.05(15)\approx 130.8$ (c) $z=(90-100)/15=-0.67$, $z=(120-100)/15=1.33$; $P\approx 0.656$

Problem 4

A service guarantees delivery within 48 hours. Delivery times are N(40, 36). What percentage fail to meet the guarantee?

Answer / solution sketch $\sigma=6$; $z=(48-40)/6\approx 1.33$; $P(X>48)=P(Z>1.33)\approx 0.0918$ → about **9.2%** fail.

Problem 5

If X ~ N(50, 100), find the value of k such that P(50-k < X < 50+k) = 0.90

Answer / solution sketch $\sigma=10$; middle 90% ↔ $z^*=1.645$. $$k=1.645\times 10=16.45$$

Common Mistakes

  1. Assuming every PA variable is normal. Skewed income or count data may not be; check shape or rely on CLT for means of large samples.
  2. Mixing left- and right-tail areas on the z-table. Most tables give P(Z ≤ z) for positive z; use symmetry for left tails and $1 - P(Z \leq z)$ for upper tails.
  3. Forgetting units when converting back. After finding z, $X = \mu + z\sigma$ restores the original scale (days, NPR, scores).
  4. Using the normal approximation to binomial when $np$ or $nq$ is small. The rule of thumb exists for a reason.

Summary

Concept Formula/Value
Z-score $Z = \frac{X - \mu}{\sigma}$
Find X from Z $X = \mu + Z\sigma$
P(Z < z) Read from table
P(Z > z) $1 - P(Z < z)$
P(a < Z < b) $P(Z < b) - P(Z < a)$
68-95-99.7 Rule Within 1, 2, 3 standard deviations
95% z-value 1.96

Unit 3 Complete!

You have completed Unit 3: Probability Theory. This foundational knowledge of probability distributions (especially the normal distribution) will be essential for the upcoming units on Estimation and Hypothesis Testing.

In Unit 4, we will study Estimation - how to estimate population parameters from sample data and construct confidence intervals.