Unit 5 · Expedition into the Normal Distribution

Chapter 1 — Section 5.1

The Binomial Bridge

Where a staircase of coin flips turns into a smooth hill.

At the edge of camp there's an old rope bridge made of planks — one plank for every possible outcome of flipping a coin 50 times. Flip it once, twice, three times: the planks are jagged and uneven. But flip it fifty times, a hundred times, and step back — the bridge's railing traces a smooth, bell-shaped curve. That's the whole secret of this chapter: enough repeated trials, and a bumpy binomial count starts looking like a normal curve.

Concept

When can a binomial be treated as normal?

Suppose \(X\) counts the number of successes in \(n\) independent trials, each with success probability \(p\) — this is the binomial setting, \(X \sim B(n,p)\). When \(n\) is large enough, the jagged bar-chart shape of \(X\) is well approximated by a smooth normal curve with the same center and spread.

Matching mean & spread

\[ \mu = np \qquad \sigma = \sqrt{np(1-p)} \]

Large Counts ConditionUse the normal approximation only when \(np \geq 10\) and \(n(1-p) \geq 10\).

Continuity CorrectionBinomial is discrete, normal is continuous — always shift the boundary by \(0.5\) before finding \(z\).

Continuity correction cheat-sheet

\[ P(X \le k) \approx P(Y \le k+0.5) \qquad P(X \ge k) \approx P(Y \ge k-0.5) \qquad P(X = k) \approx P(k-0.5 \le Y \le k+0.5) \]

Worked Examples

EX 1Setting up the normal shadow

Question

A school reports that 40% of students bring a packed lunch. In a random sample of \(n = 50\) students, let \(X\) be the number who bring a packed lunch. Check the Large Counts Condition, then find \(\mu\) and \(\sigma\) for the normal approximation.

Solution

1Check the condition: \(np = 50(0.4) = 20\) and \(n(1-p) = 50(0.6) = 30\). Both are \(\ge 10\), so the approximation is safe to use.

2Mean: \(\mu = np = 20\).

3Spread: \(\sigma = \sqrt{50(0.4)(0.6)} = \sqrt{12} \approx 3.46\).

Answer: \(X\) behaves like \(N(20,\ 3.46)\).

EX 2Crossing the bridge with continuity correction

Question

Using the same setting as Example 1 (\(n=50,\ p=0.4\)), estimate \(P(X \le 25)\).

Solution

1Apply the continuity correction: \(P(X \le 25) \approx P(Y \le 25.5)\).

2Standardize: \(z = \dfrac{25.5 - 20}{3.46} \approx 1.59\).

3Read the standard normal table: \(\Phi(1.59) \approx 0.9441\).

Answer: \(P(X \le 25) \approx 0.9441\) — about a 94% chance.

EX 3A fair-coin experiment

Question

A fair coin is tossed 100 times. Estimate the probability of getting at least 60 heads.

Solution

1Here \(n=100,\ p=0.5\): \(\mu = 50\), \(\sigma = \sqrt{100(0.5)(0.5)} = \sqrt{25} = 5\).

2Continuity correction: \(P(X \ge 60) \approx P(Y \ge 59.5)\).

3Standardize: \(z = \dfrac{59.5-50}{5} = 1.9\), so \(\Phi(1.9) \approx 0.9713\).

4Take the upper tail: \(P(Z \ge 1.9) = 1 - 0.9713 = 0.0287\).

Answer: about a 2.9% chance of 60 or more heads.

Chapter 2 — Section 5.2

Z-Score Valley

Where every value gets measured the same way: steps away from base camp.

Deep in the valley, no two mountains are marked in the same units — one trail is measured in meters, another in the number of switchbacks. So the rangers invented one universal ruler: how many standard deviations away from the mean are you? That number is the z-score, and once everything is converted to it, any two valleys can finally be compared side by side.

Concept

The z-score: a universal ruler

For a value \(x\) from a normal distribution with mean \(\mu\) and standard deviation \(\sigma\), the z-score tells you how many standard deviations \(x\) sits above (positive) or below (negative) the mean.

Standardizing

\[ z = \frac{x - \mu}{\sigma} \qquad\Longleftrightarrow\qquad x = \mu + z\sigma \]

Once a value is turned into \(z\), you can look up its position on the standard normal curve \(N(0,1)\) — the same table works for test scores, heights, reaction times, anything normal.

68 – 95 – 99.7 RuleAbout 68% of values fall within 1 SD of the mean, 95% within 2 SD, and 99.7% within 3 SD.

Reading z-negativeA negative z simply means the value is below the mean — the size still tells you the distance.

Worked Examples

EX 1From score to z

Question

Exam scores are normally distributed with \(\mu = 75\) and \(\sigma = 8\). A student scores \(x = 91\). Find the z-score and interpret it.

Solution

1\(z = \dfrac{91-75}{8} = \dfrac{16}{8} = 2\).

Answer: \(z = 2\) — the student scored exactly 2 standard deviations above average.

EX 2From z back to score

Question

IQ scores follow \(N(100, 15)\). What score corresponds to \(z = -1.5\)?

Solution

1Use \(x = \mu + z\sigma\).

2\(x = 100 + (-1.5)(15) = 100 - 22.5 = 77.5\).

Answer: \(x = 77.5\).

EX 3Turning z into a probability

Question

Heights of students at a school follow \(N(165\text{ cm}, 6\text{ cm})\). Estimate \(P(X < 172)\).

Solution

1Standardize: \(z = \dfrac{172-165}{6} = \dfrac{7}{6} \approx 1.17\).

2Look up the table: \(\Phi(1.17) \approx 0.8790\).

Answer: about an 88% chance a student is shorter than 172 cm.

Chapter 3 — Section 5.3

The Summit — Mixed Applications

Where you decide, on your own, which tool the trail is asking for.

At the summit there's no more signage telling you which trail you're on — every problem just looks like a plain question. The real skill up here isn't new math at all. It's pattern recognition: is this a count coming from many repeated trials (Chapter 1's tool), or a single value from a known normal population (Chapter 2's tool)? Get that one decision right, and the rest is just plugging in.

Concept

Choosing the right tool

Ask: what is being counted?A number of "successes" out of many repeated trials (like "how many of 200 shoppers used a coupon")?

↓ if yes

Use the Binomial → Normal route (5.1)Find \(\mu = np\), \(\sigma=\sqrt{np(1-p)}\), check the Large Counts Condition, then apply the continuity correction.

Otherwise: is it one measured value from a known normal population?Like a single test score, height, or weight, where \(\mu\) and \(\sigma\) of the population are already given?

↓ if yes

Use the Z-score route (5.2)Standardize with \(z = \dfrac{x-\mu}{\sigma}\) directly — no continuity correction needed, because there's no discrete count involved.

Worked Examples

EX 1A count problem, fully worked

Question

A store finds that 40% of shoppers use a coupon. In a random sample of \(n=200\) shoppers, estimate \(P(X > 90)\), the probability that more than 90 used a coupon.

Solution

1\(\mu = np = 200(0.4)=80\), \(\sigma = \sqrt{200(0.4)(0.6)} = \sqrt{48} \approx 6.93\). Check: \(np=80\ge10\), \(n(1-p)=120\ge10\). ✓

2Continuity correction: \(P(X>90) \approx P(Y>90.5)\).

3\(z = \dfrac{90.5-80}{6.93} \approx 1.52\), and \(\Phi(1.52)\approx 0.9357\).

4\(P(Z>1.52) = 1-0.9357 = 0.0643\).

Answer: about a 6.4% chance more than 90 shoppers used a coupon.

EX 2A single-value problem, fully worked

Question

National test scores follow \(N(500, 100)\). What score marks the 90th percentile?

Solution

1Find the z-score whose area to the left is 0.90: from the table, \(z \approx 1.28\).

2Convert back: \(x = \mu+z\sigma = 500+1.28(100)=628\).

Answer: a score of about 628 marks the 90th percentile.

EX 3Method-check drill (spot the route)

Question — for each scenario, name the route before reading the answer

"Of 300 registered voters surveyed, 55% support a proposal. Find the probability at least 180 support it."

Binomial → Normal

"A patient's temperature is 101.2°F. The population of healthy adults has \(\mu=98.6\), \(\sigma=0.8\). Find how unusual this reading is."

Z-score

"A factory ships parts in boxes of 500; 2% are defective. Estimate the probability that a box has more than 15 defective parts."

Binomial → Normal

You reached basecamp

Field Notes — Unit 5 Summary

Everything from the trail, condensed onto one page for quick review before a quiz.

5.1 — Binomial → Normal

\[ \mu = np \qquad \sigma=\sqrt{np(1-p)} \]

Condition: \(np \ge 10\) and \(n(1-p)\ge 10\).
Continuity correction: shift the boundary by \(0.5\) before standardizing.
\(P(X\le k)\approx P(Y\le k+0.5)\), \(\ P(X\ge k)\approx P(Y\ge k-0.5)\).

5.2 — Z-scores

\[ z=\frac{x-\mu}{\sigma} \qquad x=\mu+z\sigma \]

68–95–99.7 rule for 1, 2, 3 standard deviations.
Positive \(z\) = above the mean; negative \(z\) = below the mean.
Standard normal table gives area (probability) to the left of \(z\).

5.3 — Which route do I take?

Repeated trials, counting successes → Binomial → Normal (use \(\mu=np\), \(\sigma=\sqrt{np(1-p)}\), continuity correction).
One value from a known normal population → Z-score directly (no continuity correction — nothing discrete is being counted).
Always check the Large Counts Condition before trusting a binomial-to-normal estimate.