Jecy/Lessons

JECY LESSONS / STATISTICS 01

Normal
distribution

正态分布 · 期末考前串讲

一份订单比平均时间慢 5 分钟，算慢吗？移动图上的时间线，看有多少订单比它更慢。

开始串讲 ↓15 min 主讲 · 5 min 选练
FIG. 01　/　TAIL AREA配送时间实验
2540 min55
15.9%
模型 T ∼ N(40, 5²)　·　右侧面积 P(T > t)

Part 1 / 知识速讲

Read the normal curve

0–4 min

Center and spread

T ∼ N(40, 5²)

μ = 40 min · σ = 5 min

A normal curve is continuous and symmetric about its mean. The mean μ sets the center. A larger standard deviation σ produces a wider, lower curve.

均值决定中心，标准差决定宽窄。本课用配送时间 T 举例。

Quick check

If the SD grows from 5 to 8 minutes, where is the center?

Check answer

Still at 40 minutes. The two curves share the same center and total area.

Same center, different spread
Same center, different spreadTwo normal curves share a mean of 40 minutes. The dashed curve with SD 8 is wider and lower than the solid curve with SD 5.Density0.020.040.06μ = 4025303540455055min
SD 5 minSD 8 min

Read probability as area

P(35 ≤ T ≤ 45)

An interval is a region under the curve.

The total area is 1. Each side of the mean holds one half. A single exact value has zero area in a continuous model.

看横轴上的区间，再看它对应的阴影面积。曲线的高度表示密度。

Quick check

Does including the endpoint 45 change this probability?

Check answer

No. P(T = 45) = 0, so < and ≤ give the same result at an endpoint.

An interval has area
An interval has areaThe central interval between 35 and 45 minutes is shaded on a normal density with mean 40 and SD 5.Density0.020.040.06μ = 4025303540455055min

Selected interval · 35 to 45 min

Count standard deviations

z = (t − μ) / σ

To return to minutes, t = μ + zσ.

The sign tells you which side of the mean a value lies on. The magnitude counts standard deviations. A z-score has no units.

先减均值，再除以标准差。负号表示在均值左边。

Quick check

What does a delivery time of 35 minutes mean on the z scale?

Check answer

z = (35 − 40)/5 = −1. The delivery took one standard deviation less time than the mean, or five minutes less.

Two scales, one position
Two scales, one positionThe marked time of 35 on the upper time axis lines up with minus one on the standardized axis.Density0.020.040.06μ = 4035 min25−330−235−140045+150+255+3minz

A time of 35 min corresponds to z = −1.

Start with the left tail

Φ(z) = P(Z ≤ z)

Z follows the standard normal distribution N(0, 1).

In a cumulative left-tail table, each entry is the area up to z. For an upper tail, take the complement. Between two cutoffs, subtract the two cumulative areas.

本课的 z 表给左侧累计面积。拿到考试用表，先看表头阴影。

Quick check

If Φ(1) ≈ 0.8413, what percentage lies above z = 1?

Check answer

About 15.87%. The right tail is 1 − 0.8413 = 0.1587.

Cumulative area to the left
Cumulative area to the leftAll times to the left of 45 minutes are shaded. The area extends into the full lower tail.Density0.020.040.06μ = 4045 min25303540455055min

Φ(1) ≈ 0.8413 · area up to 45 min

Estimate central coverage

μ ± kσ

About 68%, 95% and 99.7% for k = 1, 2 and 3.

The normal curve leaves equal areas in the two tails outside a central interval. These rounded percentages are useful for a quick reasonableness check.

经验法则用作估算。题目要求指定的精确覆盖率时，查相应分位数。

Quick check

About 95% lies within two SDs. Roughly how much is above the upper limit?

Check answer

About 2.5% by the empirical rule. Using Φ(2), the one-tail probability is about 2.28%.

How much lies in the middle?
Within 1 standard deviationsThe central region from 35 to 45 minutes is shaded.Density0.020.040.06μ = 4025303540455055min

68.27% from 35 to 45 min

Density formula & notation

A continuous variable X is normal with mean μ and standard deviation σ when its density follows this formula. Here x and μ can be any real numbers, and σ > 0.

本课写作 N(μ, σ²)，第二个参数是方差。OpenStax 这本教材写作 N(μ, σ)，第二个参数是标准差。先看教材约定，再读参数。

f of x equals exp[−(x − μ)² / (2σ²)] divided by [σ √(2π)].

Part 2 / 题型识别

What are you solving for?

4–7 min

Read each stem. Name the unknown before doing any arithmetic. 先说要交什么答案，再展开解法。

Two course examsWho performed better relative to their class?
Relative standing

Given · Two scores and a mean and SD for each group.

zₐ = (a − μₐ)/σₐ, zᵦ = (b − μᵦ)/σᵦ

Compare the two z-scores in context. For exam marks, higher is better.

把每个成绩放回自己的组比较。若题目改成比赛用时，更短的时间代表更快。

Delivery reliabilityWhat proportion arrives late or within a specified time window?
One probability

Given · A normal model and one or two numerical limits.

Below: Φ(z) · Above: 1 − Φ(z) · Between: Φ(zᵦ) − Φ(zₐ)

For a < b, standardize both limits before subtracting. A percentage is 100 times the probability.

已知边界，求比例。区间概率用右端的累计面积减左端的累计面积。

A service benchmarkWhat time separates the fastest 10% of deliveries?
One cutoff

Given · A fraction and the direction of the desired tail.

c = μ + σ Φ⁻¹(p)

Use the area to the left of the cutoff for p. Faster deliveries take fewer minutes.

已知比例，反求有单位的数。先按横轴的含义决定看左侧还是右侧。

Bottling tolerancesWhich two fill volumes enclose the middle 90% of bottles?
Two cutoffs

Given · A central coverage p and a normal model.

Left-tail probabilities: (1 − p)/2 and (1 + p)/2.

For the central 90%, use the 5th and 95th percentiles. Convert both z-values back to mL.

先把剩余 10% 平分给两端，再把两个分位数换回原单位。

Part 3 / 考题与纠错

Work the exam questions

7–15 min
T ∼ N(40, 5²)

Assume delivery times are approximately normal. Mean 40 minutes · SD 5 minutes.

Try each question aloud or on paper. You can reveal one hint at a time or open the full working.

Exam aid · normal table values

Φ(z) is the cumulative area to the left. Use the forward table for an area and the inverse table for a cutoff. 给 z 查面积，给左侧面积反查 z。

Given a z-score
zΦ(z)
−1.000.1587
0.800.7881
1.000.8413
1.200.8849
1.500.9332
2.000.9772
Given a left-tail area
pΦ⁻¹(p)
0.025−1.9600
0.05−1.6449
0.10−1.2816
0.90+1.2816
0.95+1.6449
0.975+1.9600
Core problem 1 / 3½ min

An order is late if delivery takes more than 46 minutes. What percentage of orders are late?

先写事件，再估计答案会大于还是小于 50%。

%

One decimal place is enough.

Mark the late orders
Mark the late ordersThe curve marks 46 minutes to the right of the mean. Select the region corresponding to late orders.Density0.020.040.06μ = 4046 min25303540455055min

Mark the event before calculating.

Get a hint
  1. Locate the cutoff. Put 46 on the time axis. Do late orders lie to its left or right? 46 比平均慢。符合延误条件的时间在哪边？
  2. Write the event. Write P(T > 46). Convert 46 minutes to z using (46 − 40)/5. 这条边界离均值有几个标准差？
  3. Read the other tail. Φ(1.20) ≈ 0.8849 gives the area to the left. Use the remaining area, then convert it to a percentage. 表里的 0.8849 包含准时送达的订单。
Full solution
Event
P(T > 46)The cutoff is above 40, so the answer should be below 50%.
Standardize
z = (46 − 40)/5 = 1.2046 minutes is 1.20 standard deviations above the mean.
Find the area
P(Z > 1.20) = 1 − Φ(1.20) ≈ 1 − 0.8849 = 0.1151The table value is cumulative from the left.
Report
About 11.51% of orders are late.Express the probability as a percentage of deliveries.
接住学生的回答

若学生说 88.49%，请他把 46 右边的部分用手指圈出来，再判断那块会不会超过一半。若回答 1.2，追问这个数的单位以及题目所要的单位。

Core problem 2 / 3½ min

What is the longest delivery time that still belongs to the fastest 10% of orders?

题目给了一个比例。你最后要交的答案用什么单位？

min

One decimal place is enough.

Locate the time threshold
Locate the time thresholdA neutral time distribution with mean 40 minutes. Locate a cutoff for the fastest orders.Density0.020.040.06μ = 4025303540455055min

Place the unknown cutoff on the time axis.

Get a hint
  1. Name the unknown. Call the cutoff c minutes. Would a fast delivery take more or less time than 40 minutes? 越快，用时越少。先定分界线在哪一边。
  2. Shade the fastest 10%. The shortest times occupy the left tail. Write P(T ≤ c) = 0.10. 如果学生选了 0.90，让他重新按用时从小到大排序。
  3. Convert back to time. The 10th standard normal percentile is z ≈ −1.2816. Substitute it into c = 40 + 5z. z 没有单位，换回分钟才是题目要的答案。
Full solution
Event
P(T ≤ c) = 0.10Faster means a smaller time, so c is below the mean.
Find z
z = Φ⁻¹(0.10) ≈ −1.2816Use the 10th percentile of the standard normal.
Rescale
c = 40 + 5(−1.2816) ≈ 33.6 minThe cutoff has the same units as the original time.
Report
The fastest 10% of deliveries take about 33.6 minutes or less.The cutoff is below 40 minutes, as the sketch predicted.
接住学生的回答

若学生直接说前 10% 要查 0.90，请他先回答“最快的是用时最长还是最短的订单？”若他停在 −1.28，请他把题目末尾的单位读出来，再换回分钟。

Exit questionFinal minute

What delivery time should the company promise if it wants 90% of orders to arrive by that time?

同一模型。先判断承诺时间在 40 分钟的哪边，再解释理由。

Check the reasoning

P(T ≤ c) = 0.90   ⇒   z ≈ +1.2816   ⇒   c = 40 + 5z ≈ 46.4 min (to 1 decimal place)

Under this model, about 90% arrive within 46.4 minutes. A deadline covering 90% must be above the mean. If the company must promise whole minutes and cover at least 90%, round up to 47 minutes.

Five more minutes? Choose one transfer problem.

15 分钟试讲可在上面的出口题收尾。若有 5 分钟余量，按学生薄弱点从下列两题中选一题。

Optional A Compare relative performance

Maya scored 82 on Exam A (mean 70, SD 8). Leo scored 88 on Exam B (mean 80, SD 10). Both groups are approximately normal. Who performed better relative to their own class?

先选一个人，再用各自班级的数据说明理由。

Who would you choose?
Get a hint
  1. Choose a common scale. The two exams have different centers and spreads. How far is each mark above its own mean? 先各自减去本班均分。
  2. Use each exam’s SD. For Maya, divide 82 − 70 by 8. For Leo, divide 88 − 80 by 10. 分母也要对应各自班级。
  3. Compare the positions. A higher z-score corresponds to a higher relative position in these normal score models. 现在比较两个没有单位的数。
Full solution
Maya
zₐ = (82 − 70)/8 = 1.50Her mark is 1.50 SDs above her class mean.
Leo
zᵦ = (88 − 80)/10 = 0.80His mark is 0.80 SDs above his class mean.
Report
Maya performed better relative to her own class.Her standardized position is higher, although Leo’s raw mark is higher.
接住学生的回答

若学生选 Leo，问：“两场考试同样难、分数也同样分散吗？把每个人放回自己的班里看看。”

Optional B Find a central range

Bottle fill volumes are approximately normal with a mean of 500 mL and an SD of 5 mL. Which two volumes enclose the middle 90%?

答案需要两个边界。先说剩余的 10% 去了哪里。

mL
mL

One decimal place is enough.

Get a hint
  1. Split the remainder. 10% is outside the central interval. Divide that equally between the two tails. 左右对称，两尾各拿一半。
  2. Locate two percentiles. The lower cutoff has 0.05 to its left. How much lies to the left of the upper cutoff? 上界左侧包括左尾和中间的 90%。
  3. Return to mL. The z cutoffs are about −1.6449 and +1.6449. Apply 500 + 5z to each. 两个结果应当关于 500 对称。
Full solution
Tails
(1 − 0.90)/2 = 0.05 in each tail.The endpoint percentiles are 5% and 95%.
Find z
Φ⁻¹(0.05) ≈ −1.6449 and Φ⁻¹(0.95) ≈ +1.6449.The z cutoffs are equally far from zero.
Rescale
500 ± 5(1.6449) ≈ 491.8 and 508.2 mL.The endpoints are equally far from 500 mL.
Report
The middle 90% of bottle fills lie between about 491.8 and 508.2 mL.Each outside tail contains about 5% of bottles.
接住学生的回答

若学生用 ±1.28，先让他说清楚“中间 90% 之外还剩多少，两边各有多少”。若只报一个数，问最后应交一条线还是两条线。

Field notes / 选讲

Change the question

任选一题 · 5 min

同一组配送数据，换一个词，式子就会变。先用英文写出事件，再展开核对。

TRANSFER　/　01

Between two times

What proportion of deliveries take between 35 and 47.5 minutes?

这道题要一块面积，为什么需要查两次表？

核对解法

P(35 < T < 47.5) = Φ(1.5) − Φ(−1) ≈ 0.9332 − 0.1587 = 0.7745.

About 77.5% of deliveries fall in this window.

若只写 93.32%，你的面积还包含所有早于 35 分钟的订单。

TRANSFER　/　02

One order or four?

Four independent deliveries follow the same normal model. What is the chance their average exceeds 45 minutes?

同样的 45 分钟，这次横轴上的一个点代表什么？

核对解法

For four independent normal times, the mean is normal with mean 40 and standard error 5/√4 = 2.5 min. P(T̄ > 45) = 1 − Φ(2) ≈ 0.0228.

Only about 2.3% of four-order averages exceed 45 minutes.

用 5 作分母会得到 15.9%：那是单份订单的概率。平均数的波动较小。

TRANSFER　/　03

Which guarantee holds?

A manager promises that at least 95% of orders arrive within 48 minutes. Does the model support the claim?

“至少 95%”是验收标准。你用概率回答，还是用时间回答？

核对解法

P(T ≤ 48) = Φ((48 − 40)/5) = Φ(1.6) ≈ 0.9452 < 0.95.

No. The model predicts about 94.5% on time. The 95th percentile is about 48.22 minutes.

先用未取整的概率比较。若承诺时间只能取整分钟，要向上取到 49 分钟才能达标。