Statistics — Distribution of the Sample Mean and the Central Limit Theorem

Statistics — Distribution of the Sample Mean and the Central Limit Theorem. Practice questions to deepen understanding of the distribution of the sample mean and the Central Limit Theorem. Online statistics practice with full solutions and step-by-step explanations.

Sampling distribution and CLT practice — 50 questions: Central Limit Theorem, standard error, confidence intervals, normal approximation, Bootstrap. Advanced statistics.

Part A — Theory.

100 questions

Question 1

2.00 pts

🎯 :

(Central Limit Theorem)?

x̄ (-n large), No

Other

always equal

Explanation:

💡 :

! 🎯

📚 :

(CLT):

X₁, X₂, ..., Xₙ μ σ²,

n → ∞:

Question 2

10.00 pts

🎯 The most important theorem in statistics:

What does the Central Limit Theorem (CLT) state?

The sample mean x̄ is approximately normally distributed (when n is large), even if the population is not normal.

Every population is normally distributed.

The sample is identical to the population.

The mean always equals the median.

Explanation:

💡 Detailed explanation:

The Central Limit Theorem! 🎯

📚 Statement of the theorem:

Central Limit Theorem (CLT):

If X₁, X₂, ..., Xₙ is a random sample from a population with mean μ and variance σ²,

then as n → ∞:

x̄ ~ N(μ, σ²/n)

Or in standardized form:

Z = (x̄ - μ)/(σ/√n) ~ N(0, 1)

⭐ The magic:

This works regardless of the original distribution!

• The population can be:
- Uniform
- Exponential
- Binomial
- Asymmetric
- With outliers
- Any distribution!

• But x̄ always (when n is large) will be distributed normally!

💡 Conditions:

1️⃣ n large enough
• Rule of thumb: n ≥ 30
• If the population is symmetric: n=15 is enough
• If the population is normal: any n is enough!

2️⃣ The sample is random

3️⃣ μ and σ² exist and are finite

🎲 Intuitive example:

Rolling a die:
• Distribution: uniform (1,2,3,4,5,6)
• Not normal at all!

But:
• Average of 30 rolls
• Average of another 30 rolls
• Etc...

→ The averages will be distributed normally!

Question 3

10.00 pts

📊 Parameters:

If X has any distribution with E(X) = μ and Var(X) = σ², what is the distribution of x̄?

E(x̄) = μ, Var(x̄) = σ²/n, SD(x̄) = σ/√n

E(x̄) = μ/n, Var(x̄) = σ²

E(x̄) = 0, Var(x̄) = 1

E(x̄) = nμ, Var(x̄) = nσ²

Explanation:

💡 Detailed explanation:

Parameters of x̄! 📊

📐 The complete formulas:

Parameters of the distribution of x̄:

1️⃣ Expectation (mean):
E(x̄) = μ

2️⃣ Variance:
Var(x̄) = σ²/n

3️⃣ Standard deviation (standard error):
SD(x̄) = SE = σ/√n

🔍 Proofs:

Proof 1: E(x̄) = μ

x̄ = (X₁ + X₂ + ... + Xₙ) / n

E(x̄) = E[(X₁ + X₂ + ... + Xₙ) / n]

= [E(X₁) + E(X₂) + ... + E(Xₙ)] / n

= [μ + μ + ... + μ] / n

= nμ / n = μ ✓

Proof 2: Var(x̄) = σ²/n

x̄ = (X₁ + X₂ + ... + Xₙ) / n

Var(x̄) = Var[(X₁ + ... + Xₙ) / n]

= (1/n²) × Var(X₁ + ... + Xₙ)

Since the variables are independent:

= (1/n²) × [Var(X₁) + ... + Var(Xₙ)]

= (1/n²) × [σ² + σ² + ... + σ²]

= (1/n²) × nσ²

= σ²/n ✓

⭐ Insights:

1️⃣ x̄ is an unbiased estimator:
E(x̄) = μ → "on average we hit the target"

2️⃣ The spread shrinks with n:
Var(x̄) = σ²/n → as n grows, the variance shrinks

3️⃣ Rate of convergence:
SE = σ/√n → to halve SE you need 4 times the sample!

Question 4

2.00 pts

📊 :

X ~ E(X)=μ, Var(X)=σ², x̄?

E(x̄)=μ, Var(x̄)=σ²/n, SD(x̄)=σ/√n

E(x̄)=μ/n, Var(x̄)=σ²

E(x̄)=0, Var(x̄)=1

E(x̄)=nμ, Var(x̄)=nσ²

Explanation:

💡 :

x̄! 📊

📐 No:

x̄:

1️⃣ ():
E(x̄) = μ

Question 5

2.00 pts

📏 :

?

n ≥ 30

n ≥ 10

n ≥ 100

n ≥ 5

Explanation:

💡 :

-CLT! 📏

📊 :

Question 6

10.00 pts

📏 Rule of thumb:

What is the commonly used minimum sample size for x̄ to be approximately normally distributed?

n ≥ 30

n ≥ 10

n ≥ 100

n ≥ 5

Explanation:

💡 Detailed explanation:

Sample size for CLT! 📏

📊 Rules of thumb:

Population state	Minimum n	Explanation
Normal	any n!	x̄ is always normal
Symmetric (no outliers)	n ≥ 15	Fast convergence
General (any distribution)	n ≥ 30	The general rule!
Highly asymmetric (outliers)	n ≥ 50-100	More data needed

⭐ The golden rule:

n ≥ 30

This is the safest value
that works in most cases

💡 Why 30?

• It is not a physical law, but a practical rule
• Found empirically to work well
• In scientific research: n=30 is considered a "significance threshold"

🎯 Examples:

📊 Case 1: Uniform distribution
• Already at n=10 a good approximation
• At n=30 excellent

📊 Case 2: Exponential distribution
• Need n≥40-50
• Highly asymmetric

📊 Case 3: Normal
• Even n=5 is enough
• x̄ is always normal!

⚠️ Important:

If n < 30 and we don't know the population is normal
→ we should use the t distribution (not Z)

Question 7

10.00 pts

🎲 Independence:

Does the CLT depend on the shape of the population distribution?

No, it works for any distribution (when n is large enough).

Yes, only for the normal distribution.

Yes, only for symmetric distributions.

Depends on the sample.

Explanation:

💡 Detailed explanation:

Universality of CLT! 🎲

🌟 The magic of CLT:

The amazing principle:

The Central Limit Theorem works for any distribution!

• Uniform ✓
• Exponential ✓
• Binomial ✓
• Poisson ✓
• Beta ✓
• Gamma ✓
• Asymmetric ✓
• With outliers ✓
• Mixed ✓
• Any distribution that has finite μ and σ²!

📊 Visual illustration:

The population:
Can be any shape
⬜ ▅ ⬛ ▂ ▄ (asymmetric)

After sampling:
x̄ is distributed normally
🔔 (symmetric bell!)

🎯 Concrete examples:

Example 1: Uniform distribution

🎲 Rolling a die (1-6):
• Distribution: flat rectangle ▭
• Not normal at all!

But average of 30 rolls:
→ Distributed normally! 🔔

Example 2: Exponential distribution

⏱️ Waiting time (highly asymmetric):
• Most times short /
• Few times long

But average of 50 customer times:
→ Distributed normally! 🔔

⭐ Why is this so important?

1️⃣ No need to know the original distribution!

2️⃣ We can always use the Z table

3️⃣ Allows statistical inference almost always

That's why CLT is called
"the most important theorem in statistics"

Question 8

2.00 pts

🎲 :

?

No, (-n large )

Yes,

Other

Explanation:

💡 :

-CLT! 🎲

🌟 CLT:

:

!

• ✓
• be

Question 9

2.00 pts

✅ :

?

, μ -σ² , n large

Other

large

Other

Explanation:

💡 :

CLT! ✅

📋 No:

:

1️⃣
X₁, X₂, ..., Xₙ
( )

2️⃣
E(X) = μ
Var(X) = σ²

Question 10

10.00 pts

✅ Conditions:

What are the conditions required for the CLT?

Random sample, finite μ and σ², n sufficiently large.

Only that the population is normal.

Only that the sample is large.

No special conditions.

Explanation:

💡 Detailed explanation:

CLT conditions! ✅

📋 The full conditions:

Conditions of the Central Limit Theorem:

1️⃣ Random sample
X₁, X₂, ..., Xₙ is a random sample
(independent and identically distributed)

2️⃣ Existence of mean and variance
E(X) = μ exists and is finite
Var(X) = σ² exists and is finite

3️⃣ Sufficient sample size
n large enough
(usually n ≥ 30)

🔍 Detailed explanation of each condition:

1️⃣ Random sample (i.i.d):

i.i.d = independent and identically distributed

• independent:
The choice of one does not affect the other

• identically distributed:
All from the same population

Example: sampling with replacement ✓
Counter-example: dependent sampling ✗

2️⃣ Finite μ and σ²:

Why needed?
• If μ = ∞ → no defined mean
• If σ² = ∞ → infinite variance

When is this a problem?
In rare cases:
• Cauchy distribution (no mean!)
• Distributions with "heavy tails"

But in most practical distributions:
μ and σ² exist and are finite ✓

3️⃣ n large enough:

We saw in question 3:
• Safe: n ≥ 30
• If symmetric: n ≥ 15
• If normal: any n

The more "unusual" the population
→ the larger n needed

⭐ Summary:

The conditions are quite easily met in practical cases!

The theorem is very robust and powerful

Question 11

10.00 pts

➕ Important property:

If X₁ ~ N(μ₁, σ₁²) and X₂ ~ N(μ₂, σ₂²) are independent,

what is the distribution of X₁ + X₂?

N(μ₁ + μ₂, σ₁² + σ₂²)

N(μ₁ × μ₂, σ₁² × σ₂²)

N(μ₁ + μ₂, σ₁ + σ₂)

Not normal.

Explanation:

💡 Detailed explanation:

Normal additivity! ➕

📐 The additivity property:

Independent normal variables:

If X₁ ~ N(μ₁, σ₁²) and X₂ ~ N(μ₂, σ₂²)

and X₁, X₂ are independent

then:

X₁ + X₂ ~ N(μ₁+μ₂, σ₁²+σ₂²)

📊 Generalization:

If X₁, X₂, ..., Xₙ are independent

where Xᵢ ~ N(μᵢ, σᵢ²)

then:

ΣXᵢ ~ N(Σμᵢ, Σσᵢ²)

🎯 Special case — sum of n identical:

If X₁, ..., Xₙ ~ N(μ, σ²) identical and independent

then:

S = X₁ + X₂ + ... + Xₙ

S ~ N(nμ, nσ²)

💡 Connection to CLT:

From sum to mean:

x̄ = S/n = (X₁ + ... + Xₙ)/n

If S ~ N(nμ, nσ²)

then x̄ ~ N(nμ/n, nσ²/n²)

= N(μ, σ²/n) ✓

⭐ Important to remember:

Means add: μ₁+μ₂

Variances add: σ₁²+σ₂²

(not standard deviations!)

⚠️ Common mistake:

❌ σ₁+σ₂ (wrong!)

✓ σ₁²+σ₂² (correct!)

Question 12

2.00 pts

➕ :

X₁ ~ N(μ₁, σ₁²) -X₂ ~ N(μ₂, σ₂²) ,

X₁ + X₂?

N(μ₁+μ₂, σ₁²+σ₂²)

N(μ₁×μ₂, σ₁²×σ₂²)

N(μ₁+μ₂, σ₁+σ₂)

Explanation:

💡 :

! ➕

📐 :

:

X₁ ~ N(μ₁, σ₁²) -X₂ ~ N(μ₂, σ₂²)

-X₁, X₂

:

Question 13

2.00 pts

🔢 :

x̄ ?

x̄ = (X₁+X₂+...+Xₙ)/n = (1/n)ΣXᵢ

x̄ = n(X₁+X₂+...+Xₙ)

x̄ = (X₁+X₂+...+Xₙ)²

x̄ = √(ΣXᵢ)

Explanation:

💡 :

x̄ ! 🔢

📐 :

= partial sum n:

x̄ = (X₁+X₂+...+Xₙ)/n

:

Question 14

10.00 pts

🔢 Connection:

How can x̄ be written using a sum?

x̄ = (X₁ + X₂ + … + Xₙ)/n = (1/n)ΣXᵢ

x̄ = n(X₁ + X₂ + … + Xₙ)

x̄ = (X₁ + X₂ + … + Xₙ)²

x̄ = √(ΣXᵢ)

Explanation:

💡 Detailed explanation:

x̄ as a sum! 🔢

📐 The connection between sum and mean:

Mean = sum divided by n:

x̄ = (X₁+X₂+...+Xₙ)/n

Or in short notation:

x̄ = (1/n)ΣXᵢ

🔄 From sum to mean:

Let: S = ΣXᵢ = X₁+X₂+...+Xₙ

Then: x̄ = S/n

If each Xᵢ ~ N(μ, σ²):

1️⃣ The sum:
S ~ N(nμ, nσ²)

2️⃣ The mean:
x̄ = S/n ~ N(μ, σ²/n)

📊 Effect of division on the distribution:

If Y ~ N(μ, σ²)

then Y/c ~ N(μ/c, σ²/c²)

So:

S ~ N(nμ, nσ²)

→ S/n ~ N(nμ/n, nσ²/n²)

→ x̄ ~ N(μ, σ²/n) ✓

⭐ Why is this useful?

1️⃣ Conceptual understanding:
x̄ is a linear transformation of a sum

2️⃣ Probability computation:
Sometimes easier to work with the sum

3️⃣ Proofs:
Helps prove properties of x̄

💡 Example:

Given: X ~ N(10, 4), n=9

Sum:
S = ΣXᵢ ~ N(9×10, 9×4)
= N(90, 36)
SD(S) = 6

Mean:
x̄ = S/9 ~ N(90/9, 36/81)
= N(10, 4/9)
SD(x̄) = 2/3

Question 15

10.00 pts

📊 Extension:

Does the CLT apply only to the sample mean?

No, it also applies to sums and other statistics.

Yes, only to the mean.

Only to the variance.

Only to the median.

Explanation:

💡 Detailed explanation:

General CLT! 📊

🌟 CLT applications:

CLT applies to:

1️⃣ Sample mean:
x̄ ~ N(μ, σ²/n)

2️⃣ Sum:
ΣXᵢ ~ N(nμ, nσ²)

3️⃣ Sample proportion:
p̂ ~ N(p, p(1-p)/n)

📐 Application 1: Sum

S = X₁ + X₂ + ... + Xₙ

If Xᵢ are independent with E(Xᵢ)=μ, Var(Xᵢ)=σ²

and n is large:

S is approximately distributed N(nμ, nσ²)

Example: total monthly income

📐 Application 2: Proportion

p̂ = number of successes / n

In a binomial sample with parameter p

and n is large:

p̂ is approximately distributed N(p, p(1-p)/n)

Example: proportion of supporters in a poll

📐 Application 3: Difference

x̄₁ - x̄₂

Difference between two sample means

Also distributed normally!

Example: difference between two groups

⭐ The general principle:

CLT applies to any linear function of the sample

That is:
a₁X₁ + a₂X₂ + ... + aₙXₙ

also distributed normally when n is large!

💡 Important:

CLT does not apply to:
• median
• min/max
• range

(these are not linear functions)

Question 16

2.00 pts

📊 Extension:

Does the Central Limit Theorem apply only to the sample mean?

No, it also applies to the sum and to other statistics

Yes, only to the mean

Only to the variance

Only to the median

Explanation:

💡 Detailed explanation:

General CLT! 📊

🌟 Applications of CLT:

CLT applies to:

1️⃣ Sample mean:
x̄ ~ N(μ, σ²/n)

2️⃣ Sum:
ΣXᵢ ~ N(nμ, nσ²)

3️⃣ Sample proportion:
p̂ ~ N(p, p(1-p)/n)

📐 Application 1: Sum

S = X₁ + X₂ + ... + Xₙ

If Xᵢ are independent with E(Xᵢ)=μ, Var(Xᵢ)=σ²

and n is large:

S is approximately N(nμ, nσ²)

Example: total monthly income

📐 Application 2: Proportion

p̂ = number of successes / n

In a binomial sample with parameter p

and n is large:

p̂ is approximately N(p, p(1-p)/n)

Example: proportion of supporters in a survey

📐 Application 3: Difference

x̄₁ - x̄₂

Difference between two sample means

It is also normally distributed!

Example: difference between two groups

⭐ The general principle:

CLT applies to any linear function of the sample

That is:
a₁X₁ + a₂X₂ + ... + aₙXₙ

Is also normally distributed when n is large!

💡 Important:

CLT does not apply to:
• Median
• min/max
• Range

(These are not linear functions)

Question 17

2.00 pts

🔧 Correction:

What is a continuity correction?

A correction when approximating a discrete distribution (e.g. binomial) by a normal distribution.

A correction for variance.

A correction for the mean.

A correction for small n.

Explanation:

💡 Detailed explanation:

Continuity correction! 🔧

🎯 The problem:

• Discrete distribution: separate values (0, 1, 2, 3, ...)
• Continuous distribution: any value possible

When approximating discrete by continuous → there is a small inaccuracy

Continuity correction:

When computing P(X = k) in a binomial
using a normal:

P(k-0.5 < Y < k+0.5)

where Y ~ N(μ, σ²)

📊 Correction rules:

Discrete	→ Continuous
P(X = k)	P(k-0.5 < Y < k+0.5)
P(X ≤ k)	P(Y < k+0.5)
P(X ≥ k)	P(Y > k-0.5)
P(X < k)	P(Y < k-0.5)
P(X > k)	P(Y > k+0.5)

💡 Example:

X ~ Binomial(100, 0.5)

We want: P(X = 55)

Without correction:
μ = 50, σ = 5
Z = (55-50)/5 = 1
P(Z = 1) = 0 (continuous normal!)

With correction:
P(54.5 < Y < 55.5)
= P((54.5-50)/5 < Z < (55.5-50)/5)
= P(0.9 < Z < 1.1)
≈ 0.027

⭐ When to use?

✓ When approximating binomial by normal
✓ When approximating Poisson by normal
✓ Any discrete→continuous approximation

✗ When n is very large (>100) - the effect is negligible

💡 Remember:

The correction adds/subtracts ±0.5

Direction: per the rule in the table above

Question 18

10.00 pts

🔧 Correction:

What is the continuity correction (Continuity Correction)?

A correction when approximating a discrete distribution (e.g. binomial) by a normal distribution.

A correction for variance.

A correction for the mean.

A correction for small n.

Explanation:

💡 Detailed explanation:

Continuity correction! 🔧

Continuity correction:

When computing P(X = k) in a binomial
using a normal:

P(k-0.5 < Y < k+0.5)

where Y ~ N(μ, σ²)

📊 Correction rules:

Discrete	→ Continuous
P(X = k)	P(k-0.5 < Y < k+0.5)
P(X ≤ k)	P(Y < k+0.5)
P(X ≥ k)	P(Y > k-0.5)
P(X < k)	P(Y < k-0.5)
P(X > k)	P(Y > k+0.5)

The correction adds/subtracts ±0.5

Direction: per the rule in the table above

Question 19

10.00 pts

📚 Theoretical summary:

What is the central insight of the Central Limit Theorem?

Means behave "nicely" (normally) even when the data do not.

Everything is normal.

The sample is identical to the population.

There is no need for statistics.

Explanation:

💡 הסבר מפורט:

התובנה המרכזית! 📚

🌟 סיכום משפט הגבול המרכזי:

התובנה המדהימה:

סכומים וממוצעים של משתנים מקריים מתפלגים נורמלית

גם אם המשתנים עצמם לא נורמליים!

זו תופעת "הסדר מתוך כאוס"

🎯 המסר המרכזי:

הנתונים הבודדים:
יכולים להיות מבולגנים, לא סימטריים, משונים
🔀 📊 📈 ⚡

הממוצעים:
מסודרים, צפויים, נורמליים!
🔔 📊 🎯

💡 למה זה חשוב?

1️⃣ אוניברסליות:
אין צורך לדעת את ההתפלגות המקורית!

2️⃣ פשטות:
תמיד אפשר להשתמש בטבלת Z

3️⃣ הסקה:
מאפשר רווחי סמך ובדיקות השערות

4️⃣ חיזוי:
יודעים מה לצפות מממוצעים

⭐ הנוסחאות המרכזיות:

1. ממוצע המדגם:
x̄ ~ N(μ, σ²/n)

2. סכום:
ΣXᵢ ~ N(nμ, nσ²)

3. תקנון:
Z = (x̄-μ)/(σ/√n) ~ N(0,1)

🎓 המסר לקח הביתה:

ממוצעים הם יציבים ויותר צפויים מנתונים בודדים

זו הסיבה שאנחנו משתמשים בממוצעים בסטטיסטיקה!

וזו הסיבה ש-CLT נקרא:
"המשפט החשוב ביותר בסטטיסטיקה"

Question 20

2.00 pts

📚 :

?

Sample means behave "nicely" (normally distributed) even when the underlying data do not.

Everything is always normal.

The sample is identical to the population.

Statistics is not needed.

Explanation:

💡 :

! 📚

🌟 :

:

No !

Question 21
2.00 pts

🧮 Exercise 1:

Given: σ=20, n=64

Compute SE (the standard error).

2.5

20

5

10

Explanation:

💡 Detailed explanation:
Computing SE! 🧮
📐 Solution:

Formula:
SE = σ / √n

Given:
• σ = 20
• n = 64

Step 1: √n
√64 = 8

Step 2: Compute SE
SE = 20 / 8 = 2.5

💡 Interpretation:
Sample means of size 64 will be distributed with a standard deviation of 2.5 around μ

Question 22
10.00 pts

🧮 Exercise 1:

Given: σ = 20, n = 64.

Calculate SE (standard error).

2.5

20

5

10

Explanation:

💡 הסבר מפורט:
חישוב SE! 🧮
📐 פתרון:

נוסחה:
SE = σ / √n

נתונים:
• σ = 20
• n = 64

שלב 1: √n
√64 = 8

שלב 2: חישוב SE
SE = 20 / 8 = 2.5

💡 פירוש:
ממוצעי מדגמים בגודל 64 יתפלגו עם סטיית תקן של 2.5 סביב μ

Question 23
10.00 pts

🧮 Exercise 2:

Given: μ = 100, σ = 15, n = 25, x̄ = 106.

Calculate Z.

2

0.4

6

1

Explanation:

💡 הסבר מפורט:
חישוב Z! 🧮
📐 פתרון:

נוסחה:
Z = (x̄ - μ) / (σ/√n)

נתונים:
• μ = 100
• σ = 15
• n = 25
• x̄ = 106

שלב 1: SE
SE = σ/√n = 15/√25 = 15/5 = 3

שלב 2: Z
Z = (106 - 100) / 3
Z = 6 / 3
Z = 2

💡 פירוש:
x̄=106 נמצא 2 סטיות תקן מעל μ

Question 24
2.00 pts

🧮 Exercise 2:

Given: μ=100, σ=15, n=25, x̄=106

Compute Z.

2

0.4

6

1

Explanation:

💡 Detailed explanation:
Computing Z! 🧮
📐 Solution:

Formula:
Z = (x̄ - μ) / (σ/√n)

Given:
• μ = 100
• σ = 15
• n = 25
• x̄ = 106

Step 1: SE
SE = σ/√n = 15/√25 = 15/5 = 3

Step 2: Z
Z = (106 - 100) / 3
Z = 6 / 3
Z = 2

💡 Interpretation:
x̄=106 lies 2 standard deviations above μ

Question 25
10.00 pts

🧮 Exercise 3:

Given: μ = 50, σ = 12, n = 36.

What is P(48 < x̄ < 52)?
(Hint: SE = 2, range is ±1 SE.)

≈0.68

≈0.95

≈0.50

≈0.99

Explanation:

💡 הסבר מפורט:
הסתברות בטווח! 🧮
📐 פתרון:

נתונים:
• μ = 50
• σ = 12
• n = 36
• רוצים: P(48 < x̄ < 52)

שלב 1: SE
SE = σ/√n = 12/√36 = 12/6 = 2

שלב 2: תקנון לZ

Z₁ = (48 - 50) / 2 = -2/2 = -1

Z₂ = (52 - 50) / 2 = 2/2 = +1

שלב 3: הסתברות

P(48 < x̄ < 52) = P(-1 < Z < 1)

כלל 68-95-99.7:

P(-1 < Z < 1) ≈ 0.68

(68% של הנורמלית בטווח ±1σ)

💡 פירוש:

יש 68% סיכוי שממוצע המדגם יהיה בין 48 ל-52

זכור:
• ±1σ → 68%
• ±2σ → 95%
• ±3σ → 99.7%

Question 26
2.00 pts

🧮 3:

given: μ=50, σ=12, n=36

P(48 < x̄ < 52)?
(: SE=2, ±1 )

≈0.68

≈0.95

≈0.50

≈0.99

Explanation:

💡 :
! 🧮
📐 :

:
• μ = 50
• σ = 12
• n = 36
• : P(48 < x̄ < 52)

1: SE
SE = σ/√n = 12/√36 = 12/6 = 2

2: Z

Z₁ = (48 - 50) / 2 = -2/2 = -1<

Question 27
2.00 pts

🧮 4:

given: μ=200, σ=40, n=100

P(x̄ > 208)?
(: SE=4, Z=2)

≈0.025

≈0.05

≈0.16

≈0.50

Explanation:

💡 :
-! 🧮
📐 :

:
• μ = 200
• σ = 40
• n = 100
• : P(x̄ > 208)

1: SE
SE = σ/√n = 40/√100 = 40/10 = 4

2: Z
Z = (208 - 200) / 4 = 8/4 = 2

Question 28
10.00 pts

🧮 Exercise 4:

Given: μ = 200, σ = 40, n = 100.

What is P(x̄ > 208)?
(Hint: SE = 4, Z = 2.)

≈0.025

≈0.05

≈0.16

≈0.50

Explanation:

💡 הסבר מפורט:
הסתברות חד-צדדית! 🧮
📐 פתרון:

נתונים:
• μ = 200
• σ = 40
• n = 100
• רוצים: P(x̄ > 208)

שלב 1: SE
SE = σ/√n = 40/√100 = 40/10 = 4

שלב 2: Z
Z = (208 - 200) / 4 = 8/4 = 2

שלב 3: הסתברות

P(x̄ > 208) = P(Z > 2)

מטבלת Z:

P(Z > 2) ≈ 0.025

(2.5% בזנב הימני)

💡 הגיון:

Z=2 זה די רחוק מהממוצע

רק 2.5% מהממוצעים יהיו כל כך גבוהים

זכור:
• P(Z > 1) ≈ 0.16
• P(Z > 1.96) ≈ 0.025
• P(Z > 2) ≈ 0.025
• P(Z > 3) ≈ 0.0013

Question 29
10.00 pts

🧮 Exercise 5:

Given: μ = 100, SE = 5.

Find c such that P(x̄ < c) = 0.95.
(Hint: Z₀.₉₅ = 1.645.)

108.225

105

110

100

Explanation:

💡 הסבר מפורט:
מציאת ערך קריטי! 🧮
📐 פתרון:

נתונים:
• μ = 100
• SE = 5
• P(x̄ < c) = 0.95

שלב 1: מציאת Z

רוצים P(Z < z) = 0.95

מטבלת Z: z = 1.645

שלב 2: המרה חזרה ל-x̄

נוסחה:

Z = (x̄ - μ) / SE

→ x̄ = μ + Z × SE

c = 100 + 1.645 × 5

c = 100 + 8.225

c = 108.225

💡 פירוש:

95% מממוצעי המדגמים יהיו מתחת ל-108.225

רק 5% יהיו מעל לזה

Question 30
2.00 pts

🧮 5:

given: μ=100, SE=5

find c -P(x̄ < c) = 0.95
(: Z₀.₉₅ = 1.645)

108.225

105

110

100

Explanation:

💡 :
! 🧮
📐 :

:
• μ = 100
• SE = 5
• P(x̄ < c) = 0.95

1: Z

P(Z < z) = 0.95

Z: z = 1.645

2: -x̄

Question 31
2.00 pts

🧮 Exercise 6:

Given: σ=20

If n₁=25 then SE₁=4
If n₂=100 then SE₂=?

2

4

8

1

Explanation:

💡 Detailed explanation:
Comparing SE! 🧮
📐 Solution:

Given:
• σ = 20
• n₁ = 25 → SE₁ = 4
• n₂ = 100 → SE₂ = ?

Formula:
SE = σ / √n

Method 1: Direct calculation

SE₂ = 20 / √100
SE₂ = 20 / 10
SE₂ = 2

Method 2: Ratio

n₂/n₁ = 100/25 = 4

→ √(n₂/n₁) = √4 = 2

→ SE₂/SE₁ = 1/2

→ SE₂ = SE₁/2 = 4/2 = 2

💡 Insight:

Increasing n by a factor of 4
→ SE shrinks by a factor of 2

Relation: SE ∝ 1/√n

Question 32
10.00 pts

🧮 Exercise 6:

Given: σ = 20.

If n₁ = 25 then SE₁ = 4.
If n₂ = 100 then SE₂ = ?

2

4

8

1

Explanation:

💡 הסבר מפורט:
השוואת SE! 🧮
📐 פתרון:

נתונים:
• σ = 20
• n₁ = 25 → SE₁ = 4
• n₂ = 100 → SE₂ = ?

נוסחה:
SE = σ / √n

דרך 1: חישוב ישיר

SE₂ = 20 / √100
SE₂ = 20 / 10
SE₂ = 2

דרך 2: יחס

n₂/n₁ = 100/25 = 4

→ √(n₂/n₁) = √4 = 2

→ SE₂/SE₁ = 1/2

→ SE₂ = SE₁/2 = 4/2 = 2

💡 תובנה:

הגדלת n פי 4
→ SE קטן פי 2

הקשר: SE ∝ 1/√n

Question 33
10.00 pts

🧮 Exercise 7:

Given: X ~ N(10, 4), n = 9.

What are E(ΣXᵢ) and Var(ΣXᵢ)?

E = 90, Var = 36

E = 10, Var = 4

E = 90, Var = 4

E = 10, Var = 36

Explanation:

💡 הסבר מפורט:
סכום משתנים! 🧮
📐 פתרון:

נתונים:
• X ~ N(10, 4)
• כלומר: μ = 10, σ² = 4
• n = 9
• S = ΣXᵢ = X₁+...+X₉

נוסחאות לסכום:

E(ΣXᵢ) = n × μ

Var(ΣXᵢ) = n × σ²

חישוב תוחלת:

E(S) = E(ΣXᵢ)
= n × μ
= 9 × 10
= 90

חישוב שונות:

Var(S) = Var(ΣXᵢ)
= n × σ²
= 9 × 4
= 36

💡 השוואה:

משתנה בודד:
E(X) = 10
Var(X) = 4
סכום:
E(S) = 90 = 9×10
Var(S) = 36 = 9×4
ממוצע:
E(x̄) = 10 = 90/9
Var(x̄) = 4/9 = 36/81

Question 34
2.00 pts

🧮 Exercise 7:

Given: X ~ N(10, 4), n=9

What are E(ΣXᵢ) and Var(ΣXᵢ)?

E=90, Var=36

E=10, Var=4

E=90, Var=4

E=10, Var=36

Explanation:

💡 Detailed explanation:
Sum of variables! 🧮
📐 Solution:

Given:
• X ~ N(10, 4)
• That is: μ = 10, σ² = 4
• n = 9
• S = ΣXᵢ = X₁+...+X₉

Formulas for the sum:

E(ΣXᵢ) = n × μ

Var(ΣXᵢ) = n × σ²

Computing the expectation:

E(S) = E(ΣXᵢ)
= n × μ
= 9 × 10
= 90

Computing the variance:

Var(S) = Var(ΣXᵢ)
= n × σ²
= 9 × 4
= 36

💡 Comparison:

Single variable:
E(X) = 10
Var(X) = 4
Sum:
E(S) = 90 = 9×10
Var(S) = 36 = 9×4
Mean:
E(x̄) = 10 = 90/9
Var(x̄) = 4/9 = 36/81

Question 35
2.00 pts

🧮 8:

: E(X)=5, Var(X)=3
n=50

x̄ ?

Yes, because n=50≥30

No, because No

n≥100

No known

Explanation:

💡 :
CLT No No ! 🧮
📊 :

:
• : (No !)
• E(X) = 5
• Var(X) = 3
• n = 50

Question 36
10.00 pts

🧮 Exercise 8:

Uniform population: E(X) = 5, Var(X) = 3.
n = 50.

Can the normal distribution be used for x̄?

Yes, because n = 50 ≥ 30.

No, because the population is not normal.

Only if n ≥ 100.

Unknown.

Explanation:

💡 הסבר מפורט:
CLT לאוכלוסייה לא נורמלית! 🧮
📊 ניתוח:

נתונים:
• אוכלוסייה: אחידה (לא נורמלית!)
• E(X) = 5
• Var(X) = 3
• n = 50

משפט הגבול המרכזי:

אם n ≥ 30:

x̄ מתפלג נורמלית בקירוב

גם אם האוכלוסייה לא נורמלית!

בדיקת תנאים:

✓ n = 50 ≥ 30 ✓
✓ יש E(X) וVar(X) סופיים ✓
✓ דגימה אקראית ✓

→ כל התנאים מתקיימים!

ההתפלגות של x̄:

E(x̄) = 5
Var(x̄) = 3/50 = 0.06
SD(x̄) = √0.06 ≈ 0.245

x̄ ~ N(5, 0.06)

בקירוב טוב!

💡 הקסם:

האוכלוסייה אחידה (מלבן)

אבל x̄ מתפלג נורמלית (פעמון)!

זה הכוח של CLT

Question 37
10.00 pts

🧮 Exercise 9:

Non-normal population, n = 12.

Can the CLT be applied?

No, n is too small (n ≥ 30 is required).

Yes, always.

Yes, if the variance is small.

Depends on the mean.

Explanation:

💡 הסבר מפורט:
n קטן מדי! 🧮
⚠️ בעיה:

נתונים:
• אוכלוסייה: לא נורמלית
• n = 12

n=12 < 30

קטן מדי ל-CLT!

לא ניתן להניח ש-x̄ נורמלי

מה עושים?

אם יודעים שהאוכלוסייה נורמלית:
→ x̄ נורמלי לכל n!
→ אפשר להמשיך
אם האוכלוסייה לא נורמלית:
→ n=12 קטן מדי
→ צריך שיטות אחרות
→ או להגדיל את המדגם

אלטרנטיבות:

1️⃣ התפלגות t:
אם האוכלוסייה נורמלית אבל σ לא ידוע

2️⃣ שיטות לא פרמטריות:
לא מניחות התפלגות מסוימת

3️⃣ Bootstrap:
שיטה מודרנית מבוססת סימולציה

💡 הכלל:

אוכלוסייה לא נורמלית:

• n < 30 → לא CLT
• n ≥ 30 → כן CLT

אוכלוסייה נורמלית:

• כל n → כן

Question 38
2.00 pts

🧮 9:

No , n=12

-CLT?

No, n small ( n≥30)

Yes, always

Yes,

Other

Explanation:

💡 :
n small ! 🧮
⚠️ :

:
• : No
• n = 12

n=12 < 30
<

Question 39
2.00 pts

🧮 10:

given: x̄=50, s=10, n=100

95% -μ.

[48.04, 51.96]

[48, 52]

[40, 60]

[49, 51]

Explanation:

💡 :
CLT! 🧮
📐 :

95%:

x̄ ± 1.96 × SE

:
• x̄ = 50
• s = 10 ()
• n

Question 40
10.00 pts

🧮 Exercise 10:

Given: x̄ = 50, s = 10, n = 100.

Build a 95% confidence interval for μ.

[48.04, 51.96]

[48, 52]

[40, 60]

[49, 51]

Explanation:

NULL

Question 41
10.00 pts

📊 Binomial–normal approximation:

When can a binomial distribution be approximated by a normal distribution?

When np ≥ 5 and n(1 − p) ≥ 5.

Only when n ≥ 30.

Only when p = 0.5.

Always.

Explanation:

NULL

Question 42
2.00 pts

📊 -:

?

np≥5 n(1-p)≥5

n≥30

p=0.5

always

Explanation:

💡 :
! 📊
📚 binomial distribution:

X ~ Binomial(n, p)

:

np ≥ 5

Question 43
2.00 pts

🧮 Binomial exercise:

X ~ Binomial(100, 0.4)

What are the normal approximation parameters?

μ=40, σ²=24, σ≈4.9

μ=100, σ²=40

μ=0.4, σ²=0.24

μ=40, σ²=40

Explanation:

💡 Detailed explanation:
Approximation parameters! 🧮
📐 Solution:

Given:
X ~ Binomial(n=100, p=0.4)

Binomial formulas:

E(X) = np

Var(X) = np(1-p)

SD(X) = √[np(1-p)]

Step 1: Check conditions

np = 100 × 0.4 = 40 ≥ 5 ✓
n(1-p) = 100 × 0.6 = 60 ≥ 5 ✓

→ Approximation is valid!

Step 2: Mean

μ = np = 100 × 0.4 = 40

Step 3: Variance

σ² = np(1-p)
= 100 × 0.4 × 0.6
= 40 × 0.6
= 24

Step 4: Standard deviation

σ = √24 ≈ 4.9

The normal approximation:

X ~ N(40, 24)

or: X ~ N(40, 4.9²)

Question 44
10.00 pts

🧮 Binomial exercise:

X ~ Binomial(100, 0.4).

What are μ, σ², and σ?

μ = 40, σ² = 24, σ ≈ 4.9

μ = 100, σ² = 40

μ = 0.4, σ² = 0.24

μ = 40, σ² = 40

Explanation:

NULL

Question 45
10.00 pts

🧮 Exercise:

X ~ Binomial(100, 0.5).

Use the normal approximation to calculate P(X = 55).

P(54.5 < Y < 55.5) where Y ~ N(50, 25).

P(Y = 55) where Y ~ N(50, 25).

P(Y < 55) where Y ~ N(50, 25).

P(Y > 55) where Y ~ N(50, 25).

Explanation:

NULL

Question 46
2.00 pts

🧮 :

X ~ Binomial(100, 0.5)

calculate P(X = 55) .

P(54.5 < Y < 55.5) Y ~ N(50, 25)

P(Y = 55) Y ~ N(50, 25)

P(Y < 55) Y ~ N(50, 25)

P(Y > 55) Y ~ N(50, 25)

Explanation:

💡 :
! 🧮
📐 :

given:
X ~ Binomial(100, 0.5)
: P(X = 55)

1:

μ = np = 100 × 0.5 = 50

σ² = np(1-p) = 100 × 0.5 × 0.5 = 25

σ = √25 = 5

→ Y ~ N(50, 25)

Question 47
2.00 pts

📊 :

p̂ = /n, p̂ -n large?

p̂ ~ N(p, p(1-p)/n)

p̂ ~ N(p, p²/n)

p̂ ~ N(0, 1)

No

Explanation:

💡 :
! 📊
📚 :

:

p̂ = X/n

X ~ Binomial(n, p)

Question 48
10.00 pts

📊 Proportion:

If p̂ = number of successes/n, how is p̂ distributed when n is large?

p̂ ~ N(p, p(1 − p)/n) approximately.

p̂ ~ N(p, p²/n)

p̂ ~ N(0, 1)

Not normally distributed.

Explanation:

NULL

Question 49
10.00 pts

🧮 Exercise:

Survey: n = 900, p̂ = 0.55.

Build a 95% confidence interval for p.

[0.518, 0.582]

[0.5, 0.6]

[0.45, 0.65]

[0.53, 0.57]

Explanation:

NULL

Question 50
2.00 pts

🧮 Exercise:

In a survey: n=900, p̂=0.55

Build a 95% confidence interval for p.

[0.518, 0.582]

[0.5, 0.6]

[0.45, 0.65]

[0.53, 0.57]

Explanation:

💡 Detailed explanation:
Confidence interval for a proportion! 🧮
📐 Solution:

Formula:

p̂ ± 1.96 × √[p̂(1-p̂)/n]

Given:
• n = 900
• p̂ = 0.55

Step 1: SE

SE = √[p̂(1-p̂)/n]

= √[0.55 × 0.45 / 900]

= √[0.2475 / 900]

= √0.000275

≈ 0.0166

Step 2: Margin of error

ME = 1.96 × 0.0166

≈ 0.0325

Step 3: Interval

Lower bound: 0.55 - 0.0325 = 0.5175

Upper bound: 0.55 + 0.0325 = 0.5825

95% confidence interval:

[0.518, 0.582]

or: [51.8%, 58.2%]

💡 Interpretation:

We are 95% confident that the true proportion in the population is between 51.8% and 58.2%

Question 51
2.00 pts

🧮 Sample design:

We want a confidence interval for a proportion with width ±3% (95% confidence)

What n is required? (Assume p≈0.5)

n ≈ 1068

n = 100

n = 500

n = 30

Explanation:

💡 Detailed explanation:
Survey sample size! 🧮
📐 Solution:

Sample size formula:

n = (Z² × p(1-p)) / E²

E = desired Margin of Error

Given:
• E = 0.03 (3%)
• Z = 1.96 (95%)
• p = 0.5 (conservative assumption)

Why p=0.5?

p(1-p) is maximized when p=0.5

0.5 × 0.5 = 0.25 ← maximum!

This gives the largest required sample (conservative)

Calculation:

n = (1.96² × 0.5 × 0.5) / 0.03²

= (3.8416 × 0.25) / 0.0009

= 0.9604 / 0.0009

= 1067.1

→ n ≈ 1068

💡 Insight:

For a margin of ±3%, a large sample is required!

This is why surveys typically use n≈1000-1200

Question 52
10.00 pts

🧮 Sample planning:

We want a confidence interval for a proportion with width ±3% (at 95%).

What sample size is needed? (Assume p ≈ 0.5.)

n ≈ 1068

n = 100

n = 500

n = 30

Explanation:

NULL

Question 53
10.00 pts

📊 Two samples:

If x̄₁ ~ N(μ₁, σ₁²/n₁) and x̄₂ ~ N(μ₂, σ₂²/n₂) are independent,

what is the distribution of x̄₁ − x̄₂?

N(μ₁ − μ₂, σ₁²/n₁ + σ₂²/n₂)

N(μ₁ − μ₂, σ₁² − σ₂²)

N(0, 1)

Not normal.

Explanation:

NULL

Question 54
2.00 pts

📊 :

x̄₁ ~ N(μ₁, σ₁²/n₁) -x̄₂ ~ N(μ₂, σ₂²/n₂) ,

x̄₁ - x̄₂?

N(μ₁-μ₂, σ₁²/n₁ + σ₂²/n₂)

N(μ₁-μ₂, σ₁²-σ₂²)

N(0, 1)

No

Explanation:

💡 :
! 📊
📚 :

:

:
x̄₁ ~ N(μ₁, σ₁²/n₁)
x̄₂ ~ N(μ₂, σ₂²/n₂)

-x̄₁, x̄₂

:

Question 55
2.00 pts

🧮 :

H₀: μ=100 vs H₁: μ≠100
given: x̄=106, s=15, n=25

calculate Z.

2

0.4

6

1

Explanation:

💡 :
! 🧮
📐 :

Z:

Z = (x̄ - μ₀) / (s/√n)

μ₀ H₀

:
• H₀: μ = 100

Question 56
10.00 pts

🧮 Exercise:

H₀: μ = 100 vs H₁: μ ≠ 100.
Given: x̄ = 106, s = 15, n = 25.

Calculate Z.

2

0.4

6

1

Explanation:

NULL

Question 57
10.00 pts

🧮 Exercise:

One-tailed test: H₀: μ ≤ 100 vs H₁: μ > 100.
Z = 2.5.

What is the p-value?

≈0.006

≈0.012

≈0.05

≈0.025

Explanation:

NULL

Question 58
2.00 pts

🧮 :

-: H₀: μ≤100 vs H₁: μ>100
Z=2.5

-p-value ?

≈0.006

≈0.012

≈0.05

≈0.025

Explanation:

💡 :
p-value! 🧮
📐 :

:
• - ()
• H₁: μ > 100
• Z = 2.5

p-value:

p-value = P(Z > 2.5)

Z large
-H₀

Question 59
2.00 pts

📊 :

(Power) n ?

Increases (easier to detect a real difference).

Decreases.

Does not change.

Depends on alpha.

Explanation:

💡 :
-n! 📊
📚 :

:

Power = 1 - β

= H₀
H₀

= " "

Question 60
10.00 pts

📊 Power:

What happens to the power of the test when n increases?

Increases (easier to detect a real difference).

Decreases.

Does not change.

Depends on α.

Explanation:

NULL

Question 61
10.00 pts

🔧 Finite population:

When sampling without replacement from a finite population of size N,
and n/N > 0.05,

how is SE adjusted?

SE = (σ/√n) × √[(N − n)/(N − 1)]

SE = σ/√n (unchanged)

SE = (σ/√n) × (N/n)

SE = σ/N

Explanation:

NULL

Question 62
2.00 pts

🔧 :

No N,
large (n/N > 0.05),

SE?

SE = (σ/√n) × √[(N-n)/(N-1)]

SE = σ/√n (No )

SE = (σ/√n) × (N/n)

SE = σ/N

Explanation:

💡 :
! 🔧
📚 FPC:

Finite Population Correction:

SE = (σ/√n) × √[(N-n)/(N-1)

Question 63
2.00 pts

📊 :

CLT -X₁, X₂, ..., Xₙ
No ?

Yes, ( -)

No,

n>1000

Other

Explanation:

💡 :
CLT ! 📊
📚 CLT:

-:

X₁, X₂, ..., Xₙ

( No !)

E(Xᵢ)=μᵢ, Var(Xᵢ)=σᵢ²

,

:

Question 64
10.00 pts

📊 Generalisation:

Does the CLT apply even when X₁, X₂, …, Xₙ are not identically distributed?

Yes, under certain conditions (Lindeberg–Lévy theorem).

No, they must be identically distributed.

Only if n > 1000.

Only if all are normal.

Explanation:

NULL

Question 65
10.00 pts

📈 Rate:

At what rate does the normal approximation improve as n increases?

At rate 1/√n (Berry–Esseen).

At rate 1/n.

At rate 1/n².

At a constant rate.

Explanation:

NULL

Question 66
2.00 pts

📈 :

-n ?

1/√n (Berry-Esseen)

1/n

1/n²

Other

Explanation:

💡 :
! 📈
📚 Berry-Esseen:

:

:

O(1/√n)

: -1/√n

Question 67
2.00 pts

🔗 :

CLT -Xᵢ No ?

Yes, (mixing conditions)

No, - No

n>10000

Other

Explanation:

💡 :
CLT ! 🔗
📚 :

:

CLT

: Xᵢ -Xⱼ ,

Question 68
10.00 pts

🔗 Dependence:

Can the CLT apply even when Xᵢ are not independent?

Yes, in cases of weak dependence (mixing conditions).

No, full independence is required.

Only if n > 10000.

Depends on the distribution.

Explanation:

NULL

Question 69
10.00 pts

🎯 Vectors:

Does the CLT apply to random vectors as well?

Yes, there is a multivariate CLT — the vector converges to a multivariate normal distribution.

No, only for scalar variables.

Only in dimension 2.

Depends on the variance.

Explanation:

NULL

Question 70
2.00 pts

🎯 vector:

CLT vector ?

Yes, CLT - - vector -

No,

2

Other

Explanation:

💡 :
CLT -! 🎯
📚 CLT vector:

-:

X₁, X₂, ..., Xₙ
vector

μ

Question 71
2.00 pts

🔧 :

x̄ ~ N(μ, σ²/n) -g ,

g(x̄) ?

N(g(μ), [g′(μ)]²σ²/n) — Delta method

x̄

No

-g

Explanation:

💡 :
! 🔧
📚 Delta Method:

:

x̄ ~ N(μ, σ²/n)

-g ()

:

g

Question 72
10.00 pts

🔧 Transformations:

If x̄ ~ N(μ, σ²/n) and g is a differentiable function,

what is the approximate distribution of g(x̄)?

N(g(μ), [g′(μ)]²σ²/n) — Delta method.

Exactly the same as x̄.

Not normal.

Depends on g.

Explanation:

NULL

Question 73
10.00 pts

💻 Bootstrap:

What is the connection between Bootstrap and the CLT?

Bootstrap resamples to approximate the sampling distribution guaranteed by the CLT.

There is no connection.

Bootstrap replaces the CLT.

Bootstrap works only without the CLT.

Explanation:

NULL

Question 74
2.00 pts

💻 Bootstrap:

Bootstrap -CLT?

Bootstrap " CLT

Other

Bootstrap CLT

Bootstrap No CLT

Explanation:

💡 :
Bootstrap -CLT! 💻
📚 Bootstrap:

Bootstrap:

1️⃣ n

2️⃣ n <

Question 75
2.00 pts

📈 :

CLT ?

because

Other

calculate R²

Other

Explanation:

💡 :
CLT ! 📈
📚 :

:

Y = β₀ + β₁X + ε

ε ~ N(0, σ²)

Question 76
10.00 pts

📈 Regression:

What is the role of the CLT in linear regression?

It ensures that the estimated regression coefficients are approximately normally distributed.

It determines the best-fit line.

It calculates R².

There is no connection.

Explanation:

NULL

Question 77
10.00 pts

📊 Practical significance:

With a very large n, why is it important to check practical significance and not only statistical significance?

Because a large n makes everything statistically significant, even trivial differences.

It is not important; statistical significance is sufficient.

Only for academic research.

Depends on α.

Explanation:

NULL

Question 78
2.00 pts

📊 :

n large , No ?

because n large ,

No ,

Other

-α

Explanation:

💡 :
vs ! 📊
⚠️ n large:

:

n → ∞

SE = σ/√n → 0

→ , ,

Question 79
2.00 pts

🌟 :

calculate " "?

Enables inference for any distribution, opens up all of modern statistics.

The easiest to remember.

Only for exams.

Nothing special.

Explanation:

💡 :
CLT! 🌟
🎯 CLT ?

10 :

1️⃣
!

2️⃣
always -

3️⃣
,

Question 80
10.00 pts

🌟 Summary:

Why is the Central Limit Theorem considered "the most important theorem" in statistics?

It enables inference for any distribution, and opens up all of modern statistics.

It is the easiest to remember.

Only for exams.

There is nothing special about it.

Explanation:

NULL

Question 81
10.00 pts

🧮 Comprehensive exercise:

A factory produces bolts. Diameter: μ = 10 mm, σ = 0.5 mm.
A sample of n = 100 bolts is drawn.

What is P(9.9 < x̄ < 10.1)?

≈0.95

≈0.68

≈0.99

≈0.50

Explanation:

NULL

Question 82
2.00 pts

🧮 :

. : μ=10mm, σ=0.5mm
n=100

9.9 -10.1?

≈0.95

≈0.68

≈0.99

≈0.50

Explanation:

💡 :
! 🧮
📐 No:

1:
• μ = 10
• σ = 0.5
• n = 100
• : P(9.9 < x̄ < 10.1)

2: x̄

CLT:
x̄ ~ N(10, 0.5²/100)
= N(10, 0.0025)

SE = 0.5/10 = 0.05

3:

Question 83
2.00 pts

🧮 :

: μ=165, σ=6
36

find c -P(x̄ > c) = 0.05

166.645

165

170

167

Explanation:

💡 :
! 🧮
📐 :

1: SE
SE = 6/√36 = 6/6 = 1

2: Z
P(x̄ > c) = 0.05
→ P(Z > z) = 0.05
→ z = 1.645

3:
Z = (c - μ) / SE
1.645 = (c - 165) / 1

Question 84
10.00 pts

🧮 Comprehensive exercise:

Height of women: μ = 165, σ = 6.
A sample of 36 women is drawn.

Find c such that P(x̄ > c) = 0.05.

166.645

165

170

167

Explanation:

NULL

Question 85
10.00 pts

🧮 Comprehensive exercise:

X ~ Binomial(200, 0.3).

Use the normal approximation to calculate P(X ≥ 70).
(Use continuity correction.)

≈0.025

≈0.05

≈0.16

≈0.50

Explanation:

NULL

Question 86
2.00 pts

🧮 :

X ~ Binomial(200, 0.3)

P(X ≥ 70)
( )

≈0.025

≈0.05

≈0.16

≈0.50

Explanation:

💡 :
! 🧮
📐 :

1:
μ = np = 200×0.3 = 60
σ² = np(1-p) = 200×0.3×0.7 = 42
σ = √42 ≈ 6.48

2:
P(X ≥ 70)
= P(X > 69.5)

3:
Z

Question 87
2.00 pts

🧮 :

A: x̄₁=85, s₁=10, n₁=50
B: x̄₂=80, s₂=12, n₂=60

95% μ₁-μ₂

[1.12, 8.88]

[3, 7]

[0, 10]

[2, 8]

Explanation:

💡 :
! 🧮
📐 :

1:
x̄₁ - x̄₂ = 85 - 80 = 5

2: SE()
SE = √(s₁²/n₁ + s₂²/n₂)
= √(100/50 + 144/60)
= √(2 + 2.4)
= √4.4
≈ 2.1

3: ME
ME = 1.9

Question 88
10.00 pts

🧮 Comprehensive exercise:

Group A: x̄₁ = 85, s₁ = 10, n₁ = 50.
Group B: x̄₂ = 80, s₂ = 12, n₂ = 60.

Build a 95% confidence interval for μ₁ − μ₂.

[1.12, 8.88]

[3, 7]

[0, 10]

[2, 8]

Explanation:

NULL

Question 89
10.00 pts

✅ Test exercise:

Given: x̄ = 52, μ₀ = 50, SE = 2.

Is H₀: μ = 50 rejected at the 5% level?

Yes, because |Z| = 1 < 1.96.

No.

Depends on n.

More information is needed.

Explanation:

NULL

Question 90
2.00 pts

✅ :

given: x̄=52, μ₀=50, SE=2

H₀: μ=50 5%?

Yes, because |Z|=1 < 1.96

No

-n

Other

Explanation:

💡 :
! ✅
📐 :

Z = (52-50)/2 = 1

: 1.96

|1| < 1.96

→ No H₀

Question 91
2.00 pts

📚 Concept:

What is called the "Law of Large Numbers"?

x̄ converges to μ as n→∞

Same as CLT

Only for large numbers

No such law exists

Explanation:

💡 Detailed explanation:
Law of Large Numbers! 📚
Difference from CLT:

Law of Large Numbers:
x̄ → μ (convergence)

CLT:
describes the distribution of x̄

Question 92
10.00 pts

📚 Concept:

What is the "Law of Large Numbers"?

x̄ converges to μ as n → ∞.

Identical to the CLT.

Only for large numbers.

No such law exists.

Explanation:

NULL

Question 93
10.00 pts

🎯 Practical:

In which field is the CLT most critical?

Almost every statistical field: inference, regression, machine learning.

Only in physics.

Only in surveys.

Only in pure mathematics.

Explanation:

NULL

Question 94
2.00 pts

🎯 :

CLT because ?

Almost every statistical field: inference, regression, machine learning.

Only in physics.

Only in surveys.

Only in pure mathematics.

Explanation:

💡 :
CLT! 🎯
:

•
•
•
•
• AI/ML
•
•

!

Question 95
2.00 pts

🔍 :

"" ""?

Convergence = a single value; distribution = a full distribution.

No difference.

Only terminology.

Depends on context.

Explanation:

💡 :
vs ! 🔍
(LLN):
x̄ → μ ( )

(CLT):
x̄ ~ N(μ, σ²/n) ()

Question 96
10.00 pts

🔍 Understanding:

What is the difference between "converges" and "is distributed"?

Converges = a single value; distributed = a full distribution.

No difference.

Only terminology.

Depends on context.

Explanation:

NULL

Question 97
10.00 pts

⚠️ Common error:

What is wrong with the statement: "The CLT says that the data are normally distributed"?

Not the data, but the mean of the data is approximately normally distributed.

Nothing — it is correct.

A sample of n > 100 is needed.

Depends on the distribution.

Explanation:

NULL

Question 98
2.00 pts

⚠️ :

: "CLT "?

No , No

, Correct

n>100

Other

Explanation:

💡 :
! ⚠️
❌ :
" "

✓ Correct:
" x̄ "

!

Question 99
2.00 pts

🎓 :

147?

x̄ ~ N(μ, σ²/n) -n large, !

Other

n 30

Other

Explanation:

💡 :
! 🎓

x̄ ~ N(μ, σ²/n)

n → ∞

Question 100
10.00 pts

🎓 Final summary:

What is the key result to remember from this topic?

x̄ ~ N(μ, σ²/n) when n is large, for any distribution!

Everything is normal.

n must equal 30.

Only for the normal distribution.

Explanation:

NULL

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