Step 1: State the Central Limit Theorem For a random sample of size n from a population with mean μ and standard deviation σ:

As n increases, the sampling distribution of x̄ (sample mean) approaches a normal distribution with:

• Mean: μₓ̄ = μ
• Standard deviation: σₓ̄ = σ/√n

This happens REGARDLESS of the shape of the population distribution!

Step 2: Conditions that must be met

1. RANDOMNESS:

   • Sample must be randomly selected
   • Each member has equal chance of selection
   • Ensures sample is representative

2. INDEPENDENCE:

   • Observations must be independent
   • 10% condition: n ≤ 0.10N (sample ≤ 10% of population)
   • If sampling without replacement from small population, need correction

3. SAMPLE SIZE:

   • Larger n → more normal
   • Rule of thumb: n ≥ 30 usually sufficient
   • If population is normal: works for ANY n
   • If population is skewed: need larger n (maybe 40+)
   • If population has extreme outliers: need even larger n

Step 3: Why randomness matters Non-random samples:

• Convenience sample → biased
• Voluntary response → biased
• Cannot apply CLT to biased samples

Random selection ensures:

• Sample represents population
• Results can be generalized

Step 4: Why independence matters Independence violated when:

• Sampling without replacement from small population (>10%)
• Clustered sampling (family members, same class)
• Time series data (measurements over time)

Effects of dependence:

• Standard formulas don't apply
• Need special methods

Step 5: Why sample size matters Small n:

• x̄ distribution looks like population distribution
• If population is skewed, x̄ is skewed
• If population is bimodal, x̄ is bimodal

Large n:

• CLT "kicks in"
• x̄ distribution becomes normal
• Averaging smooths out population shape

Step 6: Examples of CLT conditions check VALID: n = 50 from random digit table ✓ Random ✓ Independent (infinite population) ✓ n = 50 ≥ 30

INVALID: n = 100 from class of 200 students without replacement ✓ Could be random ✗ Not independent (100 > 0.10 × 200 = 20) ✓ n = 100 ≥ 30 Conclusion: 10% condition violated

VALID: n = 20 from normal population ✓ Random (assumed) ✓ Independent (assumed) ✓ Population is normal (works for any n)

Answer: CENTRAL LIMIT THEOREM: The sampling distribution of x̄ approaches Normal(μ, σ/√n) as n increases, regardless of population shape.

CONDITIONS:

1. Random sample from population
2. Independent observations (10% condition: n ≤ 0.10N)
3. Large enough sample (n ≥ 30, or any n if population is normal)

All three conditions must be met to apply CLT and use normal probability calculations.

Step 1: Check CLT conditions Population: Right-skewed (not normal) Sample size: n = 64

Is n ≥ 30? Yes, 64 ≥ 30 ✓ By CLT: x̄ is approximately normally distributed

Step 2: Find parameters of sampling distribution Mean: μₓ̄ = μ = 25

Standard deviation (standard error): σₓ̄ = σ/√n = 8/√64 = 8/8 = 1

Step 3: Describe the sampling distribution x̄ ~ Normal(μ = 25, σ = 1) approximately

Key points:

• Shape: approximately normal (due to CLT)
• Center: μₓ̄ = 25 (same as population)
• Spread: σₓ̄ = 1 (much less than population σ = 8)

Even though population is right-skewed, x̄ is approximately normal!

Step 4: Calculate P(24 < x̄ < 26) Standardize to z-scores:

z₁ = (24 - 25)/1 = -1/1 = -1 z₂ = (26 - 25)/1 = 1/1 = 1

P(24 < x̄ < 26) = P(-1 < Z < 1)

Step 5: Use empirical rule or table From empirical rule: About 68% of normal distribution is within 1 SD of mean

P(-1 < Z < 1) ≈ 0.68

More precisely from table: P(Z < 1) = 0.8413 P(Z < -1) = 0.1587 P(-1 < Z < 1) = 0.8413 - 0.1587 = 0.6826

Step 6: Interpret the result About 68.3% of all samples of size 64 will have sample means between 24 and 26.

This range is μ ± 1σₓ̄ = 25 ± 1 Very common for x̄ to fall in this range!

Step 7: Compare to individual values For individual value X from population:

• Can't use normal (population is skewed)
• Can't easily find P(24 < X < 26)
• Would need actual population distribution

For sample mean x̄:

• CAN use normal (CLT applies)
• Easy to calculate probabilities
• CLT is powerful!

Step 8: Effect of sample size If we used n = 16 instead: σₓ̄ = 8/√16 = 2 P(24 < x̄ < 26) = P(-0.5 < Z < 0.5) ≈ 0.38 Less likely to be close to μ with smaller sample

If we used n = 256 instead: σₓ̄ = 8/√256 = 0.5 P(24 < x̄ < 26) = P(-2 < Z < 2) ≈ 0.95 More likely to be close to μ with larger sample

Answer: Sampling distribution: x̄ ~ Normal(μ = 25, σ = 1) approximately

Despite the right-skewed population, the large sample size (n = 64) allows CLT to apply, making x̄ approximately normal.

P(24 < x̄ < 26) ≈ 0.683 or 68.3%

About 68% of samples will have means within 1 unit of the population mean.

Step 1: Set up the problem Population (luggage weights):

• Heavily right-skewed
• μ = 18 lbs
• σ = 6 lbs

Sample:

• n = 100 passengers
• Find: P(x̄ > 19)

Step 2: Check CLT conditions Random: Assume passengers are representative sample ✓ Independent: 100 passengers << all passengers (10% rule) ✓ Sample size: n = 100 ≥ 30, even with heavy skew ✓

CLT applies!

Step 3: Find sampling distribution parameters μₓ̄ = μ = 18 lbs

σₓ̄ = σ/√n = 6/√100 = 6/10 = 0.6 lbs

x̄ ~ Normal(18, 0.6) approximately

Step 4: Calculate P(x̄ > 19) Standardize: z = (19 - 18)/0.6 = 1/0.6 = 5/3 ≈ 1.67

P(x̄ > 19) = P(Z > 1.67)

Step 5: Look up probability From standard normal table: P(Z < 1.67) ≈ 0.9525

Therefore: P(Z > 1.67) = 1 - 0.9525 = 0.0475

Step 6: Interpret Only about 4.75% chance (less than 5%) that average luggage weight exceeds 19 lbs.

Even though population is heavily skewed:

• Individual bags vary a lot (σ = 6)
• Average of 100 bags is much more stable (σₓ̄ = 0.6)
• Large sample "averages out" the skewness

Step 7: Why this matters for airlines Airline might set weight limit based on average:

• If average > 19 lbs, might have safety concerns
• Probability < 5% means this is rare
• Can plan accordingly

Individual approach would be harder:

• Individual weights range widely
• Many over 19 lbs (population is skewed right)
• But average is more predictable!

Step 8: Compare to individual luggage For one random bag: P(X > 19) = ?

Can't easily calculate - population is skewed, not normal. But probably much higher than 4.75%! Maybe 30-40% of bags exceed 19 lbs.

But average of 100 bags rarely exceeds 19 lbs.

Step 9: Check reasonableness 19 lbs is 1 lb above mean In terms of SE: 19 = 18 + 1(0.6) = 18 + 1.67σₓ̄ About 1.67 SE above mean Should be fairly unlikely ✓

Answer: P(x̄ > 19) ≈ 0.048 or 4.8%

There's only about a 4.8% chance that the average luggage weight for 100 passengers exceeds 19 lbs. The Central Limit Theorem allows us to treat the sample mean as approximately normal despite the heavily skewed population, and the large sample size (n = 100) makes the sample mean much less variable than individual values.

Step 1: Translate the question "More than 25 hours away from true mean" means: Either x̄ < 475 or x̄ > 525

Find: P(|x̄ - μ| > 25) = P(x̄ < 475) + P(x̄ > 525)

Step 2: Set up sampling distribution μ = 500 hours σ = 100 hours n = 50

Check CLT: n = 50 ≥ 30 ✓

Step 3: Find sampling distribution parameters μₓ̄ = μ = 500

σₓ̄ = σ/√n = 100/√50 = 100/7.07 ≈ 14.14 hours

x̄ ~ Normal(500, 14.14) approximately

Step 4: Use symmetry By symmetry of normal distribution: P(x̄ < 475) = P(x̄ > 525)

So: P(x̄ < 475 or x̄ > 525) = 2 × P(x̄ > 525)

Step 5: Calculate P(x̄ > 525) Standardize: z = (525 - 500)/14.14 = 25/14.14 ≈ 1.77

P(x̄ > 525) = P(Z > 1.77)

Step 6: Look up probability From table: P(Z < 1.77) ≈ 0.9616

Therefore: P(Z > 1.77) = 1 - 0.9616 = 0.0384

Step 7: Find total probability P(more than 25 away) = 2 × 0.0384 = 0.0768 ≈ 0.077

Step 8: Interpret About 7.7% of samples will have means more than 25 hours from the true mean.

This means:

• 92.3% of samples within 25 hours of μ = 500
• Quality control can use this to set thresholds
• If x̄ is more than 25 away, might indicate problem

Step 9: Express in terms of standard errors 25 hours = 1.77 × 14.14 ≈ 1.77 SE

So we're asking: P(more than 1.77 SE from mean)

From 68-95-99.7 rule:

• Within 1 SE: ≈68%
• Within 2 SE: ≈95%
• Within 1.77 SE: ≈92.3%

Outside 1.77 SE: ≈7.7% ✓

Step 10: Decision rule for quality control Factory might use rule: "Flag sample if x̄ < 475 or x̄ > 525"

False alarm rate: 7.7% About 1 in 13 good samples will be flagged Reasonable tradeoff for quality control

Answer: P(|x̄ - μ| > 25) ≈ 0.077 or 7.7%

There's about a 7.7% probability that a sample mean will be more than 25 hours away from the true mean of 500 hours. This represents being more than 1.77 standard errors from the mean. Quality control can use this threshold to identify unusual samples that might indicate production problems.

Step 1: Understand what we're finding For n adults, total weight = n × x̄ Want: P(total weight > 2000) Equivalently: P(n × x̄ > 2000) Or: P(x̄ > 2000/n)

Step 2: Set up for n = 10 Maximum average weight: 2000/10 = 200 lbs per person

Find: P(x̄ > 200) when n = 10

Step 3: Sampling distribution for n = 10 Population is normal, so x̄ is normal for ANY n (don't need CLT!)

μₓ̄ = μ = 180 lbs

σₓ̄ = σ/√n = 30/√10 = 30/3.16 ≈ 9.49 lbs

x̄ ~ Normal(180, 9.49)

Step 4: Calculate P(x̄ > 200) for n = 10 Standardize: z = (200 - 180)/9.49 = 20/9.49 ≈ 2.11

P(x̄ > 200) = P(Z > 2.11)

From table: P(Z < 2.11) ≈ 0.9826

P(Z > 2.11) = 1 - 0.9826 = 0.0174

Step 5: Interpret n = 10 result About 1.74% chance that 10 adults exceed 2000 lbs Fairly safe - less than 2% risk

Step 6: Set up for n = 12 Maximum average weight: 2000/12 ≈ 166.67 lbs per person

Find: P(x̄ > 166.67) when n = 12

Step 7: Sampling distribution for n = 12 μₓ̄ = 180 lbs

σₓ̄ = σ/√n = 30/√12 = 30/3.46 ≈ 8.66 lbs

x̄ ~ Normal(180, 8.66)

Step 8: Calculate P(x̄ > 166.67) for n = 12 Standardize: z = (166.67 - 180)/8.66 = -13.33/8.66 ≈ -1.54

P(x̄ > 166.67) = P(Z > -1.54)

From table: P(Z < -1.54) ≈ 0.0618

P(Z > -1.54) = 1 - 0.0618 = 0.9382

Step 9: Interpret n = 12 result About 93.8% chance that 12 adults exceed 2000 lbs! Very risky - almost certain to exceed limit

Step 10: Why such a big difference? n = 10: Need average > 200 lbs (20 lbs above μ) = 2.11 SE above mean Unlikely!

n = 12: Need average > 166.67 lbs (13.33 lbs below μ) = 1.54 SE below mean
Very likely!

Step 11: Find maximum safe capacity At what n does P(exceed) = 0.05 (5% risk)?

Need: P(x̄ > 2000/n) = 0.05 P(Z > z) = 0.05 means z = 1.645

(2000/n - 180)/(30/√n) = -1.645

Solving: 2000/n = 180 - 1.645(30/√n) 2000 = 180n - 49.35√n

Approximately n ≈ 10.6

So maximum safe capacity is about 10 adults for 5% risk level.

Answer: n = 10: P(exceed 2000 lbs) ≈ 0.017 or 1.7% n = 12: P(exceed 2000 lbs) ≈ 0.938 or 93.8%

With 10 adults, there's only about 1.7% chance of exceeding the limit (relatively safe). With 12 adults, there's about 93.8% chance of exceeding the limit (very dangerous!). The maximum average weight needed drops from 200 lbs (n=10) to 166.67 lbs (n=12), and 166.67 is well below the population mean of 180, making it very likely to exceed.