Problem Of N=10000 Has Sigma=25 In Each Case

A Problem Of N10000 Has Sigma25 In Each Of The Following Cases

A Problem Of N10000 Has Sigma25 In Each Of The Following Cases

Analyze several statistical problems involving sampling distributions, normal distributions, and probabilities related to quality control, income distributions, distances, and timing. For each problem, select and apply the appropriate formulas to calculate means, standard deviations, probabilities, and interpret the shape of distributions.

Paper For Above instruction

Introduction

The following paper addresses a series of statistical problems involving large sample sizes, normal distributions, and probability calculations. Each problem requires understanding sampling distribution concepts, selecting appropriate formulas, and performing calculations to interpret real-world scenarios such as quality control, income analysis, distances, and timing. This comprehensive analysis demonstrates mastery of statistical methodologies in practical contexts.

Problem 1: Standard Deviation of Sample Mean for Large n

Given a population with size n=10,000 and population standard deviation sigma=25, the standard deviation of the sample mean depends on the sample size n. The formula used is:

  • Standard error (SE) = sigma / √n

For (a) n=2000:

SE = 25 / √2000 ≈ 25 / 44.721 ≈ 0.559

For (b), if n were a different value, say n, then the formula remains the same, substituting the specific n value.

Problem 2: Setting Machine Mean to Achieve 99% Package Content

A cheese packaging machine produces 10 oz packages with a standard deviation of 0.1 oz. To ensure 99% of packages contain at least 10 oz, find the mean (μ) to set on the machine.

Using the properties of the normal distribution, we determine the cutoff point:

  • Let X be the amount of cheese, X ~ N(μ, 0.1²)
  • We want P(X ≥ 10) = 0.99

Corresponding z-value for 99% one-sided confidence:

z = 2.33 (from standard normal tables)

Solving for μ:

10 = μ - z σ = μ - 2.33 0.1

μ = 10 + 2.33 * 0.1 = 10 + 0.233 = 10.233 oz

Problem 3: Income Distribution in America

Given the population mean income of hybrid vehicle owners as $58,000 with a standard deviation of $8,300, and a sample size n=50, determine the sampling distribution's mean, standard deviation, shape, and probability that the sample mean exceeds $60,000.

Calculations:

  • Mean of sampling distribution: μ_x̄ = μ = $58,000
  • Standard deviation (Standard error): σ_x̄ = σ / √n = 8300 / √50 ≈ 8300 / 7.07 ≈ $1174
  • Shape: Approximately normal (by Central Limit Theorem, since n=50 is adequate)

To find P( x̄ > 60,000 ):

Calculate the z-score:

z = (60,000 - 58,000) / 1174 ≈ 2000 / 1174 ≈ 1.703

Using standard normal tables, P(z > 1.703) ≈ 0.0446

Therefore, there is approximately a 4.46% probability that the sample mean income exceeds $60,000.

Problem 4: Quality Control for CDs

The machine produces 6% defective CDs. Every day, a sample of 100 CDs is inspected. If 8% or more are defective, the process is stopped.

Using the sample proportion p̂, which follows approximately a normal distribution for large n:

  • Mean of p̂: μ_p̂ = p = 0.06
  • Standard deviation: σ_p̂ = √(p(1-p)/n) = √(0.06 * 0.94 / 100) ≈ √(0.0564 / 100) ≈ 0.0238

Probability that process stops (p̂ ≥ 0.08):

Calculate z:

z = (0.08 - 0.06) / 0.0238 ≈ 0.02 / 0.0238 ≈ 0.84

Using z-tables, P(z ≥ 0.84) ≈ 1 - 0.7995 = 0.2005

Thus, approximately 20.05% chance that the process will be stopped on any given day when the true defective rate is 6%.

Problem 5: Distance from College to Home

The distance distribution has a mean of 15 miles and standard deviation of 8 miles, skewed right.

  • a) Probability that the sample mean distance of 50 students exceeds 18 miles:

See if normal approximation applies: Since n=50, CLT suggests using normal approximation.

Standard error: SE = 8 / √50 ≈ 8 / 7.07 ≈ 1.131

Z-score:

z = (18 - 15) / 1.131 ≈ 3 / 1.131 ≈ 2.65

P(X̄ > 18) = P(z > 2.65) ≈ 0.004

  • b) Probability that distances are between 14 and 17 miles:

Calculate z-scores:

z_14 = (14 - 15) / 1.131 ≈ -0.885

z_17 = (17 - 15) / 1.131 ≈ 1.77

P(14

  • c) Shape if sample size reduces to 10:

For n=10, the standard error would be SE = 8 / √10 ≈ 8 / 3.16 ≈ 2.53. The sampling distribution would still be approximately normal by CLT, but less reliable for normality with smaller n. The mean remains 15 miles; standard deviation of the sampling distribution (standard error) is 2.53 miles.

Problem 6: Timeliness of Freddy's Arrival

Let the population distribution of travel time be normal with mean 15 minutes, standard deviation 3 minutes.

a) To ensure Freddy arrives on time (say, within 15 min) with 99% confidence:

Find z for 99%:

z = 2.33

Calculate the time Freddy should leave:

Travel time: mean + z std deviation = 15 + 2.33 3 ≈ 15 + 6.99 ≈ 21.99 minutes

Since the mean travel time is 15 mins, and he needs to arrive within this duration to get bonus points, he must leave early enough to compensate for variability.

Time to leave = scheduled class start time - (mean travel time + z * standard deviation). Assuming class starts at 6 pm, Freddy should leave by approximately:

6:00 pm - 22 mins ≈ 5:38 pm.

b) If Freddy leaves at 5:40 pm, the probability he arrives on time (i.e., within 6 pm) is:

P(T ≤ 6:00 pm) where T is total travel time.

Calculate z:

z = (scheduled arrival time - departure time - mean) / std dev

Departure at 5:40 pm, so travel time varies, and arrival time is:

Departure time + travel time

Probability that travel time ≤ 20 minutes (since 5:40 pm to 6:00 pm is 20 mins):

z = (20 - 15) / 3 ≈ 1.67

P(z ≤ 1.67) ≈ 0.9525

Therefore, Freddy has approximately a 95.25% chance of arriving on time if he leaves at 5:40 pm.

Conclusion

All problems demonstrate the practical applications of statistical concepts such as using the standard error to determine the variability of sample means, applying the normal distribution for probability calculations, and understanding how sample size influences the shape and reliability of sampling distributions. Proper formula selection and interpretation support decision-making in quality control, business, and daily planning.

References

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