Math Problem Statement

About 5% of employed adults in the United States held multiple jobs. A random sample of 75 employed adults is chosen. Use the TI-84 Plus calculator as needed. (a) Is it appropriate to use the normal approximation to find the probability that less than 6.5% of the individuals in the sample hold multiple jobs? If so, find the probability. If not, explain why not. (b) A new sample of 350 employed adults is chosen. Find the probability that less than 6.5% of the individuals in this sample hold multiple jobs. Round the answer to at least four decimal places. (c) Find the probability that more than 6% of the individuals in the sample of 350 hold multiple jobs. Round the answer to at least four decimal places. (d) Find the probability that the proportion of individuals in the sample of 350 who hold multiple jobs is between 0.05 and 0.10. Round the answer to at least four decimal places. (e) Would it be unusual if less than 4% of the individuals in the sample of 350 held multiple jobs?

Solution

Let's go through each part of this question step by step.

Given Information

  • About 5% of employed adults in the U.S. hold multiple jobs.
  • We will consider different samples and use the normal approximation method when possible.

Part (a)

  1. Check if normal approximation is appropriate for a sample of n=75n = 75 and p=0.05p = 0.05:
    • For normal approximation to be used, we need both npnp and n(1p)n(1-p) to be at least 10.
    • Calculate np=75×0.05=3.75np = 75 \times 0.05 = 3.75, which is less than 10.
    • Since np<10np < 10, it is not appropriate to use the normal approximation.

Part (b)

  1. Probability for a sample of n=350n = 350 that less than 6.5% of individuals hold multiple jobs:
    • Here, p=0.05p = 0.05 and p^=0.065\hat{p} = 0.065.
    • We first need to calculate the mean and standard deviation of the sampling distribution:
      • Mean μ=p=0.05\mu = p = 0.05
      • Standard deviation σ=p(1p)n=0.05×0.95350\sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.05 \times 0.95}{350}}
    • Convert p^=0.065\hat{p} = 0.065 to a z-score and use the z-table or calculator to find the probability.

Part (c)

  1. Probability for a sample of n=350n = 350 that more than 6% of individuals hold multiple jobs:
    • Use p^=0.06\hat{p} = 0.06 and convert it to a z-score with the same mean and standard deviation calculated in Part (b).
    • Find the probability corresponding to the z-score.

Part (d)

  1. Probability for the proportion to be between 0.05 and 0.10:
    • Calculate z-scores for p^=0.05\hat{p} = 0.05 and p^=0.10\hat{p} = 0.10 using the mean and standard deviation from Part (b).
    • Find the probability between these two z-scores.

Part (e)

  1. Determine if it would be unusual for less than 4% to hold multiple jobs:
    • Calculate the probability for p^=0.04\hat{p} = 0.04 using the same mean and standard deviation.
    • If this probability is very low (typically less than 5%), it would be considered unusual.

Would you like detailed calculations for each part or have any specific questions?

Related Questions

  1. What does the requirement np10np \geq 10 and n(1p)10n(1 - p) \geq 10 ensure in using the normal approximation?
  2. How can you interpret the z-score in the context of sampling distributions?
  3. How would the calculations change if the percentage of adults holding multiple jobs was different, say 8%?
  4. How would you interpret a very high probability for one of these proportions in a practical sense?
  5. How does increasing the sample size improve the accuracy of the normal approximation?

Tip

Remember, the larger the sample size, the closer the sample proportion will be to a normal distribution, allowing more accurate use of z-scores for probability estimates.

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Math Problem Analysis

Mathematical Concepts

Probability
Sampling distributions
Normal approximation to binomial distribution
Z-scores

Formulas

np ≥ 10 and n(1 - p) ≥ 10 for normal approximation
Mean (μ) of sampling distribution: μ = p
Standard deviation (σ) of sampling distribution: σ = √(p(1 - p) / n)
Z-score: z = (x - μ) / σ

Theorems

Central Limit Theorem
Conditions for Normal Approximation to Binomial Distribution

Suitable Grade Level

College level (Statistics)