Math Problem Statement

MSNBC recently reported that the mean annual cost of auto insurance is 981 dollars. Assume the standard deviation is 259 dollars. You take a simple random sample of 85 auto insurance policies. Find the probability that a single randomly selected value is at least 970 dollars. Find the probability that a sample of size n = 85 is randomly selected with a mean that is at least 970 dollars.

Solution

Let's go through the problem step-by-step.

Given Information

  • Mean (μ\mu) of annual auto insurance cost: 981 dollars.
  • Standard deviation (σ\sigma) of annual auto insurance cost: 259 dollars.
  • Sample size (nn) for part of the question: 85.

We need to answer two questions:

1. Finding P(x>970)P(x > 970)

This is the probability that a single randomly selected value is at least 970 dollars.

  • First, calculate the z-score: z=xμσ=970981259z = \frac{x - \mu}{\sigma} = \frac{970 - 981}{259}

  • Compute the result for zz and find the probability P(x>970)P(x > 970) using a z-table or normal distribution calculator.

2. Finding P(xˉ>970)P(\bar{x} > 970) for a sample size n=85n = 85

This is the probability that the sample mean of a sample size of 85 is at least 970 dollars.

  • First, calculate the standard error (σxˉ\sigma_{\bar{x}}) for the sample mean: σxˉ=σn=25985\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{259}{\sqrt{85}}

  • Calculate the z-score for the sample mean: z=xˉμσxˉ=970981σxˉz = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{970 - 981}{\sigma_{\bar{x}}}

  • Find P(xˉ>970)P(\bar{x} > 970) using the z-score and a z-table or normal distribution calculator.

Let's calculate each of these values for a more accurate answer. I'll proceed with the calculations.Here are the calculated probabilities:

  1. Probability that a single randomly selected value is at least 970: P(x>970)0.5169P(x > 970) \approx 0.5169

  2. Probability that the sample mean (with n=85n = 85) is at least 970: P(xˉ>970)0.6523P(\bar{x} > 970) \approx 0.6523

These values are rounded to four decimal places as required.

Would you like a deeper explanation of the calculations, or do you have further questions?


Related Questions

  1. What is the significance of using the standard error in sample mean calculations?
  2. How does increasing the sample size affect the probability of P(xˉ>970)P(\bar{x} > 970)?
  3. Why does the probability for a single value differ from that of a sample mean?
  4. How would the probabilities change if the standard deviation were different?
  5. What is the role of the z-score in normal distribution probability calculations?

Tip

When working with probabilities involving sample means, always remember to use the standard error instead of the original standard deviation. This accounts for the reduced variability due to the larger sample size.

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Math Problem Analysis

Mathematical Concepts

Probability
Normal Distribution
Z-score
Central Limit Theorem

Formulas

Z = (X - μ) / σ
σ_x̄ = σ / √n

Theorems

Central Limit Theorem

Suitable Grade Level

College Level