SOLUTION: CNNBC recently reported that the mean annual cost of auto insurance is 1025 dollars. Assume the standard deviation is 252 dollars. You take a simple random sample of 57 auto insura
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-> SOLUTION: CNNBC recently reported that the mean annual cost of auto insurance is 1025 dollars. Assume the standard deviation is 252 dollars. You take a simple random sample of 57 auto insura
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Question 1177956: CNNBC recently reported that the mean annual cost of auto insurance is 1025 dollars. Assume the standard deviation is 252 dollars. You take a simple random sample of 57 auto insurance policies.
Find the probability that a single randomly selected value is less than 989 dollars.
P(X < 989) =
Find the probability that a sample of size
n
=
57
is randomly selected with a mean less than 989 dollars.
P(M < 989) =
Enter your answers as numbers accurate to 4 decimal places. Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**Given Information**
* Population Mean (μ): $1025
* Population Standard Deviation (σ): $252
* Sample Size (n): 57
**1. Probability for a Single Randomly Selected Value (X < 989)**
* We need to find P(X < 989) where X is a normally distributed random variable.
1. **Calculate the Z-score:**
* Z = (X - μ) / σ
* Z = (989 - 1025) / 252
* Z = -36 / 252 ≈ -0.1429
2. **Find the Probability using the Z-table or Calculator:**
* P(Z < -0.1429) ≈ 0.4431
Therefore, P(X < 989) ≈ 0.4431
**2. Probability for a Sample Mean (M < 989)**
* We need to find P(M < 989) where M is the sample mean.
1. **Calculate the Standard Error (SE):**
* SE = σ / √n
* SE = 252 / √57
* SE ≈ 252 / 7.5498 ≈ 33.377
2. **Calculate the Z-score for the Sample Mean:**
* Z = (M - μ) / SE
* Z = (989 - 1025) / 33.377
* Z = -36 / 33.377 ≈ -1.0786
3. **Find the Probability using the Z-table or Calculator:**
* P(Z < -1.0786) ≈ 0.1403
Therefore, P(M < 989) ≈ 0.1403
**Answers**
* P(X < 989) ≈ 0.4431
* P(M < 989) ≈ 0.1403
