Suppose The Age Of Buyers Was Collected And The Data Followe
Suppose The Age Of Buyers Were Collected And The Data Follows The Bell
Suppose the age of buyers were collected and the data follows the bell curve (a normal distribution). The buyers had an average age of 41 years and a standard deviation of 7. You plan to take a random sample of 68 buyers and calculate a sample mean.
Directions: Fill in the blanks and click the button to have your answers graded. A.
Do not round intermediate calculations. B. For questions 1-3 enter a number between 0 and 1 rounding it to four places. C. Round your answer to question 4 to two decimal places.
- What is the probability of finding a sample mean less than 40.96 years? =
- What is the chance of finding a sample average with a value greater than 41.84 years? =
- What is the probability of finding a sample average within 2.16 years of the population mean? =
- What two values of the sample mean cut off the center 16% of the sampling distribution? Lower Value: = Upper Value: =
Paper For Above instruction
The assessment of probabilities concerning sample means in a normal distribution setup involves understanding the behavior of the sampling distribution, especially when employing the Central Limit Theorem. Given the data—population mean (μ) of 41 years, standard deviation (σ) of 7, and a sample size (n) of 68—scientific calculations can be performed to answer the specific questions about probability and distribution cut-offs.
First, the standard error (SE) of the mean is critical in translating population parameters into sampling distribution characteristics. It is calculated as:
SE = σ / √n
Substituting the given values:
SE = 7 / √68 ≈ 7 / 8.246 ≈ 0.8496
This standard error reflects the dispersion of the sample mean around the population mean, assuming the sample is randomly chosen.
Question 1: Probability the sample mean is less than 40.96 years
To find this, calculate the z-score for 40.96:
z = (X̄ - μ) / SE = (40.96 - 41) / 0.8496 ≈ -0.04 / 0.8496 ≈ -0.047
Using standard normal distribution tables or calculator, the probability corresponding to z ≈ -0.047 is approximately 0.4810.
Answer: 0.4810
Question 2: Probability the sample mean is greater than 41.84 years
Calculate the z-score for 41.84:
z = (41.84 - 41) / 0.8496 ≈ 0.84 / 0.8496 ≈ 0.989
Looking up z ≈ 0.989 yields a cumulative probability of about 0.8374.
Since we're interested in the probability of being greater than 41.84, subtract from 1:
1 - 0.8374 = 0.1626
Answer: 0.1626
Question 3: Probability the sample mean is within 2.16 years of μ
Calculate the z-scores for μ ± 2.16:
Lower bound: (41 - 2.16) = 38.84, z = (38.84 - 41) / 0.8496 ≈ -2.16 / 0.8496 ≈ -2.543
Upper bound: (41 + 2.16) = 43.16, z = (43.16 - 41) / 0.8496 ≈ 2.16 / 0.8496 ≈ 2.543
Find cumulative probabilities for z = ±2.543:
Φ(2.543) ≈ 0.9947, Φ(-2.543) ≈ 0.0053
Probability within bounds: 0.9947 - 0.0053 = 0.9894
Answer: 0.9894
Question 4: Values of sample mean that cut off the center 16% of the sampling distribution
We seek the z-values corresponding to the middle 84% (since 16% is split equally in tails). Thus, the lower and upper z-values are at the 8th and 92nd percentiles.
From z-tables, the z-values corresponding to cumulative probabilities of 0.08 and 0.92 are approximately -1.41 and +1.41 respectively.
Convert these z-scores back to sample means:
Lower value: μ + z SE = 41 + (-1.41) 0.8496 ≈ 41 - 1.20 ≈ 39.80
Upper value: 41 + 1.41 * 0.8496 ≈ 41 + 1.20 ≈ 42.20
Answer: Lower Value = 39.80; Upper Value = 42.20
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