Math Problem Statement
The acceptable level for insect filth in a certain food item is 33 insect fragments (larvae, eggs, body parts, and so on) per 10 grams. A simple random sample of 5050 ten-gram portions of the food item is obtained and results in a sample mean of x overbarxequals=3.73.7 insect fragments per ten-gram portion. Complete parts (a) through (c) below. Question content area bottom Part 1 (a) Why is the sampling distribution of x overbarx approximately normal? A. The sampling distribution of x overbarx is approximately normal because the population is normally distributed. B. The sampling distribution of x overbarx is approximately normal because the population is normally distributed and the sample size is large enough. C. The sampling distribution of x overbarx is assumed to be approximately normal. D. The sampling distribution of x overbarx is approximately normal because the sample size is large enough. Your answer is correct. Part 2 (b) What is the mean and standard deviation of the sampling distribution of x overbarx assuming mu equals 3μ=3 and sigmaσequals=StartRoot 3 EndRoot3? mu Subscript x overbarμxequals=33 (Round to three decimal places as needed.) Part 3 sigma Subscript x overbarσxequals=0.2450.245 (Round to three decimal places as needed.) Part 4 (c) What is the probability a simple random sample of 5050 ten-gram portions of the food item results in a mean of at least 3.73.7 insect fragments? P(x overbarxgreater than or equals≥3.73.7)equals=0.00210.0021 (Round to four decimal places as needed.) Part 5 Is this result unusual? A. This result is notis not unusual because its probability is smallsmall. B. This result is notis not unusual because its probability is largelarge. C. This result isis unusual because its probability is largelarge. D. This result isis unusual because its probability is smallsmall.
Solution
Let's go through each part of your problem step by step.
(a) Why is the sampling distribution of approximately normal?
The correct answer here is:
D. The sampling distribution of is approximately normal because the sample size is large enough.
This is due to the Central Limit Theorem (CLT), which states that for sufficiently large sample sizes (typically ), the sampling distribution of the sample mean will be approximately normal, regardless of the population's original distribution. Since the sample size is 50, it meets this criterion.
(b) What is the mean and standard deviation of the sampling distribution of ?
For this part:
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The mean of the sampling distribution of is the same as the population mean .
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The standard deviation of the sampling distribution of (also called the standard error) is calculated using the formula: where (as given) and (sample size).
Thus, the mean is 3 and the standard deviation is approximately 0.245.
(c) What is the probability a simple random sample of 50 ten-gram portions of the food item results in a mean of at least 3.7 insect fragments?
To find , we can standardize this using the z-score formula: Substituting the values:
Now, using the standard normal distribution table or a calculator, the probability of a z-score of 2.857 or higher is approximately 0.0021. Thus:
Is this result unusual?
To determine if this result is unusual, we compare the probability with a typical threshold (commonly, probabilities less than 0.05 or 5% are considered unusual). Since is much smaller than 0.05, this event is unusual.
The correct answer is: D. This result is unusual because its probability is small.
Let me know if you need further clarifications! Here are some related questions to consider:
- What is the Central Limit Theorem and why is it important?
- How is the standard error related to sample size?
- What factors affect the shape of the sampling distribution of ?
- How would changing the sample size impact the probability found in part (c)?
- What would be the z-score if the sample mean were 3.2 instead of 3.7?
Tip: When calculating probabilities from a normal distribution, always standardize the values using z-scores.
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Math Problem Analysis
Mathematical Concepts
Sampling Distribution
Normal Distribution
Central Limit Theorem
Z-score
Formulas
Standard error formula: σ_x = σ / sqrt(n)
Z-score formula: z = (x̄ - μ) / σ_x
Theorems
Central Limit Theorem
Suitable Grade Level
Grades 11-12 (Statistics/Probability)
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