Math Problem Statement
Solution
Let's solve this problem step by step.
Given Information:
- Population mean () = 800 gms
- Population standard deviation () = 140 gms
- Sample size () = 16
For a sample, the standard error of the mean (SEM) is calculated using the formula:
Since the weights are normally distributed, we can use the Z-distribution to answer the questions.
(a) What is the probability that the mean of the sample would exceed 900 gms?
We are looking for: First, calculate the Z-score for 900 gms: Using the Z-table, the probability for is approximately 0.9979.
The probability that the sample mean exceeds 900 gms is: So, the probability is 0.0021 (or 0.21%).
(b) What is the probability that the mean of the sample would be between 860 and 890 gms?
We are looking for:
-
For 860 gms: From the Z-table, the probability for is approximately 0.9564.
-
For 890 gms: From the Z-table, the probability for is approximately 0.9951.
Now, the probability that the sample mean is between 860 and 890 gms: So, the probability is 0.0387 (or 3.87%).
(c) Find the limits within which 95 percent of sample means would lie on either side of the population mean.
For 95% confidence, the Z-scores corresponding to the middle 95% of a normal distribution are approximately .
Now calculate the margin of error:
Thus, the limits are:
Therefore, 95% of the sample means would lie between 731.4 gms and 868.6 gms.
Recap:
- (a) Probability of exceeding 900 gms: 0.0021
- (b) Probability of being between 860 and 890 gms: 0.0387
- (c) 95% confidence interval for sample means: 731.4 gms to 868.6 gms
Would you like any additional explanations or details on these solutions?
Here are some related questions to explore:
- How do you interpret the result for part (a) in a practical manufacturing context?
- What happens to the probability in part (b) if we increase the sample size to 36?
- How does changing the population standard deviation affect the results?
- Why is the standard error important in the context of sample means?
- Can the sample mean ever exceed the population mean? Why or why not?
Tip: The Z-distribution is a powerful tool for solving problems related to sample means when the population is normally distributed!
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Math Problem Analysis
Mathematical Concepts
Normal Distribution
Z-Score
Sample Mean
Confidence Interval
Formulas
Standard Error of the Mean (SEM) = σ / √n
Z-Score = (X̄ - μ) / SEM
Confidence Interval = μ ± Z * SEM
Theorems
Central Limit Theorem
Suitable Grade Level
Undergraduate Statistics
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