Government regulations dictate that for any production process involving a certain toxic chemical, the water in the output of the process must not exceed 7470 parts per million (ppm) of the chemical. For a particular process of concern, the water sample was collected by a manufacturer 25 times randomly and the sample average x was 7482 ppm. It is known from historical data that the standard deviation is 60 ppm. Complete parts (a) and (b) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. (a) What is the probability that the sample average in this experiment would exceed the government limit if the population mean is equal to the limit? Use the Central Limit Theorem. The probability is (Round to four decimal places as needed.) (b) Is an observed x = 7482 in this experiment firm evidence that the population mean for the process exceeds the government limit? Answer your question by computing P(X≥7482 | μ = 7470). Assume that the distribution of the concentration is normal. Since P(X≥7482 | µ=7470) = process exceeds the government limit. Round to four decimal places as needed.) negligible, the observed x evidence that the population mean for the
Government regulations dictate that for any production process involving a certain toxic chemical, the water in the output of the process must not exceed 7470 parts per million (ppm) of the chemical. For a particular process of concern, the water sample was collected by a manufacturer 25 times randomly and the sample average x was 7482 ppm. It is known from historical data that the standard deviation is 60 ppm. Complete parts (a) and (b) below. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. (a) What is the probability that the sample average in this experiment would exceed the government limit if the population mean is equal to the limit? Use the Central Limit Theorem. The probability is (Round to four decimal places as needed.) (b) Is an observed x = 7482 in this experiment firm evidence that the population mean for the process exceeds the government limit? Answer your question by computing P(X≥7482 | μ = 7470). Assume that the distribution of the concentration is normal. Since P(X≥7482 | µ=7470) = process exceeds the government limit. Round to four decimal places as needed.) negligible, the observed x evidence that the population mean for the
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
solve the following
- [Page 2 of the Standard Normal Distribution Table](#)
---
#### Tasks:
### (a) Probability of Sample Average Exceeding the Government Limit
**Question:** What is the probability that the sample average in this experiment would exceed the government limit if the population mean is equal to the limit? Use the Central Limit Theorem.
**Solution:**
1. First, denote the population mean as \( \mu = 7470 \).
2. Using the Central Limit Theorem, find the required probability:
\[
\text{The probability is} \_\_\_\_.
\]
(Round to four decimal places as needed.)
### (b) Can We Conclude the Population Mean Exceeds the Government Limit?
**Question:** Is an observed \( \bar{X} = 7482 \) in this experiment firm evidence that the population mean for the process exceeds the government limit? Answer your question by computing \( P(\bar{X} \geq 7482 \mid \mu = 7470) \). Assume that the distribution of the concentration is normal.
**Solution:**
1. Compute the probability \( P(\bar{X} \geq 7482 \mid \mu = 7470) \):
\[
P(\bar{X} \geq 7482 \mid \mu = 7470) = \_\_\_\_.
\]
2. Analyze the computed probability:
\[
\text{Since } P(\bar{X} \geq 7482 \mid \mu = 7470) = \_\_\_\_ \text{ is} \]
\[ \ \ \ \
\]
\[ \ \ \ \
\text{ negligible, the observed } \bar{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d1e5150-1781-42a0-9e07-3d8cfcc0113b%2F9ac479f0-6425-4fc4-bc6f-3fc98050c5ad%2Fsz5qxvk_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Probability in Production Processes Using Central Limit Theorem
Government regulations dictate that for any production process involving a certain toxic chemical, the water in the output of the process must not exceed 7470 parts per million (ppm) of the chemical. For a particular process of concern, the water sample was collected by a manufacturer 25 times randomly and the sample average \( \bar{X} \) was 7482 ppm. It is known from historical data that the standard deviation \( \sigma \) is 60 ppm. Complete parts (a) and (b) below.
#### Click to Access Standard Normal Distribution Tables:
- [Page 1 of the Standard Normal Distribution Table](#)
- [Page 2 of the Standard Normal Distribution Table](#)
---
#### Tasks:
### (a) Probability of Sample Average Exceeding the Government Limit
**Question:** What is the probability that the sample average in this experiment would exceed the government limit if the population mean is equal to the limit? Use the Central Limit Theorem.
**Solution:**
1. First, denote the population mean as \( \mu = 7470 \).
2. Using the Central Limit Theorem, find the required probability:
\[
\text{The probability is} \_\_\_\_.
\]
(Round to four decimal places as needed.)
### (b) Can We Conclude the Population Mean Exceeds the Government Limit?
**Question:** Is an observed \( \bar{X} = 7482 \) in this experiment firm evidence that the population mean for the process exceeds the government limit? Answer your question by computing \( P(\bar{X} \geq 7482 \mid \mu = 7470) \). Assume that the distribution of the concentration is normal.
**Solution:**
1. Compute the probability \( P(\bar{X} \geq 7482 \mid \mu = 7470) \):
\[
P(\bar{X} \geq 7482 \mid \mu = 7470) = \_\_\_\_.
\]
2. Analyze the computed probability:
\[
\text{Since } P(\bar{X} \geq 7482 \mid \mu = 7470) = \_\_\_\_ \text{ is} \]
\[ \ \ \ \
\]
\[ \ \ \ \
\text{ negligible, the observed } \bar{
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