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A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 3364 with a mean life of 530 minutes. If the claim is true, in a sample of 75 batteries, what is the probability that the mean battery life would be greater than 533.2 minutes? Round your answer to four decimal places.

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 3364 with a mean life of 530 minutes. If the claim is true, in a sample of 75 batteries, what is the probability that the mean battery life would be greater than 533.2 minutes? Round your answer to four decimal places.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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**Text:** A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 3364 with a mean life of 530 minutes. If the claim is true, in a sample of 75 batteries, what is the probability that the mean battery life would be greater than 533.2 minutes? Round your answer to four decimal places. **Answer** How to enter your answer (opens in new window) **Explanation:** In this scenario, a quality control expert is interested in testing a claim made by a design engineer about the battery life. The engineer suggests that the batteries have a mean life of 530 minutes and a variance of 3364. When dealing with statistical questions like this, it's important to know how to calculate probabilities, particularly for sample means. 1. **Sample Size (n):** 75 batteries. 2. **Mean (\(\mu\)):** 530 minutes. 3. **Variance (\(\sigma^2\)):** 3364. 4. **Standard Deviation (\(\sigma\)):** The square root of 3364. 5. **Target Mean:** 533.2 minutes. **Goal:** Calculate the probability that the sample mean exceeds 533.2 minutes, using standard deviation and Z-scores for a sample distribution. This often involves using the Z-score formula and statistical tables or software to find the probability.
Transcribed Image Text:**Text:** A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of 3364 with a mean life of 530 minutes. If the claim is true, in a sample of 75 batteries, what is the probability that the mean battery life would be greater than 533.2 minutes? Round your answer to four decimal places. **Answer** How to enter your answer (opens in new window) **Explanation:** In this scenario, a quality control expert is interested in testing a claim made by a design engineer about the battery life. The engineer suggests that the batteries have a mean life of 530 minutes and a variance of 3364. When dealing with statistical questions like this, it's important to know how to calculate probabilities, particularly for sample means. 1. **Sample Size (n):** 75 batteries. 2. **Mean (\(\mu\)):** 530 minutes. 3. **Variance (\(\sigma^2\)):** 3364. 4. **Standard Deviation (\(\sigma\)):** The square root of 3364. 5. **Target Mean:** 533.2 minutes. **Goal:** Calculate the probability that the sample mean exceeds 533.2 minutes, using standard deviation and Z-scores for a sample distribution. This often involves using the Z-score formula and statistical tables or software to find the probability.
### Standard Normal Distribution Table This is a table representing the Standard Normal Distribution. The values indicate the area under the curve to the left of a given Z-score. #### Key Components: - **Z-Score Column**: The Z column shows values ranging from -3.9 to -0.0. This axis represents standardized scores in a normal distribution. - **Decimals Columns**: These headers (0.00 to 0.09) detail additional specificity to the Z-score, refining to two decimal places. ### How to Read the Table: 1. **Locate the Z-score**: First, find the row that aligns with the whole number and first decimal of your Z-score. 2. **Match Decimal**: Next, follow the column with the second decimal. 3. **Cross-reference**: The intersecting cell provides the area to the left of that Z-score in the standard normal distribution curve. #### Example: For a Z-score of -1.3: - Start in the row labeled `-1.3`. - Move across to the column labeled `.08`. - The value in the intersecting cell is `0.08327`. This indicates that approximately 8.327% of the data falls below a Z-score of -1.3. ### Highlighted Sections: - Highlight at Z = -1.3, and column .08 shows value `0.08327`. This structured format allows for quick analysis of probabilities and is vital in fields requiring statistical assessments, like psychology and finance.
Transcribed Image Text:### Standard Normal Distribution Table This is a table representing the Standard Normal Distribution. The values indicate the area under the curve to the left of a given Z-score. #### Key Components: - **Z-Score Column**: The Z column shows values ranging from -3.9 to -0.0. This axis represents standardized scores in a normal distribution. - **Decimals Columns**: These headers (0.00 to 0.09) detail additional specificity to the Z-score, refining to two decimal places. ### How to Read the Table: 1. **Locate the Z-score**: First, find the row that aligns with the whole number and first decimal of your Z-score. 2. **Match Decimal**: Next, follow the column with the second decimal. 3. **Cross-reference**: The intersecting cell provides the area to the left of that Z-score in the standard normal distribution curve. #### Example: For a Z-score of -1.3: - Start in the row labeled `-1.3`. - Move across to the column labeled `.08`. - The value in the intersecting cell is `0.08327`. This indicates that approximately 8.327% of the data falls below a Z-score of -1.3. ### Highlighted Sections: - Highlight at Z = -1.3, and column .08 shows value `0.08327`. This structured format allows for quick analysis of probabilities and is vital in fields requiring statistical assessments, like psychology and finance.
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