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Replacement times for TV sets are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years. Find the probability a randomly selected TV will have a replacement time less than 5 years (round to three decimal places). 0.0018

Replacement times for TV sets are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years. Find the probability a randomly selected TV will have a replacement time less than 5 years (round to three decimal places). 0.0018

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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**Understanding Probability: Replacement Times for TV Sets** When considering replacement times for TV sets, data shows that this variable follows a normal distribution. Here we present a specific example to demonstrate this concept and guide you on how to calculate probabilities for normally distributed variables: ### Problem Statement Replacement times for TV sets are normally distributed with: - **Mean (μ)**: 8.2 years - **Standard Deviation (σ)**: 1.1 years **Question**: What is the probability that a randomly selected TV will have a replacement time less than 5 years? The answer should be rounded to three decimal places. ### Solution To find this probability, we need to calculate the Z-score, which represents the number of standard deviations a value (X) is away from the mean. The Z-score formula is: \[ Z = \frac{X - \mu}{σ} \] For this problem: - \( X = 5 \) years - \( \mu = 8.2 \) years - \( σ = 1.1 \) years ### Calculation \[ Z = \frac{5 - 8.2}{1.1} = \frac{-3.2}{1.1} \approx -2.91 \] Using the Z-score table or a normal distribution calculator, a Z-score of approximately -2.91 corresponds to a cumulative probability near **0.0018**. Therefore, the probability that a randomly selected TV will have a replacement time less than 5 years is **0.0018**. ### Explanation This result (0.0018) indicates that there is a **0.18% chance** that a TV will need to be replaced in less than 5 years. This low probability highlights that it is quite rare for TVs (under these distributions) to require replacement in such a short time frame. By understanding this calculation, you can better appreciate the applications of normal distribution and probability in real-world scenarios, such as estimating product lifespans and making informed decisions based on statistical evidence.
Transcribed Image Text:**Understanding Probability: Replacement Times for TV Sets** When considering replacement times for TV sets, data shows that this variable follows a normal distribution. Here we present a specific example to demonstrate this concept and guide you on how to calculate probabilities for normally distributed variables: ### Problem Statement Replacement times for TV sets are normally distributed with: - **Mean (μ)**: 8.2 years - **Standard Deviation (σ)**: 1.1 years **Question**: What is the probability that a randomly selected TV will have a replacement time less than 5 years? The answer should be rounded to three decimal places. ### Solution To find this probability, we need to calculate the Z-score, which represents the number of standard deviations a value (X) is away from the mean. The Z-score formula is: \[ Z = \frac{X - \mu}{σ} \] For this problem: - \( X = 5 \) years - \( \mu = 8.2 \) years - \( σ = 1.1 \) years ### Calculation \[ Z = \frac{5 - 8.2}{1.1} = \frac{-3.2}{1.1} \approx -2.91 \] Using the Z-score table or a normal distribution calculator, a Z-score of approximately -2.91 corresponds to a cumulative probability near **0.0018**. Therefore, the probability that a randomly selected TV will have a replacement time less than 5 years is **0.0018**. ### Explanation This result (0.0018) indicates that there is a **0.18% chance** that a TV will need to be replaced in less than 5 years. This low probability highlights that it is quite rare for TVs (under these distributions) to require replacement in such a short time frame. By understanding this calculation, you can better appreciate the applications of normal distribution and probability in real-world scenarios, such as estimating product lifespans and making informed decisions based on statistical evidence.
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